Is there a "First Principles" for integration?

[cmcraes]

Is there a "First Principles" for integration?

More specifically, For indefinite integrals. I've looked online and can't find anything.
Is there a method of finding indefinite integral analogous to finding the derivitive by:

Lim (f(c+h)-f(c))/h
h->0

So final question: Besides the 'inverse power rule', integration by parts or substitution, Is there a way to find the indefinite integral of a function? Thank you!

 

[Mandelbroth]

More specifically, For indefinite integrals. I've looked online and can't find anything.
Is there a method of finding indefinite integral analogous to finding the derivitive by:

Lim (f(c+h)-f(c))/h
h->0

So final question: Besides the 'inverse power rule', integration by parts or substitution, Is there a way to find the indefinite integral of a function? Thank you!

For a definite Riemann integral, $\displaystyle \int_{a}^{b}f(x) \, dx = \lim_{n\rightarrow\infty^-}\sum_{i=1}^{n}\left[f(a+\frac{b-a}{n}i)(\frac{b-a}{n})\right]$. For indefinite integrals, I am unaware of such a limit form. I think it should just be seen as the integral of a function f is a function F for which the derivative of F is f.

 

[VantagePoint72]

The "first principle" is the Fundamental Theorem of Calculus, which proves the definite integral / Riemann sum (which Mandelbroth gave) is equal to $F(b) - F(a)$ where $F'(x) = f(x)$. The indefinite integral of $f(x)$ is defined as the antiderivative of $f$ (plus a generic constant), by analogy with the Fundamental Theorem.

Perhaps this is the point of confusion. As a first principle thing, $\int_a^b f(x) dx$ is not defined as $F(b) - F(a)$, where $F'(x) = f(x)$. It's defined as the area under $f(x)$ between $x=a$ and $x=b$. It is the FToC that let's us connect this with the antiderivatives of $f(x)$. Having done this, the indefinite integral is just introduced as a shorthand for antidifferentiation.

 

[Mandelbroth]

The "first principle" is the Fundamental Theorem of Calculus, which proves the definite integral / Riemann sum (which Mandelbroth gave) is equal to $F(b) - F(a)$ where $F'(x) = f(x)$. The indefinite integral of $f(x)$ is defined as the antiderivative of $f$ (plus a generic constant), by analogy with the Fundamental Theorem.

Perhaps this is the point of confusion. As a first principle thing, $\int_a^b f(x) dx$ is not defined as $F(b) - F(a)$, where $F'(x) = f(x)$. It's defined as the area under $f(x)$ between $x=a$ and $x=b$. It is the FToC that let's us connect this with the antiderivatives of $f(x)$. Having done this, the indefinite integral is just introduced as a shorthand for antidifferentiation.

I disagree with parts of this.

The antiderivative of f is a function F such that the derivative of F is f. Thus, an arbitrary constant is implied by the nature of the derivative. However, what you said is, technically, correct, because a constant plus a constant is still a constant.

As to $\int_a^b f(x) \, dx$ being defined as the area under f(x), I completely disagree. It may have been originally defined as the area, but it is not, in general, going to give an area by modern definition.

Consider the integral $\int_{0}^{\pi}e^{ix} \, dx$. This will not be an area. If it was, good luck constructing any shape with an area of $2i$.

 

[VantagePoint72]

The antiderivative of f is a function F such that the derivative of F is f. Thus, an arbitrary constant is implied by the nature of the derivative. However, what you said is, technically, correct, because a constant plus a constant is still a constant.

This was more a case of linguistic imprecision on my part than anything. Where I said "the antiderivative of $f$ (plus a generic constant)" I should have said "an antiderivative". As in, particular solution $F$ to $F'(x) = f(x)$.

As to $\int_a^b f(x) \, dx$ being defined as the area under f(x), I completely disagree. It may have been originally defined as the area, but it is not, in general, going to give an area by modern definition.

Consider the integral $\int_{0}^{\pi}e^{ix} \, dx$. This will not be an area. If it was, good luck constructing any shape with an area of $2i$.

And I, in turn, completely disagree with your approach. Considering it's pretty clear the OP is doing calculus of one real variable, I think it's much more sensible to consider the integral as an area so that the Riemann sum is suitably motivated. Then when we generalize to cases where the area doesn't make sense, we keep the sum and throw away the area interpretation. That's typically how it goes in mathematics: consider something concrete, derive some properties of it, and then abstract the whole thing so we forget about the original concrete thing and just keep the properties. I think it's a pedagogical mistake to jump directly to the abstract definitions. The definite integral as an area definition motivates the definition of the Riemann sum, and the former should be understood before the latter is taken a definition in its own right.

 

[micromass]

And I, in turn, completely disagree with your approach. Considering it's pretty clear the OP is doing calculus of one real variable, I think it's much more sensible to consider the integral as an area so that the Riemann sum is suitably motivated. Then when we generalize, we keep the sum and throw away the area interpretation. That's typically how it goes in mathematics: consider something concrete, derive some properties of it, and then abstract the whole thing so we forget about the original concrete thing and just keep the properties. I think it's a pedagogical mistake to jump directly to the abstract definitions. The definite integral as an area definition motivates the definition of the Riemann sum, and the former should be understood before the latter is taken a definition in its own right.

