Exploring the Relationship Between Riemann and Lebesgue Integration

[quasar987]

In the course I'm taking, we are already done with Lebesgue integration on R, and while we have proven that for continuous fonctions, the Riemann integral and the Lebesgue integral give the same output, we have not investigated further the correspondance btw the two. So I have some questions...

i. Is the class of (absolutely) Riemann integrable function a subset of Lebesgue integrable ones? Or are there functions that are (absolutely) Riemann-integrable but whole Lebesgue integral diverge?

ii. For those Riemann-integrable functions that are also Lebesgue integrable, are the Riemann and Lebesgue integral equal?I would also be interested to know what were the historical motivations for creating this new theory of integration, and also, what are its use today?

 

[morphism]

Every Riemann integrable function is Lebesgue integrable and their integrals are equal.

The Lebesgue integral is really an extension of the Riemann integral, in the sense that it allows for a larger class of functions to be integrable, and it does not succumb to the shortcomings of the latter (e.g. interchanging limits and integrals behaves better under the Lebesgue integral).

The Wikipedia page on Lebesgue integration is nice.

 

[Big-T]

Does this count even for unproper Riemann integrals?

 

[Dragonfall]

Do you have an example of an interesting function over some domain that is not Riemann integrable, but Lebesgue integrable, and compute its integral? I took a course in measure theory, and it seemed that all we did was prove existence theorems and did nothing computationally.

 

[Big-T]

An example may be the function defined as 1 for rational x, and 0 elsewhere, but of course, this depends on what you consider interesting.

 

[mathwonk]

the reason for extending the definition is to have the limit of the integrals equal the integral of the limit more often.

a sequence of riemann integrable functions can have convergent integrals without the limit function being riemann integrable. lebesgue integrals repair this flaw.

 

[morphism]

Does this count even for unproper Riemann integrals?

Sadly, no. There are many examples; a couple of typical ones are (1/x)sin(1/x) on [0,1], or sin(x)/x on R.

As for computational things, http://www.math.ksu.edu/~ryabs/tar7.pdf seems to have some interesting ones.

 

[Kummer]

It is impossible to construct a non-Lebesaure measurable set which does not involve the axiom of choice. That is a nice advantage.

 

[mathwonk]

i recommend wendell fleming, calculus of several variables. i am not an expert but will recite what i just read there for you.

i consider only functions with bounded domains.

a step function is a function with a finite number of values, whose level sets are "measurable sets", i.e. sets to which a length can be asigned. for example intervals are measurable. but also much more complicated sets are measurable by approximating them by intervals and taking limits.

the integral of a step function with values a1,...an on the sets S1,...,Sn is the linear combination a1 m(S1)+...an m(Sn), where m(S) is the length (measure) of S.


now a function f is integrable if for every e>0, f admits both larger and smaller step functions, whose integrals differ by less than e.

a function is riemann integrable if the step functions can be chosen to be constant on intervals, i.e. if the sets Si can be chosen as intervals.

thus the integral is seen to be a direct generalization of the riemann integral, where the level sets of the approximating step functions are allowed to be more general than intervals.

a function is continuous if and only if the inverse image of each open interval is open. it is called measurable if the inverse image of each interval is measurable.

then a bounded function is integrable if and only if it is measurable. since open sets are measurable, in particular all continuous functions are integrable.

i hope this is correct.

 

[rbj]

from an engineer's POV, here is a cute quote from Richard Hamming:

Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane.

when i first learned about the Lebesgue integral, i noticed right away that it pointed to problems of interpretation and usage of the Dirac delta function. engineers think of it differently than mathematicians. we treat $\delta(x)$ as a function that is zero almost everywhere, yet with an integral of 1, which is contrary to what Lebesgue integration would say. Lebesgue integrations says that if f(x) agrees with g(x) almost everywhere, and if f(x) is integrable, so is g(x) and its integral is the same. in the physical sciences, i think that is a perfectly good way to look at it (and indeed there are undergraduate Electrical Engineering texts written as such), but the mathematicians do not like it.

anyway, from a point-of-view of describing real physical processes and quantities, i hadn't worried much about the difference betweent the two.

