How to prove that ##f## is integrable given that ##g## is integrable?

[Adesh]

TL;DR Summary

Let $f:[a,b] \mapsto \mathbb R$ be a function a partition $P= \{x_0, x_2 \cdots , x_n \}$ of $[a,b]$ is there such that, $g:[a,b] \mapsto \mathbb R$ , $g(x) = f(x) ~~~x\in(x_{i-1}, x_i)$. Given that $g$ is integrable prove that $f$ is integrable.

We have a function $f: [a,b] \mapsto \mathbb R$ (correct me if I'm wrong but the range $\mathbb R$ implies that $f$ is bounded). We have a partition $P= \{x_0, x_1 , x_2 \cdots x_n \}$ such that for any open interval $(x_{i-1}, x_i)$ we have
$$
f(x) =g(x)
$$
($g:[a,b] \mapsto \mathbb R$ is an integrable function in any closed interval $[x_{i-1}, x_i]$ I want to know how can we go on for proving that $f$ is also integrable.

If we can prove that if $f$ is integrable in any closed interval $[x_{i-1}, x_i]$ then we can go on for proving it is integrable on $[a,b]$. I can see that $f$ and $g$ are just unequal at the end points, but how to use their inequality in points between to prove the integrability from one to the other.

Thank you.

 

[member 587159]

Please state the definitions (characterisations) of Riemann-integrability you are allowed to use. I know three of them and the answer to your question will depend on which one you are allowed to use.

Moreover, $g$ is integrable and thus bounded. Since $f$ agrees with $g$ on all but finitely many points, $g$ is bounded as well.

Also, don't write $[a,b] \mapsto \Bbb{R}$ but rather $[a,b] \to \mathbb{R}$. In latex, this is simply writing '\to'.

Basically the question asks you: If you change an integrable function in finitely many points, does the function remain integrable?

Hint: Try to prove it for one point first. This is easier to visualise.

 

[Adesh]

Please state the definitions (characterisations) of Riemann-integrability you are allowed to use. I know three of them and the answer to your question will depend on which one you are allowed to use.

A function $f: [a,b] \to \mathbb R$ which is bounded on [a,b] is integrable on [a,b] if and only if there exists a sequence of the partition $\{P_n\}$ such that
$$\lim_{n\to \infty} [ U(f,P_n) - L(f,P_n)]=0
$$

${\Large OR}$

A function $f:[a,b]\to \mathbb R$ which is bounded on [a,b] is integrable if and only if we can find a partition $P$ for every $\epsilon \gt 0$ such that
$$
U(f,P) - L(f,P) \lt \epsilon
$$

 

[member 587159]

A function $f: [a,b] \to \mathbb R$ which is bounded on [a,b] is integrable on [a,b] if and only if there exists a sequence of the partition $\{P_n\}$ such that
$$\lim_{n\to \infty} [ U(f,P_n) - L(f,P_n)]=0
$$

${\Large OR}$

A function $f:[a,b]\to \mathbb R$ which is bounded on [a,b] is integrable if and only if we can find a partition $P$ for every $\epsilon \gt 0$ such that
$$
U(f,P) - L(f,P) \lt \epsilon
$$

Forum rules require that you make an attempt. What have you tried? Your second definition seems most convenient to me.

 

[Adesh]

Also, don't write [a,b]↦R but rather [a,b]→R. In latex, this is simply writing '\to'.

I would very much like to know why "\to" is better here (I'm not arguing with you, I just want to learn things from you coz you have experience and knowledge), I though it's a mapping so I use "\mapsto".

Basically the question asks you: If you change an integrable function in finitely many points, does the function remain integrable?

Hint: Try to prove it for one point first. This is easier to visualise.

It's awesome how you changed the question into something else, Real Analysts are very admirable.

P.S. : I got to know that Rudin wrote baby Rudin just after his PhD, so that could be a reason why it is not so good for self-studying, he didn't have enough experience in book writing for students.

 

[member 587159]

I would very much like to know why "\to" is better here (I'm not arguing with you, I just want to learn things from you coz you have experience and knowledge), I though it's a mapping so I use "\mapsto".

We write $f: A \to B: x \mapsto f(x)$. The map goes from $A$ to $B$ and maps the element $x$ to $f(x)$.

 

[Adesh]

Forum rules require that you make an attempt. What have you tried? Your second definition seems most convenient to me.

${\small \mathbf{GIVEN}}$: $f$ is an integrable function on $[a,b]$.

${\small \mathbf{TO ~PROVE}}$: If at some point the value of $f$ is changed then also the function will be integrable.