While I agree that the area interpretation is a very nice one, it's not the most useful one. An integral is just a way to sum infinitely many elements which are infinitely small (very roughly). I don't think we shouldn't motivate the integral as an area, but I think we should leave the area interpretation very quickly and go to the more abstract "sum" interpretation. The way that integrals are used in mathematics and physics tend to have very little to do with areas anyway. Somebody who thinks of an integral as an area tends to have some difficulty with the fact that the arc length can also be given as an integral.

 

[VantagePoint72]

I don't think we shouldn't motivate the integral as an area, but I think we should leave the area interpretation very quickly and go to the more abstract "sum" interpretation. The way that integrals are used in mathematics and physics tend to have very little to do with areas anyway.

Fair enough.

 

[cmcraes]

My problem with the Riemann Integral is that you must know what F(x) is. Although usually do-able, its seems like something is missing from indefinite integration.

Also side question: what is the Indefinite integral of
y=0
Is it c, or 0?

 

[Office_Shredder]

It's c. And if you try to fine the area under the graph between x=a and x=b, the area is c-c=0 regardless of your choice of c

 

[epenguin]

The method is - RECOGNITION!

I think the OP's question was that whereas we can almost infallibly differentiate any reasonable function (continuous derivatives etc.) we are given, it is much harder to do the inverse operation.

Isn't the answer (almost) that the only way to do it is to know the answer? Or at least the rough kind of shape of answer. By differentiation of a list of we can get a list of functions and their derivatives to be able to go backwards. Then the so called 'methods' of integration is basically a sort of handbook of recognising a number of types and when we have the rough general type we hammer it till it hopefully comes into the shape of some thing or things we recognise on our list.

I suspect that in years to come these 'methods' and excercises will disappear from curriculae, since the problems can be handled by calculators - it will go the way of doing long arithmetical calculations by hand.

 

[HallsofIvy]

More generally, this is an example of the "inverse problem" in mathematics. It typically happens that we have some operation on a set that has a direct definition and we can determine a number of "rules" from the definition but then we have the problem of finding inverses for that operation- which typically requires knowing special examples of the direct operation.

 

[pwsnafu]

For a definite Riemann integral, $\displaystyle \int_{a}^{b}f(x) \, dx = \lim_{n\rightarrow\infty^-}\sum_{i=1}^{n}\left[f(a+\frac{b-a}{n}i)(\frac{b-a}{n})\right]$.

It's worth pointing out, that is not the same thing as Riemann sums: according to what is written in the quote the characteristic function for the rationals would be integrable.

 

[avian_swine]

I am from the future, and the future desires another answer to this question. This thread is now at the top of every search engine result for the query "first principles of integration," and it is regrettable that it has only partially been answered.

Whereas the derivative is the limit of a quotient, the definite integral $\int_a^b f(x)\ \text{d}x$ is indeed a limit of partial sums. This question asks for the first principles of an indefinite integral, and this is where a semantic war begins. The first principles of definite integration can help. For this purpose, I motivate the definite integral as it has been motivated since before Riemann - as a way to find "area".

It's worth noting that we can compute the derivative of a function at one point, or at all points. You could colloquially call finding the derivative at a single point definite differentiation, and finding the derivative at all points indefinite differentiation. There are some problems with this, but it acts as a motivation for the role of definite integration when it comes to indefinite integration. I use the notion of Darboux integrals to give one iteration of integration by first principles.

DEFINITION 1.1: Let $f$ be a bounded function on the compact interval $[a, b]$. For any subset $S$ in this interval, we write: $$M(f, S) = \sup\{f(x):x\in S\}$$ and $$m(f, S) = \inf\{f(x):x\in S\},$$ where $\sup$ is the supremum (the least upper bound) and $\inf$ is the infimum (the greatest lower bound).

This gives us the greatest value and the smallest value of a function $f$ on a subset $S\subseteq [a, b]$.

DEFINITION 1.2: A partition of a compact set $[a, b]$ is a finite sequence of numbers, $(t_k)$, with $$a = t_0 < t_1 < t_2 < \cdot\cdot\cdot < t_n = b.$$

This splits an interval up into $n$ sub-intervals, which are then used for Darboux integration.

DEFINITION 1.3: Let $P$ be a partition of a compact interval $[a, b]$. We define the upper Darboux sum, $U(f, P),$ as: $$U(f, P) = \sum_{k=1}^n M(f, [t_{k-1}, t_k])(t_k - t_{k - 1}).$$ We define the lower Darboux sum, $L(f, P),$ as $$L(f, P) = \sum_{k=1}^n m(f, [t_{k-1}, t_k])(t_k - t_{k - 1}).$$

Here, $(t_k - t_{k - 1})$ is the length of a rectangle, and $M(f, [t_{k-1}, t_k])$ or $m(f, [t_{k-1}, t_k])$ is its height. Here is a visual taken from StackExchange that may help:

Here, the green represents lower Darboux sums, and the gray (on top of the green) represents the upper Darboux sums. If we take finer and finer partitions on the x-axis, it is clear that these areas converge to the "true area" under this particular curve. Let's continue.

DEFINITION 1.4: The upper Darboux integral of a function $f$ over a compact interval $[a, b]$ is defined to be: $$U(f) = \inf\{U(f, P): P \text{ is any partition}\}.$$ The lower Darboux integral of a function $f$ over a compact interval $[a, b]$ is defined to be: $$L(f) = \sup\{U(f, P): P\text{ is any partition}\}.$$

Simply put, the upper Darboux integral is the smallest possible upper Darboux sum, and the lower Darboux integral is the greatest possible lower Darboux sum. This brings us to our final definition.