 

[mathwonk]

what a philistine, head in sand, attitude. if it doesn't immediately build a better bridge, who cares?

 

[rbj]

it's not about immediately building anything, it's just that it is very convenient in Linear System Theory (the whole sub-discipline where we learn about signals, spectra, linear and time-invariant systems, impulse responses, convolution, etc.) to simply treat the dirac delta as:

$$\delta(t) = \lim_{\Delta \rightarrow 0} \delta_{\Delta}(t)$$

where $$\delta_{\Delta}(t)$$ is any of a bunch of "nascent delta" functions all having unit area in them. the simplest example is:

$$\delta_{\Delta}(t) = \frac{1}{\Delta} \mathrm{rect} \left( \frac{t}{\Delta} \right)$$

and rect() is the unit rectangular function (i can't remember how to express it nicely in LaTeX).

anyway, it's far better to use this representation with undergrad EEs than it is to try to explain to them the difference between Lebesgue and Riemann or how the Dirac delta is a "distribution" and not a function. also, Richard Hamming was pretty well respected among academics when he was alive (the "Hamming window" is his creation), so i don't know who is the Philistine you're referring to. perhaps me, and that's fine, but all of us EEs are Philistines, then. the differentiation of the simple limit of functions definition and the more proper distribution definition of the Dirac delta has never been shown to be useful in the descriptions of physical systems all of which have some decent continuity in their properties.

 

[HallsofIvy]

from an engineer's POV, here is a cute quote from Richard Hamming:



when i first learned about the Lebesgue integral, i noticed right away that it pointed to problems of interpretation and usage of the Dirac delta function. engineers think of it differently than mathematicians. we treat $\delta(x)$ as a function that is zero almost everywhere, yet with an integral of 1, which is contrary to what Lebesgue integration would say. Lebesgue integrations says that if f(x) agrees with g(x) almost everywhere, and if f(x) is integrable, so is g(x) and its integral is the same. in the physical sciences, i think that is a perfectly good way to look at it (and indeed there are undergraduate Electrical Engineering texts written as such), but the mathematicians do not like it.

anyway, from a point-of-view of describing real physical processes and quantities, i hadn't worried much about the difference betweent the two.

Actually, the difference between Riemann and Lebesque integrals should be of great importance to engineers! When Fourier, who was himself an engineer rather than a mathematician, developed Fourier series to solve differential equations, he made two claims: first, that any integrable, periodic, function could be expanded in a Fourier series and second, that any (convergent) Fourier series gave an integrable, periodic series. The first statement was obviously true. The second just as obviously false using the Riemann integral. The Lebesque integral was developed specifically to make that second statement true- since Fourier's technique obviously worked!

 

[matt grime]

rbj - since the dirac delta isn't a function from R to R, I don't really get your point. It is a functional on a some suitable space, not a function from R to R.

 

[HallsofIvy]

matt, rbj made exactly that point- that it isn't a function. In fact, it is possible to treat distributions (or generalized functions) as equivalence classes of sequences of functions. That, in effect, is what rbj is doing.

 

[matt grime]

No, I meant 'why did rjb say that lebesgue integration implies problems for...' It doesn't imply any such thing. Whilst some may treat it as if it were a function, it isn't a function, and everyone knows that, including engineers.

Indeed one doesn't even need to invoke Lebesgue integration. If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist.