${\small \mathbf{PROOF}}$ : Let's say we're a given a very small $\epsilon \gt 0$ and for this there exist a partition $P=\{x_0, x_1 , x_2 \cdots x_n\}$ such that
$$
U(f,P) - L(f,P) \lt \epsilon
$$
Now, let's say at some point in some interval $[x_{i-1}, x_i]$ the value of $f$ is changed from $f(x)$, ($x \in (x_{i-1}, x_i)$ ) to a number $\alpha$, but taking care that $m_i \leq \alpha \leq M_i$. Now, since $M_i$ and $m_i$ are unchanged by the altering the value of the function at some $x$, therefore the upper and lower Darboux sums for the partition $P$ is unchanged and hence the inequality
$$
U(f,P) - L(f,P) \lt \epsilon
$$
remains true.

Therefore, altering the value of the function at some single point retains the integrability of the function $f$.

 

[member 587159]

Now, let's say at some point in some interval $[x_{i-1}, x_i]$ the value of $f$ is changed from $f(x)$, ($x \in (x_{i-1}, x_i)$ ) to a number $\alpha$, but taking care that $m_i \leq \alpha \leq M_i$. Now, since $M_i$ and $m_i$ are unchanged by the altering the value of the function at some $x$, therefore the upper and lower Darboux sums for the partition $P$ is unchanged and hence the inequality

I don't follow why you can take $m_i \leq \alpha \leq M_i$. For example, suppose the image of $f$ is contained in the interval $[-100,100]$. Let $\alpha = 1000$.

 

[Adesh]

I don't follow why you can take $m_i \leq \alpha \leq M_i$. For example, suppose the image of $f$ is contained in the interval $[-100,100]$. Let $\alpha = 1000$.

If $\alpha$ exceeds the $M_i$ or gets lower than $m_i$ then our Darboux sums will get affected.

OH YEAH! I proved it for the case when it is $m_i \lt \alpha \lt M_i$. Now, let's imagine if $\alpha$ is replaced for $f(x)$. Then, breakdown the interval $[x_{i-1}, x_i]$ into $[x_{i-1}, x]$ and $[x, x_{i}]$ (x is the point where we changed the value of $f$.
$$
P= \{x_0, x_1 \cdots x_n\} $$
$$P' = \{x_0, x_1 \cdots x_{i-1}, x , x_i \cdots x_n\}
$$

Since, $P'$ contains more points, therefore $L(f,P') \gt L(f,P)$ and $U(f,P') \lt U(f,P)$, hence
$$
U(f,P') - L(f,P') \lt U(f,P) - L(f,P) \lt\epsilon$$
$$
U(f,P') - L(f,P') \lt\epsilon$$

$\therefore f$ is integrable.

 

[member 587159]

If $\alpha$ exceeds the $M_i$ or gets lower than $m_i$ then our Darboux sums will get affected.

OH YEAH! I proved it for the case when it is $m_i \lt \alpha \lt M_i$. Now, let's imagine if $\alpha$ is replaced for $f(x)$. Then, breakdown the interval $[x_{i-1}, x_i]$ into $[x_{i-1}, x]$ and $[x, x_{i}]$ (x is the point where we changed the value of $f$.
$$
P= \{x_0, x_1 \cdots x_n\} $$
$$P' = \{x_0, x_1 \cdots x_{i-1}, x , x_i \cdots x_n\}
$$

Since, $P'$ contains more points, therefore $L(f,P') \gt L(f,P)$ and $U(f,P') \lt U(f,P)$, hence
$$
U(f,P') - L(f,P') \lt U(f,P) - L(f,P) \epsilon$$
$$
U(f,P') - L(f,P') \epsilon$$

$\therefore f$ is integrable.

These equalities can be non-strict. And to be really precise you have to distinguish between two cases: $x= x_i$ for some $i$ and $x_i \neq x$ for all $i$. I'm not convinced by your proof. In your latest attempt you did not take into account that $f$ was changed in $x$.

I think it is best to denote $g$ as the function that is $f$ everywhere except in $x$ and such that $g(x) = \alpha$.

 

[Adesh]

I'm not convinced by your proof. In your latest attempt you did not take into account that f was changed in x.

How to mend it? Where I did wrong, I too feel that if changed something then it must be visible but then I got this thought "yeah it should be visible but not in integrability".

I need one more help, " \\\\ " seems to be not working for line breaks. For example see this image

.

I have out " \\\\" after writing out the partition P but in preview line didn't get break, it's happening to me from three weeks.

 

[member 587159]

I will answer your questions later today with a full proof. I have a deadline to catch for now.

 

[mathman]

Using Lebesgue integration makes the question trivial.

 

[member 587159]

Using Lebesgue integration makes the question trivial.

Yes, using an advanced non-trivial integration theory makes the question trivial! Moreover, it is non-trivial to prove that Riemann-integral and Lebesgue-integral coincide.