DEFINITION 1.5: $f$ is said to be integrable over a compact interval $[a, b]$ if $L(f) = U(f),$ with the notation $$\int_a^b f(x)\text{d}x = L(f) = U(f).$$

Actually using this definition to compute a definite (or "indefinite") integral is challenging, but possible. Usually, you try to go from this definition straight to a proof of the Fundamental Theorems of Calculus, but you can find the areas (and by extension antiderivatives) of polynomials and transcendental functions using this method and a little bit of effort. Indeed, Edward Wright used a less advanced form of these methods in 1599 for the integral of the secant function, and after reviewing them 40 years later, these results were used in the conjecture that $\int_0^{\theta} \sec(\phi)\text{d}\phi = \ln|\tan(\frac{\theta}{2}+\frac{\pi}{4})|$. I'll be demonstrating how you can use these first principles to find the area under a curve.

Goal: Find the integral of $f(x) = x^3$ on the compact interval $[0, b]$ for $b > 0$. I will find the upper Darboux integral; a similar algebraic manipulation allows you to find the identical lower Darboux integral (and to conclude that $f(x) = x^3$ is integrable on the interval $[0, b]$).

I will be using the fact that $\sum_{k=1}^n k^3 = \Big(\frac{n(n+1)}{2}\Big)^2.$ Bonus homework: prove this by induction.

I choose the partition $P_n$ which splits the interval $[0, b]$ into $n$ equal pieces: $t_k = \frac{kb}{n}.$ As $x^3$ is a monotonically increasing function on the interval $[0, b],$ we observe that for all $t_k,$ $M(f, [t_{k-1}, t_k]) = (t_k)^3.$ Let us begin.

$$
\begin{align*}
U(f, P_n) &= \sum_{k=1}^n M(f, [t_{k-1}, t_{k}])(t_k - t_{k-1})\\
&= \sum_{k=1}^n (t_k)^3(t_k - t_{k-1})\\
&= \sum_{k=1}^n \frac{k^3b^3}{n^3}\Big(\frac{kb}{n} - \frac{(k-1)b}{n}\Big)\\
&= \sum_{k=1}^n \frac{k^3b^3}{n^3}\Big(\frac{b}{n}\Big)\\
&= \sum_{k=1}^n \frac{k^3b^4}{n^4}\\
&= \frac{b^4}{n^4}\sum_{k=1}^n k^3\\
&= \frac{b^4}{n^4}\Big(\frac{n(n+1)}{2}\Big)^2\\
&= \frac{b^4}{4}\Big(\frac{n^4+2n^3+n^2}{n^4}\Big)\\
&= \frac{b^4}{4}\Big(1+\frac{2}{n}+\frac{1}{n^2}\Big)\\
\end{align*}
$$

Perhaps it is clear what comes next. A theorem is required to be truly rigorous, but for what little brevity this post has left:

$$
\begin{align*}
U(f) &= \inf\{U(f, P): P\text{ is any partition}\}\\
&\leq \inf\{U(f, P_n)\}\\
&= \inf\{\frac{b^4}{4}\Big(1+\frac{2}{n}+\frac{1}{n^2}\Big)\}\\
&= \frac{b^4}{4}
\end{align*}
$$

Going from the second last line to the last line is possible as we're working with a decreasing sequence. Note the inequality - you'll need to find the lower Darboux integral to complete this with true rigor (it's a similar exercise). Also notice that $\frac{b^4}{4}$ is an antiderivative of $b^3$. This is no fluke - indeed, it follows from the Fundamental Theorem of Calculus as well, which allows for an easier time working with these things!

I hope this provides some closure to the lurkers of the world. Cheers!

 

[mathwonk]

In reference to post #2, almost exactly the same limit, but with b replaced by x, and a ≤ x ≤ b, gives the indefinite integral of any continuous function f on {a,b]. This is called the (first) fundamental theorem of calculus.

see e.g. https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

The challenge to grasping this, is the challenge of grasping what a "function" on [a,b] is: namely any well defined rule that assigns a real number to each point of [a,b]. The theorem is that whenever f is continuous on [a,b], then this limit does exist for every x in [a,b], and the resulting function of x is an indefinite integral, i.e. an antiderivative, of f. Unfortunately most of us come out of calculus thinking that a "function" must be something constructed from powers of X, and sines and cosines, or possibly exponentials and logs. The question of which functions have indefinite integrals of that type, is quite different, almost a parlor game, and much less important than understanding the fundamental theorem. That parlor game does have a strategy, and has its own interest, if somewhat narrow. It is called the theory of "integration in elementary terms".

http://math.stanford.edu/~conrad/papers/elemint.pdf

 

[AndreasC]

It's kind of surprising that Lebesgue integration was never brought up. It's another (more involved) approach to integrals that is more general than Riemann sums, and math people tend to view it as the "standard" way to define integrals.

 

[epenguin]

Have we nonmath people been doing Riemann integration like we have been speaking prose, without knowing it?

 

[AndreasC]

Have we nonmath people been doing Riemann integration like we have been speaking prose, without knowing it?

There are actually some math people who are very adamant about Lebesque integrals being the "right" integrals. One such math person is Dieudonne. I remember him being really pissed at Riemann integrals in his real analysis book for some reason, or it was pretty amusing. But really for most of physics it doesn't make a difference and the Lebesgue integral is just extra prerequisites to learn.

 

[neilparker62]

Interesting thread to read - thanks to all the contributors!