 

[rbj]

No, I meant 'why did rjb say that lebesgue integration implies problems for...' It doesn't imply any such thing. Whilst some may treat it as if it were a function, it isn't a function, and everyone knows that, including engineers.

i don't think you understand what it's like in a 4-year undergrad Electrical Engineering program. now it might be worse in Math and/or Physics (i think it's conceptually deeper in Math/Physics than EE), but in a 4-year EE program, each and every semester is packed with required courses and "required electives", if you drop even one, you must make it up in the summer or you will not graduate in 4 years. anyway, in this tight schedule, the only math courses in the Math department you can expect will be 3 semesters of Calculus and 2 semesters of Diff Eq. (and similar) following. in the Diff Eq. the EE student will see Laplace Transforms and will see the Dirac Delta presented much the same as i had mentioned (mostly so we can learn that $$\mathcal{L}\{ \delta(t) \} = 1$$), and that is it. these students have not had "Real Analysis" or "Advanced Calculus" or whatever it is when you get to the real anal definitions of continuity, limits, and the derivative/integral. no "countably infinite" vs. "uncountably infinite". no Lebesgue measure or Lebesgue anything (only Riemann integral/summation). very little "given any epsilon>0, find a delta so that [such-and-such] is less than that epsilon." certainly no functional analysis or "generalized functions" or "distributions" (but they might use the word "distribution" in a course about probability).

so no. the fact is that nearly no engineer knows that the Dirac delta function is not a limit of functions as was presented to him/her in undergraduate courses. almost every electrical engineer, if they mess with the Dirac delta at all, treats it simply as a function that is zero at all non-zero values of the argument (it is zero "almost everywhere") yet it has an integral of 1, if the limits straddle 0. and from a physical POV, that's okay. not much physical difference between a "real" Dirac delta and a rectangular approximation that is, say, a femto-second wide. and if a femto-second is too wide, make the width a Planck time. i don't think physical reality will know the difference or care.

i even asked about this a few years back on the sci.physics.research site and there were physicists who admitted to using and thinking about the Dirac delta in the same way that EEs do.

while i do not challenge the mathematical legitimacy of the pure mathematical treatment of the Dirac delta (you would have to define Lebesgue integration differently to say something other than "if f(x)=g(x) almost everywhere, and if f(x) is integrable, then g(x) must also be and the integral is the same as f(x) over the same region." I know that. but nonetheless, in Linear System Theory, in Control Systems Theory, Communications Systems, Digital (or analog) Signal Processing, any of this to any level of advancement, it is just easier (and it doesn't break anything) to just think of the Dirac Delta function as a function that agrees with f(x)=0 at all points except x=0, and yet does not have an integral of zero. that is the state of things outside the ivory tower.

Indeed one doesn't even need to invoke Lebesgue integration. If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist.

not in the limits. the Riemann integral of the "nascent delta functions" are all 1, yet, in the limit, all of the nascent delta functions will equal zero at every value of x except x=0 where it is undefined. the only questionable thing is we treat the Dirac delta as one of those nascent delta functions that is thin enough for our purposes.

 

[matt grime]

"not in the limits. the Riemann integral of the "nascent delta functions" are all 1, yet, in the limit, all of the nascent delta functions will equal zero at every value of x except x=0 where it is undefined. the only questionable thing is we treat the Dirac delta as one of those nascent delta functions that is thin enough for our purposes."

That doesn't make any sense. I have lost what it is you're attempting to say. As I am not taking limits, and only dealing with functions, just like the same statement when one replaces lebesgue with Riemann, then what you say has no bearing.

All of the applied mathematics I know (from before doing any rigorous analysis) says that the dirac delta is *like* a function. Not that it is a function from R to R. Clearly it isn't, as any engineer would be happy to agree with, as it is undefined at 0. It is a functional, though it is unnecessary to explain this in order to use it.

I will reiterate one point. One does not need to invoke Lebesgue integration at all to come up with the "problem" you mention:

"If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist".

 

[Kummer]

The delta function cannot be used for a conterexample about theorems about intergration. Because it is not a "function" eventhough it is called a function, rather it is a Schwartz distribution so the known results about integration might not apply to it.

 

[rbj]

"not in the limits. the Riemann integral of the "nascent delta functions" are all 1, yet, in the limit, all of the nascent delta functions will equal zero at every value of x except x=0 where it is undefined. the only questionable thing is we treat the Dirac delta as one of those nascent delta functions that is thin enough for our purposes."