 

[mathman]

Yes, using an advanced non-trivial integration theory makes the question trivial! Moreover, it is non-trivial to prove that Riemann-integral and Lebesgue-integral coincide.

Any Riemann integrable function is Lebesgue integrable, so g is Lebesgue integrable implying f is Lebesgue integrable. To prove f is Riemann integrable, an additional requirement is needed, that f is not infinite at the break points.

 

[member 587159]

Any Riemann integrable function is Lebesgue integrable, so g is Lebesgue integrable implying f is Lebesgue integrable. To prove f is Riemann integrable, an additional requirement is needed, that f is not infinite at the break points.

Or if you use measure theory you can just use that a function is Riemann-integrable if it is bounded and the points of discontinuity have measure 0. Then you see that changing the value of the function in finitely many points won't affect the integrability.

 

[mathman]

Or if you use measure theory you can just use that a function is Riemann-integrable if it is bounded and the point of discontinuities have measure 0. Then you see that changing the value of the function in finitely many points won't affect the integrability.

Using measure theory means using Lebesgue integral.

 

[member 587159]

Using measure theory means using Lebesgue integral.

Which is not necessary here, so why use it? I'm pretty sure the OP does not know about it.

 

[mathman]

Which is not necessary here, so why use it? I'm pretty sure the OP does not know about it.

As far as I know the development of measure theory is the first step in introducing Lebesgue integration, so as I see it, the OP would not have been exposed to one without the other. I think we're splitting hairs at this point.

 

[member 587159]

As promised, here is a complete proof using your definition. We show that if $f: [a,b] \to \Bbb{R}$ is Riemann-integrable and $c \in \mathbb{R}, y \in [a,b]$ are fixed real numbers, then the function
$$g: [a,b] \to \Bbb{R}: x \mapsto \begin{cases}f(x) \quad x \neq c \\ y \quad x = c\end{cases}$$
is Riemann-integrable as well.

I will be assuming that $a < c <b$. The proof is easily modified if $c$ is on the boundary of $[a,b]$.

Proof: Let $M$ be so large that $|f|+|g| < M$.

Let $\epsilon > 0$. Choose a partition $P$ with $U(f,P)-L(f,P) < \epsilon/3$. Then, consider the partition $Q:= P \cup \{c-\frac{\epsilon}{6M}, c+ \frac{\epsilon}{6M}\}$ (if necessary, enlargen $M$ such that this makes sense). Then $|U(f,Q)-U(g,Q)| \leq \frac{\epsilon}{3}$ and $|L(f,Q)-L(g,Q)| \leq \epsilon/3$ (Why?). Consequently,

$U(g,Q)-L(g,Q)$
$\leq |U(g,Q)-U(f,Q)| + |U(f,Q)-L(f,Q)| + |L(f,Q)- L(g,Q)|$
$\leq \epsilon/3 + U(f,Q)-L(f,Q)+ \epsilon/3$
$\leq \epsilon/3 + U(f,P)-L(f,P) + \epsilon/3$
$< \epsilon/3 + \epsilon/3 + \epsilon/3$
$=\epsilon$
which ends our proof. $\quad \square$

It is late here so I may have made some slight error. Please ask if something is unclear. Thanks for the question btw, made me review the Riemann-integral haha.

 

[Adesh]

Can you teach me how to prove the integrability of $f$ if it changed at just one point? I know you have given a very nice proof just above, but I want to learn how to do it. (Sorry if I demand too much, you have answered by original question and you may not be inclined to teach something, I just requested you).

 

[member 587159]

Can you teach me how to prove the integrability of $f$ if it changed at just one point? I know you have given a very nice proof just above, but I want to learn how to do it. (Sorry if I demand too much, you have answered by original question and you may not be inclined to teach something, I just requested you).

Well, all you are given are partitions $P$ such that $U(f,P)-L(f,P)$ becomes arbitrary small.

Somehow, you want to deduce that you can find a partition $Q$ such that $U(g,Q)-L(g,Q)$ becomes arbitrary small.

Clearly you will have to use $P$ to construct $Q$.

If you write out the expression for $U(f,P)-L(f,P)$ you see that it looks like $U(g,P)-L(g,P)$ but only one or two terms in this summation are changed, so these are the ones we have to worry about.

Basically we want to make these extra terms small, so we throw in some extra points such that the horizontal distance between the points in the partition closed to $c$ dominates the vertical difference between $f$ and $g$. Then the extra terms become small and the entire thing becomes small.

This was my idea when I wrote the proof. Maybe that helps a bit?