 

[martinbn]

There are actually some math people who are very adamant about Lebesque integrals being the "right" integrals. One such math person is Dieudonne. I remember him being really pissed at Riemann integrals in his real analysis book for some reason, or it was pretty amusing. But really for most of physics it doesn't make a difference and the Lebesgue integral is just extra prerequisites to learn.

I disagree. For most physicists it does matter, whether they realize it or not. For example the $L^2$ spaces in quantum mechanics use the Lebesgue integral. The Riemann integral is not enough.

 

[AndreasC]

I disagree. For most physicists it does matter, whether they realize it or not. For example the $L^2$ spaces in quantum mechanics use the Lebesgue integral. The Riemann integral is not enough.

Yeah I guess you're right on that aspect... I'm just not sure many physicists will "need" to formally know it and use it.

 

[FactChecker]

Yeah I guess you're right on that aspect... I'm just not sure many physicists will "need" to formally know it and use it.

We probably need to distinguish between "using" Lebesgue integration in a theoretical development versus using it to actually compute an integral value. It is the basis upon which a lot of theory has been established, but I do not think that it is used often to calculate the value of a real-world integral. But then, neither is the Riemann integral used much directly. Most integration is done using the Fundamental Theorem of calculus.

 

[FactChecker]

We probably need to distinguish between "using" Lebesgue integration in a theoretical development versus using it to actually compute an integral value. It is the basis upon which a lot of theory has been established, but I do not think that it is used often to calculate the value of a real-world integral. But then, neither is the Riemann integral used much directly. Most integration is done using the Fundamental Theorem of calculus.

Actually, I should be more careful here. There are a lot of numerical integration algorithms. As far as I know, they all approach the problem as a Riemann integral. I don't know of any of the algorithms that mimic or approximate an integral from the viewpoint of a Lebesgue integral.

 

[Mark44]

Actually, I should be more careful here. There are a lot of numerical integration algorithms. As far as I know, they all approach the problem as a Riemann integral.

Right, I was going to jump in here and say the same thing, but you beat me to it. For numerical integration, there are different techniques that are similar to, but not exactly the same as what is done in Riemann integration. For example, some techniques use interpolating polynomials to mimic the shape of the curve being integrated, while others use trapezoids or other figures in the approximations.

 

[mathwonk]

edit:

oops, the first version seems flawed. hence, after sleeping on it, i will make another attempt, but still without claim of sufficiency. here goes:

second attempt: CAVEAT: Throughout this discussion the word "integrable" means having a finite integral.

As a non expert in lebesgue integration, I am nonetheless going to go out on a limb and suggest that lebesgue integration is not as scary as it seems. Yes the definitions and proofs are complicated to go through in detail, but the idea is simple enough. The idea of lebesgue integration is to build a larger space of integrable functions (larger than riemann integrable ones) in which limits of integrable functions are again integrable. moreover the elements of this space are just limits of riemann integrable functions. In particular, for those functions that are already riemann integrable, the lebesgue integral does not change. Thus the lebesgue integral of a riemann integrable function is the same as its riemann integral, hence is evaluated by the FTC, and hence the integral of a lebesgue integrable function is a limit of integrals obtained from the FTC. the only new feature is that there exist lebesgue integrable functions that are not riemann integrable, e.g. many unbounded functions.

thus you will only encounter the lebesgue integral essentially in two situations:
i) you have a function at hand which is not riemann integrable, and you want to know if it is lebesgue integrable, and then what its lebesgue integral is.

or else:
ii) you have a sequence of riemann integrable functions whose limit function is not riemann integrable, and you want to know whether that limit is lebesgue integrable, and then what that lebesgue integral equals.

the answer to all these questions is basically the same: namely, a function is a limit of riemann integrable functions, if and only if that limit is lebesgue integrable, and then the lebesgue integral of that limiting function is equal to the limit of the integrals of the riemann integrable functions approximating it.

so what do we mean by limit? this depends on a notion introduced by riemann but usually attributed to lebesgue by people who have not actually read riemann, which is apparently almost everyone. [I had the great good fortune some years back, to be assigned to review, for Math Reviews, the english translation of riemann's works, and hence was motivated, almost forced, to read riemann, aided enormously by the excellent translation. It took me something like one day per sentence to manage even a few pages.]

riemann, in his original paper defining his integral, also proved a theorem about exactly which functions are riemann integrable. namely a bounded function is "riemann integrable" (not his terminology) if and only if, for every e>0, the set of discontinuities of that function can be covered by an infinite sequence of intervals the (infinite) sum of whose lengths is < e, i.e. iff the set of discontinuities has "measure zero".

[the reason riemann is not credited with this result may also be partly that he proved a slightly different but trivially equivalent version of it, in his paper: "On the representation of a function by trigonometric series", chapter 5. I.e. he defined the "oscillation" of a function to be a positive number which is zero exactly at points of continuity, and showed that a function is riemann integrable if and only if for every d>0 and every e>0, the set of points with oscillation > d has content < e, i.e. can be covered by a finite sequence of intervals of total length < e. Taking the union over all d = 1/n, for all n, and using compactness and the trivial fact that a countably infinite union of sets of content zero has measure zero, the lebesgue criterion immediately follows. Thus I am confident that lebesgue himself knew he was only rephrasing the result of riemann. ]

a set with the property given above for the set of discontinuities of a riemann integrable function is therefore now called a set of "(lebesgue) measure zero", and this criterion is called "lebesgue's criterion".

then the notion of limit that we want is essentially this: a sequence {fn} of functions converges to a function f, if and only if (they have the same domain and) the set of points x in the domain where {fn(x)} does not converge to f(x), has measure zero. otherwise stated: {fn} converges to f "almost everywhere". We also need a "cauchy" criterion on the sequence {fn} in terms of their (riemann) integrals.

thus f is lebegue integrable if and only if there is a sequence {fn) of riemann integrable functions (or smooth functions, or step functions) such that:
1) {fn} converges to f pointwise almost everywhere, and
2) the sequence {fn} is a cauchy sequence of functions, in the sense that the integrals of the differences {|fn-fm|} approach zero as n,m approach infinity.