That doesn't make any sense. I have lost what it is you're attempting to say.

$$\delta(x) = \lim_{a\to 0} \delta_a(x)$$

where

$$\delta_a(x) = \frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2}$$

or

$$\delta_a(x)= \frac{\mathrm{rect}(x/a)}{a}$$

$$\mathrm{rect}(x) = \begin{cases} 0 & \mbox{if } |x| > \frac{1}{2} \\[3pt] \frac{1}{2} & \mbox{if } |x| = \frac{1}{2} \\[3pt] 1 & \mbox{if } |x| < \frac{1}{2}. \end{cases}$$

or a variety of other nascent delta prototypes. in all these cases, if you think of or treat $\delta(x)$ as a function, if $x \ne 0$ then $\delta(x) = 0$ and for any $a > 0$, then

$$\int_{-\infty}^{\infty}\delta_a(x) dx = 1$$

so we Neanderthal engineers just ask "why not for $a = 0$ in the limit? that is if

$$\lim_{a\to 0} \int_{-\infty}^{\infty}\delta_a(x) dx = 1$$

why not

$$\int_{-\infty}^{\infty} \lim_{a\to 0} \delta_a(x) dx = \int_{-\infty}^{\infty} \delta(x) dx = 1$$ ?

As I am not taking limits,

but i am.

and only dealing with functions, just like the same statement when one replaces lebesgue with Riemann, then what you say has no bearing.

All of the applied mathematics I know (from before doing any rigorous analysis) says that the dirac delta is *like* a function. Not that it is a function from R to R.

i know that it is not, since the mapping from 0 is undefined.

Clearly it isn't, as any engineer would be happy to agree with, as it is undefined at 0. It is a functional, though it is unnecessary to explain this in order to use it.

I will reiterate one point. One does not need to invoke Lebesgue integration at all to come up with the "problem" you mention:

"If f=g at all but a point, then the Riemann integrals of f and g are equal if they exist".

[tex] f(x) = 0 [/itex]

$$g(x) = \lim_{a\to 0} \frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2}$$

except for x=0, g(x) is defined everywhere and it is 0. what it is at 0 is undefined. yet for any $a > 0$, any of the limit functions for g(x) has an integral of 1.

The delta function cannot be used for a conterexample about theorems about intergration. Because it is not a "function" eventhough it is called a function, rather it is a Schwartz distribution so the known results about integration might not apply to it.

yes, you are repeating the same thing.

this issue is pedagogical. (you are not going to be teaching a very large portion of undergraduate engineering students anything about "generalized functions" or "Schwartz distributions" or the fine distinctions between them and functions that behave as an extreme limit of functions that we all agree are functions.)

the issue is also practical. there is not a system in the Universe that will know the difference between a legit Dirac delta function (if it could exist somewhere in physical reality) with argument t and any nascent delta function where the width is the finite, but very small 10^-44 second. wasting our time with the language of "generalized functions" or "Schwartz distributions" has never been useful in the engineering classroom (and it's never done) nor in any practical system where we are trying to model an impulse of some sort.

Electrical Engineering texts in basic circuits, signals and systems (a.k.a. linear system theory), communications systems, control systems, filter theory, analog or digital signal processing, simply do not use the formal mathematical language of the Dirac delta as a "distribution, not a function". and we don't break anything by not using it.

 

[matt grime]

rbj, you're missing my point - you said that you spotted this problem when you were taught about lebesgue integration. Well, the same thing is true for Riemann integration, which you already knew.

 

[WWGD]

A couple of comments, hoping they will complement previous answers, and a couple
of new questions.


An example of a convergent sequence of functions {f_n}->f such that f_n is R-integrable but f is not:

Do an enumeration of the rationals in [0,1]. Define then f_n by:

f_n=1 , if x=q_1, q_2,..or q_n

f_n=0 otherwise.

Each f_n is R-integrable , but the limit function f= Char.Q in [0,1] is not,

(step functions f_1, f_2 with f_1<f<f_2 must satisfy f_2-f_1>=1 ).


