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I want to prove this statment:

Let $f:[a,b] \to \mathbb{R}$ be a bounded function.

Prove that if $f$ is integrable in $[a,b]$ then $|f|$ is also integrable in $[a,b]$ - HINT: first prove that if $M_f=\sup(f(x):x \in [a,b])$ and $m_f=\inf(f(x):x \in [a,b])$, then $M_{|f|}-m_{|f|} \le M_f-m_f$

Unfortunately I have been trying to prove the hint for over than 3 hours now with no luck.

I tried proving that $M_f-m_f=\sup(|f(x)-f(y)|:x,y \in [a,b])$ but I don't know how...

Any help will be amazing!!

Thanks!

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    $||x|-|y|| \le |x-y|$. – copper.hat Apr 23 '21 at 05:18
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    https://math.stackexchange.com/a/4092394/803927 see if this gets you started – C Squared Apr 23 '21 at 05:37
  • @CSquared Yes it is, but I am trying to prove it more shortly (with no cases), thanks anyway! –  Apr 23 '21 at 05:57
  • shortest way i know of, without any cases, would be to show that if $f:[a,b]\to[m,M]$ is bounded and integrable, and $g:[m,M]\to\mathbb{R}$ is continuous, then $g\circ f:[a,b]\to\mathbb{R}$ is bounded and integrable. This is a proof from baby Rudin and makes use of the uniform continuity of $g$ on compact sets. – C Squared Apr 23 '21 at 06:17
  • One way is to note that if $f$ is integrable it is bounded and the set if discontinuities has measure zero. Since $x \mapsto |x|$ is continuous, we see that the set of discontinuities of $|f|$ also has measure zero and since it is bounded it is Riemann integrable. – copper.hat Apr 23 '21 at 06:25
  • Unfortunately I still having trouble proving it... If someone can help me more, I will very appreciate that! –  Apr 23 '21 at 09:37
  • I had discussed this in comments to my answer : https://math.stackexchange.com/a/2359466/72031 Are you still facing an issue? – Paramanand Singh Apr 23 '21 at 11:17
  • Please don't think too hard on this and instead try to grasp the definition of sup and inf and you will find the identity $M_f-m_f=\sup, {|f(x) - f(y) |\mid x, y\in[a, b] } $ relatively straightforward. – Paramanand Singh Apr 23 '21 at 11:20
  • @ParamanandSingh Oh thank you I didn't see your reply, I will try it again now :) –  Apr 23 '21 at 11:24
  • @ParamanandSingh I just want to verify, Why is $M−m$ is an upper bound for all numbers of the form $|f(x)−f(y)|$ and not just $f(x)−f(y)$? –  Apr 23 '21 at 11:26
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    Well $M-m$ is also the supremum of numbers of the form $f(x) - f(y) $. Why?? Because by interchanging $x, y$ we can change sign of $f(x) - f(y) $ and thus the set of values of $f(x) - f(y) $ contains non-negative numbers as well. – Paramanand Singh Apr 23 '21 at 12:41
  • @ParamanandSingh Got it, thank you very much!! Now I have to use the theorem that if $f$ is integrable, than there is a partition $P$ such that $\lim_{n \to \infty}[U(f,P_n)-L(f,P_n)]=0$, but I don't realize the connection... Can you suggest a way of thinking please? Thanks a lot! –  Apr 23 '21 at 12:50
  • Use your hint in question to show that $0\leq U(|f|, P_n) - L(|f|, P_n) \leq U(f, P_n) - L(f, P_n) $ and use squeeze theorem to conclude that the middle term of above inequality also tends to $0$ and hence $|f|$ is Riemann integrable. – Paramanand Singh Apr 23 '21 at 13:07
  • @ParamanandSingh Awesome! Thank you so much for your help!! I appreciate that! :) –  Apr 23 '21 at 13:12

2 Answers2

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It can be proven without the HINT. Define

$$ f^+(x)= \begin{cases} f(x),& f(x)\geq 0\\ 0,& f(x)<0 \end{cases},\qquad f^-(x)= \begin{cases} f(x),& f(x)\leq 0\\ 0,& f(x)>0. \end{cases} $$

Then $f=f^++f^-$. If $f$ is integrable $\Rightarrow$ $f^+$ and $f^-$ are integrable $\Rightarrow$ $|f|=f^+-f^-$ is integrable.

Pavel R.
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Consider only partitions of the original interval into $n$ equal lengths. Fix $\epsilon \gt 0$ and let $M$ such that the lower and upper sums for $f$ differ by less than $\epsilon$ whenever $n\gt M.$ If zero is not between the lower and upper bounds for $f$ on a subinterval, then the absolute values of these bounds will differ by the same amount. Otherwise, zero can be the lower bound for $|f|$ and the "gap" for $|f|$ will be smaller than (or equal to) that for $f.$ So lower and upper sums for $|f|$ will also differ by less than $\epsilon,$ completing the proof.

David
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