In this case, the integrals of the {fn} form a cauchy, hence convergent, sequence of real numbers, and the integral of f is the limit of the integrals of the {fn}.

If you are familiar with metric spaces and the practice of enlarging a metric space to a complete metric space, the lebesgue integrable functions are exactly the enlargement of the metric space of riemann integrable functions to a complete metric space, in which the riemann integrable functions are dense. The metric, or norm, since it is a vector space, is that the (semi-) norm of a function is the integral of its absolute value.

One modification is that we have to introduce an equivalence relation on functions, where a lebesgue integrable function f is considered zero if the integral of |f| is zero, (i.e. if the norm of f is zero), if and only if the function f equals zero almost everywhere. In particular two lebesgue integrable functions are considered equal if and only if they differ only on a set of measure zero.

Moreover the two properties 1), 2), above are almost equivalent, in the following sense:

If f is any function which is almost everywhere the pointwise limit of a sequence {fn} of riemann integrable functions as in property 1), and if you know all the functions |fn| are bounded above by some riemann or lebesgue integrable function g, then f is lebegue integrable, the sequence {fn} is cauchy in the sense of property 2), and the lebesgue integral of f equals the limit of the integrals of the {fn}. ("Dominated convergence")

Moreover, assume you have a sequence {fn} of riemann integrable functions, which is cauchy in the sense of property 2). Then at least some subsequence of the sequence {fn} converges pointwise to some function f almost everywhere, in the sense of property 1). And in fact the subsequence can be chosen so that for every e>0, the convergence is uniform off a set which can be covered by a sequence of intervals of total length < e. The function f is, almost everywhere, independent of the choice of subsequence used to define it.

My original condition 2), which merely asked for the integrals of the {fn} to be a cauchy sequence of real numbers, is in fact sufficient for monotone sequences. I.e. if {fn} is a monotone (say increasing) sequence of riemann integrable (or lebesgue integrable) functions whose (hence also monotone) sequence of integrals is a cauchy, i.e. bounded, sequence of real numbers, then the sequence of functions {fn} is cauchy in the stronger sense of the current property 2) above, and converges almost everywhere pointwise to some lebesgue integrable function f. Moreover, the convergence is also true in the integral norm, i.e. the sequence of integrals of the differences {|f-fn|} converges to zero. ("Monotone convergence")

So lebesgue integration is nothing but the theory of limits of riemann integrals, and lebesgue - integrable functions are essentially pointwise limits (almost everywhere) of riemann integrable functions, with an extra condition so that the limit of the integrals makes sense.

There are also good Fubini theorems: i.e. a function which is integrable as a function of two variables can be integrated by repeated integration, and a function of two variables which is at least a pointwise limit almost everywhere of integrable functions, and for which the repeated integrals make sense, is itself integrable as a function of two variables, and hence the integral can be computed as a repeated integral. [I mention this since I once had a course from an expert who remarked (roughly) "if you have a problem in integration theory that won't yield either to dominated convergence or Fubini, you're probably in trouble".]

Here is an example (in one variable): let f be defined on [0,1] and have value 0 at all irrational points, and value n at rational points of form m/n in lowest terms. This function is unbounded on every interval, and discontinuous everywhere, hence nowhere near riemann integrable. However, since a countable set has measure zero, hence the rationals have measure zero, this function is equivalent in the sense above to the constant function zero, i.e. they differ only on a set of measure zero. Hence the lebesgue integral of f equals the integral of the zero function, namely zero. More generally, if your function differs from a riemann integrable function only on a set of measure zero, then your function has the same lebesgue integral as that riemann integrable function.
here is another one: the function x^(-1/2) on [0,1], but with value 0 at x=0. This cannot be made equal to a riemann integrable function by changing it only on a set of measure zero, but it is a pointwise limit of monotone increasing riemann integrable functions whose integrals converge, as in the theory of "improper integrals". So this function is lebesgue integrable and its lebesgue integral equals its value as an improper riemann integral.

Unfortunately I am not an expert, so I might again make some, possibly serious, mistakes here, but this in my opinion is the key idea. A nice clear place to read about it is in the great book "Functional Analysis" by Riesz-Nagy. I do not especially recommend Dieudonne's own book, (vol. 2 of his admittedly great Foundations of analysis), where to me at least it seems far more abstract. Indeed I take with more than a grain of salt his famous diatribe against riemann integration, which I consider somewhat misleading.

Here is a nice result from Riesz-Nagy (p.36) to show the clarity of their exposition: Suppose {fn} is a sequence of riemann-integrable (or lebesgue-integrable) functions such that the integrals of the absolute values |fn| form a convergent series of real numbers; then the series whose terms are the functions fn, itself converges pointwise almost everywhere to a lebesgue integrable function f, and the series may be integrated term by term.