Also, we can construct , for any set S of measure>0, a non-measurable

function on S : use the fact that if m(S)>0, then S has a non-measurable

subset. Then work with the characteristic function of S : The set

{x: 0<f(x)<=1} is non-measurable, so that, by def. f itself is not

measurable. Just thought that may help give insight into the two

integrals.


Now, a couple of questions: ( I am told I ask "weird" questions. I hope not,

sorry if so)

1) Do we use some type of measure in doing Riemann integration?. There is

this whole layout, in preparing for the Lebesgue integral, about Lebesgue

measurability. But there is no mention of any measure in the setup for

the Riemann integral. Why is this so?.


2) Any comment on how the Lebesgue integral repairs the convergence

problems of the Riemann integral?. Superficially, we are partitioning the

range, instead of the domain, and then , for a choice of element y_i* in

each [y_i-1,y_i) , we calculate m*(f^-1(y_i*)).

How does this help repair the convergence problems of the R-integral?.

Just wondering what may have guided Lebesgue to think that his setup

would improve the poor convergence properties?.


3) Is L^2[0,1] the metric completion of the space of Riemann square-

integrable functions?. Or is this the case for L^1[0,1] ?

Maybe I am far out on this one.

Thanks for any comments.

 

[rbj]

rbj, you're missing my point - you said that you spotted this problem when you were taught about lebesgue integration. Well, the same thing is true for Riemann integration, which you already knew.

i didn't think that Riemann integration could even deal with f=g almost everywhere in cases like the Dirichlet function. i see no theorem like that on page 80 of Royden ("Real Analysis" - i know it's an old text) regarding Riemann integration:

If f and g are bounded measurable functions defined on a set E of finite measure, then:

i. $$\int_{E} (a f + b g) = a \int_{E} f + b \int_{E} g$$

ii. If f = g a.e., then $$\int_{E} f = \int_{E} g$$

iii. If $f \le g$, then $$\int_{E} f \le \int_{E} g$$

...

i don't find concepts like "measurable" or "finite measure" or "a.e." in the context of Riemann integration. so it was in the context of learning about Lebesgue that item "ii." in the above first clued me in that there was a serious difference between how engineers (students, practicing, and every engineering prof that I've talked with about the subject) and mathematicians viewed and treated the Dirac delta. again, engineers (students, practicing, and every engineering prof that I've talked with about the subject) have no problem thinking of the Dirac delta (we call it the "unit impulse") as a "function" that is zero everywhere except at the origin, and the integral is 1, if the region of integration includes the origin. i know that mathematicians don't like it, but we do it anyway. we even put it in textbooks. we even write referreed scholarly papers with such a concept of the unit impulse. and it hasn't killed anyone. indeed, if i could paraphrase Hamming:

Does anyone believe that the difference between definitions of the Dirac delta as "Schwartz distribution" or as "function that is a limit of functions" can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane.

we (electrical engineers) say outrageous things like:

$$\sum_{n=-\infty}^{+\infty} \delta(t - n) = \sum_{k=-\infty}^{+\infty} e^{i 2 \pi k t}$$

SHAME! the Dirac deltas are naked and unclothed by integrals! avert your eyes! it must be a meaningless statement!

we say even more rubbish like: "The dimension of quantity of the dependent variable of $\delta(x)$ is the reciprocal of the dimension of quantity of the independent variable x, so that the area (or integral) of $\delta(x)$ comes out to be the dimensionless 1. when we electrical engineers deal with the convolution integral, this dimension of quantity thing is important.

regarding the Dirac delta, i think you, as a mathematician, would say that the disconnect is here:

$$\lim_{\Delta \to 0} \int_{-\infty}^{\infty}\delta_{\Delta}(x) dx = \int_{-\infty}^{\infty} \lim_{\Delta \to 0} \delta_{\Delta}(x) dx$$

given any of those definitions of $\delta_{\Delta}(x)$. if $\Delta > 0$, you would say that $\delta_{\Delta}(x)$ is a well-defined function (of x as it would also be of $\delta_{\Delta}(x)$, but i would like to call the latter a "parameter" of the function and x the argument), and has an integral of 1, but you would not say that

$$\delta(x) = \lim_{\Delta \to 0} \delta_{\Delta}(x)$$

is even a function nor should be thought of as a function. but we engineers treat it as such and no planes have crashed because of it. if you really don't like that, then define the "unit impulse" (in time) as the rectangular function of 1 Planck time in width, even the mathematicians would accept that as a function and us EEs would see no measurable difference between that and the Dirac "non-function" we use anyway.

