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I know that there exists a theorem telling us that if a real valued function $f$ defined over an interval $[a,b]$ is Riemann integrable, then |$f$| is too (moreover I believe the converse to be false). I've tried to look at the proofs, but they seem a bit beyond my ability at the moment. My first thought was to suppose there existed a Riemann integrable function, let us say $g$, such that |$g$| was not Riemann integrable, and show that this leads to some sort of contradiction.

I thought about this for a while and have not been able to think of such a function/proof. Would anybody know of such an example?

C Squared
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jackson p
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    what is your definition of Riem. integrable? – Keshav Apr 07 '21 at 04:38
  • @Keshav there exists a unique I (the integral of f from a to b) such that I is bounded between the upper and lower Riemann sums of f with respect to any partition P of [a,b]. – jackson p Apr 07 '21 at 04:46
  • Take a partition so that the upper - lower bound $< \epsilon$. show that for the corresponding |f|, upper-lower bound is even smaller. In particular, $f \leq |f|$, so the inf won't be smaller. If $f$ is always neg, then $sup-inf=0-0=0$. Otherwise, sup doesn't change. – Eric Apr 07 '21 at 04:54
  • for your upper and lower sums, are you using a tagged partition, typically seen with $x^*$'s, or are you using supremums and infimums on subintervals of your partition? – C Squared Apr 07 '21 at 05:01
  • I was using the supremums and infimums on those subintervals. – Eric Apr 07 '21 at 05:02
  • Also, I was thinking of max(f,0) since you could then do |f|=max(f,0)+max(-f,0). – Eric Apr 07 '21 at 05:32
  • @Eric oh i didn't even notice, but my question was meant for op – C Squared Apr 07 '21 at 07:15
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    A direct proof is simpler (see answer by C Squared). And you can easily change this proof into one by contradiction. – Paramanand Singh Apr 07 '21 at 09:46

1 Answers1

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Let $a,b\in\mathbb{R}$ with $a<b$. Suppose $f:[a,b]\to\mathbb{R}$ is bounded and Riemann integrable. It suffices to show that for any interval $I\subseteq[a,b]$, we have that $$\sup_{x\in I}|f|-\inf_{x\in I}|f|\leq\sup_{x\in I}f-\inf_{x\in I}f $$ where $\sup_{x\in I}g(x)$ is defined as $\sup\{g(x)\in\mathbb{R}:x\in I\}$ if $g$ is a function that is bounded on $I$.

Note that for $x\in I$, we have $f(x)\leq |f(x)|$.

Break this into three cases:

Case 1:

$\inf_{x\in I}f(x)\geq 0$. In this case, we have that $f(x)\geq 0$ for all $x\in I$. This means $|f(x)|=f(x)$ so $$ \sup_{x\in I}|f|=\sup_{x\in I}f\text{ and }\inf_{x\in I}|f|=\inf_{x\in I}f$$ which implies that $$\sup_{x\in I}|f|-\inf_{x\in I}|f|=\sup_{x\in I}f-\inf_{x\in I}f $$

Case 2:

$\sup_{x\in I}f(x)\leq 0$. In this case, we have that $f(x)\leq 0$ for all $x\in I$, thus $|f(x)|=-f(x)$. Then we get that $$\sup_{x\in I}|f|=-\inf_{x\in I}f \text{ and }\inf_{x\in I}|f|=-\sup_{x\in I}f$$ so in particular, we have $$\sup_{x\in I}|f|-\inf_{x\in I}|f|=-\inf_{x\in I}f-\big(-\sup_{x\in I}f\big)=\sup_{x\in I}f-\inf_{x\in I}f$$ If you want Case 2 to follow relatively quickly, it would suffice to show the following:

Let $A\subseteq\mathbb{R}$ be non-empty and bounded. Define $-A=\{-a\in\mathbb{R}:a\in A\}$. Then $\sup(-A)=-\inf(A)$ and $\inf(-A)=-\sup(A)$.

Case 3:

$\inf_{x\in I}f\leq 0\leq\sup_{x\in I}f$. I will leave this case to you, as it requires some sub-cases based upon the magnitudes of $\inf_{x\in I}f$ and $\sup_{x\in I}f$.

With this it should be easier to finish the proof.

C Squared
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