 

[Adesh]

@Math_QED Here is my try:

We have a given partition $P=\{x_0, x_1, x_2, \cdots x_n\}$ such that for a given small $\epsilon \gt 0$ we have
$$
U(f,P) - L(f,P) \lt \epsilon $$
$$\sum_{i=1}^{k-1} M_i (x_i -x_{i-1}) + M_k (x_k -x_{k-1}) + \sum_{i=k+1}^{n} M_i(x_i -x_{i-1}) - L(f,P) \lt \epsilon
$$
Let’s say $c \in [x_{k-1}, x_k]$ and if $g(c) =\alpha \gt M_k =sup\{f(x) : x\in[x_{k-1}, x_k]\}$, then we have
$$
U(g,P) - L(g,P) = \sum_{i=1}^{k-1}M_i (x_i - x_{i-1}) + \alpha (x_k -x_{k-1}) + \sum_{i=k+1}^{n} M_i(x_i -x_{i-1}) - L(f,P)
$$

Now, what I have to do after this to prove the integrability of $g$.

 

[member 587159]

@Math_QED Here is my try:

We have a given partition $P=\{x_0, x_1, x_2, \cdots x_n\}$ such that for a given small $\epsilon \gt 0$ we have
$$
U(f,P) - L(f,P) \lt \epsilon $$
$$\sum_{i=1}^{k-1} M_i (x_i -x_{i-1}) + M_k (x_k -x_{k-1}) + \sum_{i=k+1}^{n} M_i(x_i -x_{i-1}) - L(f,P) \lt \epsilon
$$
Let’s say $c \in [x_{k-1}, x_k]$ and if $g(c) =\alpha \gt M_k =sup\{f(x) : x\in[x_{k-1}, x_k]\}$, then we have
$$
U(g,P) - L(g,P) = \sum_{i=1}^{k-1}M_i (x_i - x_{i-1}) + \alpha (x_k -x_{k-1}) + \sum_{i=k+1}^{n} M_i(x_i -x_{i-1}) - L(f,P)
$$

Now, what I have to do after this to prove the integrability of $g$.

But I gave a complete proof above and gave the intuition how I came to this proof?

But in your attempt you should take into account that it is possible that $c=x_i$ and then you have two intervals to worry about.

 

[Adesh]

But I gave a complete proof above and gave the intuition how I came to this proof?

But in your attempt you should take into account that it is possible that $c=x_i$ and then you have two intervals to worry about.

I couldn’t understand when you took $Q$ as the union of $P$ and that set. How it came and what’s the idea behind it?

 

[member 587159]

I couldn’t understand when you took $Q$ as the union of $P$ and that set. How it came and what’s the idea behind it?

Did you fill in all details I provided in my proof? I think the idea will come through once you did that.

The $Q$ is introduced to make the difference $|y-f(c)|$ small relative to the horizontal distance.

 

[Adesh]

Did you fill in all details I provided in my proof?

Yes I did.

 

[member 587159]

Yes I did.

Can you explain me the step with "(why)" then?

 

[Adesh]

Can you explain me the step with "(why)" then?

Beacuse I couldn’t understand it.

 

[member 587159]

Beacuse I couldn’t understand it.

Then you didn't fill in all the details!

 

[pasmith]

There is a more elegant proof.

Notation-wise, you will confuse the reader and yourself if you have $x_i$ being both the points at which $f(x) \neq g(x)$ and the points of an arbitrary partition. So I will call the former $\{\xi_k : k = 1, \dots, n\}$.

We can write $$f(x) = g(x) + \sum_{k=1}^n (f(\xi_k) - g(\xi_k)) h_k(x)$$ where $$h_k : [a,b] \to \mathbb{R} : x \mapsto \begin{cases} 0 & x \neq \xi_k \\ 1 & x = \xi_k.\end{cases}$$ It suffices to show that each $h_k$ is (Darboux) integrable, since a linear combination of finitely many (Darboux) integrable functions is (Darboux) integrable. (The one with the upper and lower sums being arbitrarily close to each other is Darboux integrability; it is however completely equivalent to Riemann integrability.)

Let $\epsilon > 0$ and $D = \{x_0 = a, \dots, x_m = b\}$ a partition of $[a.b]$ with $\|D\| = \max\{\delta_i = x_i - x_{i-1}, i = 1, \dots, m\} < \frac12 \epsilon$. Then there exists a minimal $1 \leq i_1\leq m$ such that $x_{i_1-1} \leq \xi_k \leq x_{i_1}.$ We then have that $M_i - m_i = 0$ for every $i$ except for $i_1$ and, if $x_{i_1} = \xi_k$ and $i_1 < m$, also $i_1+1$. For these intervals $M_i - m_i = 1$. Thus $(M_i - m_i)\delta_i = 0$ except for at most two intervals for which $(M_i - m_i)\delta_i = \delta_i \leq \|D\|$. Thus $$U(h_k,D) - L(h_k,D) \leq 2\|D\| < \epsilon.$$