 

[matt grime]

i didn't think that Riemann integration could even deal with f=g almost everywhere in cases like the Dirichlet function. i see no theorem like that on page 80 of Royden ("Real Analysis" - i know it's an old text) regarding Riemann integration:

I didn't say almost everywhere. The case you're talking about is stronger than almost everywhere - the problem is only at a single point. It is elementary to prove (and you should do this if you don't see that it immediately obvious) that if f and g are Riemann integral, and f=g at all but a set of isolated points, then f and g have the same Riemann integral.

 

[quasar987]

A couple of comments, hoping they will complement previous answers, and a couple
of new questions.

Firstly, thanks for the 2 comments.

About your questions:

1) This is a weird question. ;) I guess the answer is the obvious one: The Riemann is not a Lebesgue integral. If there were a measure that would make the Lebesgue integral into the Riemann-integral, then the Riemann-integral would be a Lebesgue integral. Kind of circular I guess, but the fact is that the Riemann integral proposes to calculate the area under the curve in a certain way, and Lebesgue's method is another way that does not involve partitions of the domain of integration.

2) I suppose simply studying the proof of the Fatou's lemma, monotone and dominated convergence thm would shed light on this.

3) What do you mean by metric completion?

 

[mathwonk]

why don't you read riemann some time rbj. on the next page after defining the integral, he proves a function is integrable in his (riemann's) sense if and only if it is continuous almost everywhere, [although he does not use those words].

 

[Count Iblis]

we (electrical engineers) say outrageous things like:

$$\sum_{n=-\infty}^{+\infty} \delta(t - n) = \sum_{k=-\infty}^{+\infty} e^{i 2 \pi k t}$$

http://en.wikipedia.org/wiki/Poisson_summation_formula"

 

[morphism]

1) Do we use some type of measure in doing Riemann integration?. There is

this whole layout, in preparing for the Lebesgue integral, about Lebesgue

measurability. But there is no mention of any measure in the setup for

the Riemann integral. Why is this so?.

There is some measure being used, but it's taken for granted. (The measure of [a,b] is b-a...!)

2) Any comment on how the Lebesgue integral repairs the convergence

problems of the Riemann integral?. Superficially, we are partitioning the

range, instead of the domain, and then , for a choice of element y_i* in

each [y_i-1,y_i) , we calculate m*(f^-1(y_i*)).

How does this help repair the convergence problems of the R-integral?.

Just wondering what may have guided Lebesgue to think that his setup

would improve the poor convergence properties?.

In a nutshell: Simple functions aren't as restrictive as step functions. This is the main idea. For more elaboration, consult any book that proves the convergence theorems, and study their proofs.

3) Is L^2[0,1] the metric completion of the space of Riemann square-

integrable functions?. Or is this the case for L^1[0,1] ?

Yes - well, almost. The mapping induced from the 2-norm on the set of Riemann square-integrable functions isn't a metric, it's a pseudometric. But if we do the usual modding out by the kernel business, it does become a metric space, and it's dense in L^2. So L^2 is its completion in this sense. Moreover, in the same way, the completion of the space of (equivalence classes) of Riemann p-integrable functions on [a,b] (i.e. those f s.t. $(\int_a^b |f|^p)^{1/p}< \infty$) is L^p[a,b].

By the way, your formatting is kind of weird (too much whitespace), so it's sort of hard to read in the reply box.

 

[quasar987]

Also, we can construct , for any set S of measure>0, a non-measurable

function on S : use the fact that if m(S)>0, then S has a non-measurable

subset.

Could you explain how this works please?

 

[D H]

Could you explain how this works please?

