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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.

Any help will be amazing!!

Thanks!

  • Thank you! How do I connect the set $-A$ to $f$ in this case? –  Apr 22 '21 at 17:21
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    It might be helpful to show $M_g - m_g = \sup{g(x)-g(y) : x,y\in[a,b]}$ then your inequality follows from $|f(x)|-|f(y)| \leq |f(x)-f(y)|$ which is just the triangle inequality. – Brian Moehring Apr 22 '21 at 17:23
  • @BrianMoehring Thank you, I have tried doing it. Can you give me a clue on how to prove this statement? Thanks! –  Apr 22 '21 at 17:24
  • @OliverDiaz Why should I consider $-|f|$? –  Apr 22 '21 at 17:31
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    The proof would depend on what you already know. For instance, it may follow directly as $$\sup A - \inf A = \sup A + \sup(-A) = \sup(A + (-A))$$ when $A = {g(x) : x\in[a,b]}$ but this is only helpful if you've learned about suprema and sums of sets. – Brian Moehring Apr 22 '21 at 17:32
  • @BrianMoehring And $\sup(A+(-A))=\sup(0)$? What is that mean? –  Apr 22 '21 at 17:35
  • And that's why it depends on what you know. We have $A + (-A) = {0}$ if and only if $A$ is a singleton. My method may not be appropriate for you. – Brian Moehring Apr 22 '21 at 17:37
  • @BrianMoehring Perhaps not. Thanks! –  Apr 22 '21 at 17:39
  • this link: https://math.stackexchange.com/questions/4092347/proving-f-riemann-integrable-implies-f-riemann-integrable-by-contradiction/4092394#4092394 This link:https://math.stackexchange.com/questions/4113202/prove-if-f-is-integrable-than-f-is-integrable#comment8504469_4113202 and this link: https://math.stackexchange.com/questions/451146/if-f-is-integrable-then-f-is-also-integrable?noredirect=1&lq=1 may be helpful – C Squared Apr 28 '21 at 06:03

2 Answers2

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The triangle inequality gives $$|f(x)|\le|f(x)-f(x')|+|f(x')|$$ with $x,x' \in [a,b]$.

Case 1: $\;f(x) \ge f(x')$

Then $\;|f(x)|\le f(x)-f(x')+|f(x')|\le M_f-f(x')+|f(x')|\;$ for every $x\;\; $ ($x'$ fixed)

So $\;M_{|f|}\le M_f-f(x')+|f(x')|$

and then $\;|f(x')|\ge M_{|f|}-M_f+f(x')\ge M_{|f|}-M_f+m_f\;$ for every $x'$

It follows that $\;m_{|f|}\ge M_{|f|}-M_f+m_f$

Case 2: $\;f(x) < f(x')$

Then $\;|f(x)|\le f(x')-f(x)+|f(x')|\le f(x')-m_f+|f(x')|\;$ for every $x\;\;$ ($x'$ fixed)

So $\;M_{|f|}\le f(x')-m_f+|f(x')|$

and then $\;|f(x')|\ge M_{|f|}-f(x')+m_f\ge M_{|f|}-M_f+m_f\;$ for every $x'$

It follows that $\;m_{|f|}\ge M_{|f|}-M_f+m_f$

Tony Piccolo
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$(I)$ If $M_f \geq 0$ and $m_f \geq 0$ then $M_{|f|}=M_f$ and $m_{|f|}=m_f$. We have a equality.

$(II)$ If $M_f \geq 0$ but $m_f < 0$ then $M_{|f|}=M_f$ and $m_{|f|}\geq m_f$. In this case $M_{|f|}-m_{|f|} \leq M_f-m_f$.

$(III)$ If $M_f,~m_f<0$ note that $|m_f|=M_{|f|}$ and $|M_f|=m_{|f|}$. We get $M_{|f|}-m_{|f|}=|m_f|-|M_f|=-m_f-(-M_f)=M_f-m_f$.

absolute0
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