Moreover I believe Riesz-Nagy show that every lebesgue - integrable function is obtained either as a limit almost everywhere of a monotone sequence of ordinary step functions with bounded integrals (as in the monotone convergence theorem above), or as a difference of two such limits.

Experts I know in this subject seem to me mostly to enjoy the abstract theory of measure, and measurable sets, measurable functions, and like to develop everything from scratch, without taking any advantage of the theory of riemann integration. To be sure Dieudonne' does not do this, but his version is to me even more abstract, defining lebesgue - integrable functions as limits of smooth functions, somewhat as I do here, but in a way that looked to me a bit forbidding. Of course it might look clearer to me now, (indeed I believe he generalizes in some sense the Riesz-Nagy approach), but I cannot look back there now, since I rashly gave away most of my set of Dieudonne' when moving.

Again, since I am not expert, you will do yourself a favor by consulting an expert source. Books I like include Riesz-Nagy (Functional analysis), or Berberian (Fundamentals of real analysis) , or Lang (Analysis II), or Wendell Fleming (Functions of several variables), or (I don't have this last one, but some expert friends recommend it): Wheeden and Zygmund (Measure and Integral). Many people also like especially part I of the classic Real analysis, by H.L. Royden. These books may be harder to read than this sketch, but the advantage is that what they say will actually be correct, and will include proofs and (counter) examples. Oh yes, the book Counterexamples in analysis, by Gelbaum and Olmsted is fun too.

Here is an article going at things in the opposite order to here. I.e. after having developed lebesgue integration and measure theory in the usual way, he proves afterwards that all lebesgue integrable functions are limits of continuous functions in the integral norm, in particular they are all limits of riemann integrable functions.
https://www.math.utah.edu/~savin/L3_5210.pdf
Hence he could have defined lebesgue integrable functions as such limits.

 

[Svein]

Think of Lebesgue integration as a Riemann integration rotated 90 degrees - as in this figure:

Riemann principle on the left, Lebesgue on the right.

Instead of dividing the integration interval in small portions and figure out what the value of the integrand is in each interval, we divide the function values in small portions and figure out what the "measure" of the corresponding input values is.

The figures makes it seem irrelevant what type of integration you use - and that is the case for continuous functions. Extremely discontinuous functions (like f(x)=1 for x rational and 0 otherwise) cannot be solved by the Riemann method, but the Lebesgue integral gives the answer straight away. Check out
https://math.stackexchange.com/questions/2556232/any-simple-example-of-lebesgue-integration

 

[martinbn]

Think of Lebesgue integration as a Riemann integration rotated 90 degrees - as in this figure:
View attachment 280376
Riemann principle on the left, Lebesgue on the right.

Instead of dividing the integration interval in small portions and figure out what the value of the integrand is in each interval, we divide the function values in small portions and figure out what the "measure" of the corresponding input values is.

The figures makes it seem irrelevant what type of integration you use - and that is the case for continuous functions. Extremely discontinuous functions (like f(x)=1 for x rational and 0 otherwise) cannot be solved by the Riemann method, but the Lebesgue integral gives the answer straight away. Check out
https://math.stackexchange.com/questions/2556232/any-simple-example-of-lebesgue-integration

That is not how Legesgue integral is. You don't take horizontal slabs, you take verical ones and you group those of the same height.

 

[mathwonk]

I think he meant to look at the graph of a function and intersect it with a horizontal slab. Then look at all those points on the x-axis lying below those points. This gives, as you said, a subset of the domain where the values of the function are almost the same height. Then however the challenge is to define or measure the size of those sets, which can be very exotic. This leads to the idea of measurable sets, and "simple" functions which have only finitely many values, and are constant on certain, possibly very complicated, measurable sets. The integral is then defined using those functions. I.e. an approximation to the integral of a function is defined as the (e.g. lower) height of a horizontal slab, multiplied by the measure of the set of domainpoints where the values lie in that slab, and then summed over all slabs.

I agree this gives a lovely intuitive sense of lebesgue integration. Unfortunately for me at least, this process of taking as the fundamental objects "simple" functions, whose domains consist of disjoint unions of "measurable" sets, leaves me with no good intuition as to which sets are measurable. If I can't intuit a measurable set, I can't begin to decide in practice which functions are measurable, or integrable. I.e. this process begins with a very exotic class of "simple" functions, and then takes limits to define lebesgue integrable functions. So I am lost from the start. The approach I outlined above at least starts from a very familiar class of functions, the riemann integrable ones. In fact in the approach of Riesz-Nagy, where just monotone limits, and then differences are allowed, one can start from even easier functions, i.e. step functions on intervals, or continuous, or riemann integrable ones if you prefer. I admit that in applying the usual measurable function version of lebesgue integration, I am left just assuming all sets I meet are measurable, since non measurable ones are so hard to construct (it seems to require the axiom of choice), that one is unlikely to encounter one.

You may find the linked discussion interesting:
https://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/53532#53532

 

[FactChecker]

That is not how Legesgue integral is. You don't take horizontal slabs, you take verical ones and you group those of the same height.

You really group those of similar integrand height, which is where the description of "horizontal slabs" comes from. You are hinting at the concept of the Lebesque measure when you talk about the vertical ones. That is important, but it does not really contradict @Svein's post.