Its called a Vitali set. The proof of their existence is non-constructive and invokes the axiom of choice. This is one of those goofy results that gives credence to the mathematical purists who reject both non-constructive proofs and the axiom of choice.

 

[Count Iblis]

I think that the http://en.wikipedia.org/wiki/Banach-Tarski_paradox" led Banach and Tarski to argue against the Axiom of Choice.

 

[D H]

Physicists have an answer for the measure of the Vitali set.

If V is measurable, then so is V+q for all q, and they are all pairwise disjoint; by the sigma-additivity of Lebesgue measure, the measure of the union of the V+q's would be aleph_0 times the measure of V.

Mathematicians just don't know how to count. The argument hinges on

$$\sum_{n=1}^{\infty}1 = \aleph_0$$

Every physicist worth their salt knows

$$\sum_{n=1}^{\infty}1 = \sum_{n=1}^{\infty}1/n^0 = \zeta(0) = -\,\frac 1 2$$Edited to add
Please don't take the above too seriously. Sometimes renormalization stikes me as a tad too much hand waving, kinda like the axiom of choice. Pick whatever answer you want ...

 

[quasar987]

what's a non-constructive proof?

 

[D H]

what's a non-constructive proof?

Suppose you want to prove some hypothesis regarding the existence of some thing. A constructive proof explicitly constructs the thing in question using previously accepted axioms. A non-constructive proof does not show how to construct the thing. It simply proves that it must exist. For example, one might show that the lack of existence of the thing leads to a contradiction. The thing in question must exist.

Note: The above was a non-constructive description of a non-constructive proof. I would have had to have shown you a non-constructive proof to make the description constructive.

 

[WWGD]

Firstly, thanks for the 2 comments.

About your questions:


I apologize for the ASCII characters. Someone gave me a link for help
with Latex formatting , and I am already practicing it.

3) What do you mean by metric completion?

I think I was imprecise when I mentioned this. I was mixing two different concepts:

1)Turning a semi-metric into a metric:

A semi-metric d on a space X is a function that satisfies all the properties
of the metric except that we have x,y in X with x=/y and d(x,y)=0

We turn a semi-metric into a metric by identifying x with y if d(x,y)=0,
and define d'([a],)=d(a,b) , then (X,d') is a full metric space.

A standard example, I think, is the one Morphism gave in his reply to me:
consider the space of R-integrable functions with the norm of L^2, then
you can have different functions f,g have distance 0 , with distance
given by ||f-g||, with ||.|| the norm in L^2 . This is because, as was
mentioned, if f=g at all but finitely many points, then their respective
R-integrals are equal (consider the function h=f-g and define a Riemann
sum so that each partition element containing a non-zero value is
indefinitely small, so that the sum goes to zero.). Then, as in the above
paragraph, we identify these two functions, and use the metric d' above.





2)The more common, I think, meaning of metric completion:

If (X,d) is a metric space and it is not complete, i.e, there are Cauchy
seqs. in X that do not converge in X , we can embed, in the topological
sense, (X,d) in a space (X',d') such that X is dense in X' and (X,d') is
complete. Also, the embedding e(x) is an isometry, meaning that
d(x,y)=d'(e(x),e(y)).

Now, to illustrate/ give another example of a question you asked,
this proof is constructive: we actually create a space with these
properties. We identify all sequences in (X,d) that "should" converge,
(meaning that d(x_n,y_n)->0 as n tends to oo. ) as a single class,
then we define d'([a],)=d(a,b), as with the semi-metric. It is
a good exercise to show that the space defined this way is complete,
and that the original space is dense in its completion. A lot of nasty
book-keeping, but still a good exercise.


The example that I would imagine would be closer to most
people's experience is that of the Real numbers with the usual
metric as a completion of the Rationals Q. You know Q is not
complete, e.g, take sqr2 , and a sequence of its decimal expansion.
Then we embed Q isometrically in IR , by the process above.
Notice too, that Q is dense in IR, which agrees with the claim
that IR with its standard metric is the completion of Q with this
same metric.


Hope I did not mess up here.