 

[BWV]

A real mathematician can tell the difference when y=2x is integrated with a Riemann or Lebesgue integral ;)

We probably need to distinguish between "using" Lebesgue integration in a theoretical development versus using it to actually compute an integral value. It is the basis upon which a lot of theory has been established, but I do not think that it is used often to calculate the value of a real-world integral. But then, neither is the Riemann integral used much directly. Most integration is done using the Fundamental Theorem of calculus.

FWIW numerical methods for stochastic differential equations, which are not Riemann- integrable, use a modification of Euler’s method

https://en.m.wikipedia.org/wiki/Euler–Maruyama_method

 

[mathwonk]

Here's another, admittedly amateurish, contribution to evaluation of lebesgue integrals.

We have said so far that a function f has a lebesgue integral if it is almost everywhere the pointwise limit of riemann integrable functions which form a cauchy sequence in the sense of the integral norm, and we can integrate term by term in this case to evaluate the integral of f. Moreover if the riemann integrable functions in the sequence are all bounded above (in absolute value) by some one riemann or lebesgue integrable function, we don't even have to check the cauchyness, as it will always hold. So for this method, given a lebesgue integrable function, to evaluate its integral we look for a sequence of riemann integrable functions that converge to it pointwise a.e., and then try to verify integral norm cauchyness, which at worst requires evaluating the integrals of riemann integrable functions, hence may be susceptible to the usual fundamental theorem of calculus.

There is also a theory for evaluating lebesgue integrals directly using antiderivatives i.e. a fundamental theorem of calculus for lebesgue integration. It uses one new concept, that of "absolute continuity", which is a generalization of uniform continuity.

Recall that uniform continuity for F, means for every e>0 there is a d>0 such that for every subinterval [s,t] of the domain of length less than d, we have |F(t)-F(s)| < e. The generalization is to allow several subintervals of total length less than d. I.e. F is absolutely continuous if for every e>0, there is a d>0, such that for every finite collection of non overlapping subintervals [si,ti] in the domain, with total length less than d, we have the sum of the (absolute) differerences |F(ti)-F(si)| less than e. [The key property of these functions is a generalization of (a corollary of) the mean value theorem, namely an absolutely continuous function whose derivative is zero almost everywhere is constant.]

Lebesgue’s fundamental theorem of calculus says that every Lebesgue integrable function f on [a,b] is the derivative, almost everywhere, of an absolutely continuous function F, which is unique up to a constant, and that for any such antiderivative F, the integral of f on [a,b] equals F(b)-F(a). Conversely, every absolutely continuous function F on [a,b] is differentiable a.e. and the derivative f has a lebesgue integral on [a,b] equal to F(b)-F(a).

[Remark: For perspective, continuous differentiability on [a,b] implies lipschitz continuity, which implies absolute continuity, which implies uniform continuity, which implies continuity.]

This gives us another way to evaluate the lebesgue integral of some functions. E.g. in the case of the function f given above on [0,1], which equals zero at irrationals, and equals n at a fraction of form m/n in lowest terms, this function equals, at all irrationals, the derivative of a constant function c, which is certainly absolutely continuous. Hence this function f is lebesgue integrable, and its integral equals c-c = 0.

On the other hand, this method also let's us see that some functions are not lebesgue integrable. if we take the function f(x) = (1/x).cos(1/x) on (0,1] and f(0) = 0, then f is unbounded on every nbhd of 0, hence not riemann integrable, but we will soon see that it is not even lebesgue integrable. Namely, [up to sign and a difference term sin(1/x) which is bounded and continuous a.e. on [0,1], hence riemann integrable], f has an antiderivative of form F(x) = x.sin(1/x), except at 0.

This F is continuous, but using the fact that the harmonic series has infinite sum, one can check that this F is not absolutely continuous, hence its derivative, namely essentially f, is not lebesgue integrable! (Since F is however continuously differentiable, hence absolutely continuous on every subinterval of form [e,1], there cannot be any absolutely continuous antiderivative of f on all of [0,1], since any such would have to agree, up to a constant, with this one on (0,1], hence also on [0,1].)
To me this example is somewhat analogous to the fact that the alternating series 1 - 1/2 + 1/3 - 1/4 + 1/5 - +... is convergent but not absolutely convergent, i.e. apparently even lebesgue integrals are not allowed to exist solely by virtue of cancellation of positive and negative parts. I.e. a function f is lebesgue integrable if and only the same is true for |f|.

 

[mathwonk]

One more comment on when functions are not integrable even using lebesgue integration.
Intuitively, a function is lebesgue integrable if it is not too big and not too rough. Since the lebesgue integral is a monotone function of its integrand, a lebesgue integrable function cannot be approximated from below by integrable functions whose integrals are unbounded above. E.g. the function f(x) = 1/x on [0,1], with say f(0) = 0, is too big in this sense, because the usual approximating Riemann integrable functions obtained by restricting this function to the subintervals [e,1], as e—>0+, have unbounded integrals. So this function is not only not riemann integrable, but also not lebesgue integrable.

A function is too rough, if it is not the pointwise limit a.e. of any sequence of riemann integrable functions. It takes more sophistication to produce a function that is too rough, but the axiom of choice allows such a construction. Define an equivalence relation on the interval [0,1], so that two numbers are equivalent iff their difference is rational, and then choose exactly one element from each equivalence class to form a subset S of [0,1].

Then each number in [0,1] is equivalent to exactly one element of S. Thus [0,1] can be covered by a sequence of disjoint rational translates of S. (Given a rational number, add it to each element of S, and if the sum is > 1, chop it off at 1 and slide it back to start at 0, i.e. just subtract 1 from it. Or if you prefer, consider [0,1] essentially as the quotient space R/Z where rational translation is defined.] Now write this countable collection of translates of S as a sequence: S1, S2, S3,….and define functions as follows: gj = 1 at exactly all points of Sj, and zero elsewhere. Then f 1 = g1, f2 = g1+g2, f3 = g1+g2+g3,…..

Now the sequence fj converges upward pointwise everywhere to the constant function 1 on [0,1]. These functions fj are all bounded between 0 and 1, hence not too big. But 1 is integrable with integral 1 on [0,1]. So if the functions fj are integrable, they must have integrals bounded by 1, and their sequence of integrals must converge upward to 1. Now lebesgue measure is translation - invariant, so since the sets Sj are all essentially translations of S, the functions gj all must have the same integral, equal to some number between zero and 1. But what is the value of that integral? It cannot be zero since then the integral of all the functions fj would also be 0, and then 1 would be the limit of a sequence of zeroes, contradiction. But also the integral of the gj cannot be equal to a positive number e>0, since then the integral of fn would equal the sum of the integrals of g1,…,gn, hence equal to ne, which eventually becomes larger than 1, also a contradiction. So the functions gj and fj, although not too big, are too “rough” to be lebesgue integrable.

I'll probably quit here. Thanks for your indulgence. I myself think I learned something by doing and summarizing the reading needed for this discussion.

One more comment gleaned from that long mathoverflow discussion linked above. Someone there claimed that a great Harvard mathematician, maybe Brauer, remarked that after one reads the abstract book Measure Theory, by Halmos, one will know very well how to integrate the function 1, but will not know anything about how to integrate the function x.

 

[FactChecker]

One more comment on when functions are not integrable even using lebesgue integration.
Intuitively, a function is lebesgue integrable if it is not too big and not too rough. Since the lebesgue integral is a monotone function of its integrand, a lebesgue integrable function cannot be approximated from below by integrable functions whose integrals are unbounded above. E.g. the function f(x) = 1/x on [0,1], with say f(0) = 0, is too big in this sense, because the usual approximating Riemann integrable functions obtained by restricting this function to the subintervals [e,1], as e—>0+, have unbounded integrals. So this function is not only not riemann integrable, but also not lebesgue integrable.

The function is both Riemann and Lebesgue integrable. The definite integral of both on [0,1] are infinite.

 

[mathwonk]

Forgive me, but of course my assertions use the definitions I gave, and my definition of lebesgue integrable includes that it has finite integral. and my definition of riemann integrable (which is the one given by riemann) requires the function to be bounded, hence the riemann integral is also finite. I know some people admit infinity as a value of an integral, but the sources I use, such as Berberian and Lang define "integrability" as meaning the integral is finite. Cf. page 165 of Berberian, or page 237 of Lang Analysis II. (The same is true in Rudin, Riesz-Nagy, and Fleming, hence I suspect this was probably Lebegue's own usage, or he may have used the word "summable" or a French version of that.) It may be a little confusing since Rudin e.g. does write down integrals which are infinite, but he calls a function "Lebesgue integrable" only when its integral is finite. Fleming also writes such integrals, calling them "divergent". So in these books a function whose integral is infinite is not "integrable".

Of course this all depends on your definition of "integrable". Fortunately I gave my definition, and it included that the approximating sequence of riemann integrable funcions is cauchy in the sense of the integral norm. Thus, with my definition of " lebesgue integrable" it is a consequence that the lebesgue integral is finite.
...
I have had some difficulty tracking down a copy of Lebesgue's own work, but finally found an online copy of his "Lecons sur l'integration" where on p. 115, he defines an unbounded function to be "sommable" if the infinite series of its approximating sums is convergent, hence finite. He of course does not use the term "Lebesgue integrable". Interestingly, he states repeatedly that he is unaware of any examples of bounded functions which are not summable, or measurable, apparently being unaware (in 1903) of the example in post #31, using the axiom of choice.

 

[FactChecker]

Forgive me, but of course my assertions use the definitions I gave, and my definition of lebesgue integrable includes that it has finite integral. and my definition of riemann integrable (which is with the one given by riemann) requires the function to be bounded, hence the riemann integral is also finite. I know some people admit infinity as a value of an integral, but the sources I use, such as Berberian and Lang define "integrability" as meaning the integral is finite. Cf. page 165 of Berberian, or page 237 of Lang Analysis II.

Of course this all depends on your definition of "integrable". Fortunately I gave my definition, and it included that the approximating sequence of riemann integrable funcions is cauchy in the sense of the integral norm. Thus, with my definition of " lebesgue integrable" it is a consequence that the lebesgue integral is finite.

Ok. I don't like it, but I guess I can accept it.

 

[mathwonk]

I don't claim these are the only possible defitinions, (even if, in the case of riemann's, they have the historical weight of being riemann's own). You are perfectly free to give, or cite, different ones, but these are the ones that appear in the sources I am aware of. I will try to find Lebesgue's own definition.

Or you are free to consider my discussion as a characterization of functions that are not only integrable, but that also have finite integral. As such the discusion should have value to you even if you choose to use the word "integrable" in a different sense.

Here's a potential fix. Introduce the word "summable" to mean integrable with finite integral, and then interpret this whole discussion as about summable functions.

And thank you for reminding me of the possible confusion in terminology.