# How do I complete this proof that the absolute value of an integral function is an integrable function?

I'm trying to complete the proof in this answer that if $f: [a, b] \to \mathbb{R}$ is a Riemann integrable function, then $|f|$ is an integrable function also.

I understand the proof that $$\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x) + \epsilon$$

but I need to translate this to $$\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$$ because otherwise the summations break down (see below). Can I just say that because $\epsilon > 0$ can be made as small as possible, then $$\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$$ in the limit? Once I have that, it's easy, because then I just sum the inequality across every interval to get $$\sum_{i=0}^{n-1} (\sup |f| - \inf |f|)(x_{i+1} - x_i) \leq \sum_{i=0}^{n-1} (\sup f - \inf f + \epsilon)(x_{i+1} - x_i)$$ which gives me the correct upper and lower sums. If I can't just remove the $\epsilon$, then I get \begin{align} U(|f|, P) - L(|f|, P) &= \sum_{i=0}^{n-1} (\sup |f| - \inf |f|)(x_{i+1} - x_i) \\ &< \sum_{i=0}^{n-1} (\sup f - \inf f + \epsilon)(x_{i+1} - x_i) \\ \end{align}

where the supremum and infimum are taken over the interval $[x_i, x_{i+1}]$. I somehow need to prove that $U(|f|, P) - L(|f|, P) < \epsilon$, and I can't see how to manipulate that summation to do so.

EDIT: To clarify, I know that once I have $$U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$$ the proof is complete because for any $\epsilon > 0$, I can pick a partition $P$ such that $U(f,P) - L(f,P) < \epsilon$. My question is about using the supremum and infimum properties above to show that $$U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$$ That's where I'm stuck.

• If $a \le \epsilon$ for all $\epsilon>0$, then $a \le 0$. Commented May 27, 2015 at 15:23
• @copper.hat Ok, that's what I was missing then, since that's what lets me remove the $\epsilon > 0$. That $\epsilon$ comes from the proof in the other answer that led me to ask this question.
– M T
Commented May 27, 2015 at 15:25
• The proof will work either way, see my comment below. Commented May 27, 2015 at 15:26

You get $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$, and since $f$ is integrable, you can choose a partition so the right hand side is as small as you want. (Also, we have $0 \le U(|f|,P) -L(|f|,P)$.)

You have $\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$ for all intervals in a partition $P$. Hence $\sup_{x\in I}|f(x)|l(I)-\inf_{x\in I}|f(x)|l(I)\le \sup_{x\in I}f(x)l(I)-\inf_{x\in I}f(x)l(I)$.

Since (abusing notation a little, as in $I \in P$), and similarly for $L(g,P)$, we have $U(g,P) = \sum_{I \in P} \sup_{x\in I}g(x)l(I)$.

Hence $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$.

• Right, but my question is about how to show that $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$. How does that follow from the supremum and infimum properties proved in my question and the one I linked to? That's what I'm stuck on. Once I have that, the rest is trivial (as shown in your answer).
– M T
Commented May 27, 2015 at 15:13
• Just sum the inequality (multiplied by the length of the interval) over all intervals in the partition. Commented May 27, 2015 at 15:16
• That's what I tried to do, but since the inequality is $\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x) + \epsilon$, I still get that $\epsilon$ inside the summation, which is what's causing trouble. How do I get rid of that $\epsilon$ so I just have $\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$?
– M T
Commented May 27, 2015 at 15:18
• I am not sure why you are introducing an $\epsilon$ at that stage. Even if you do, when you sum it will be multiplied by $(b-a)$, and so can be taken to be arbitrarily small. Commented May 27, 2015 at 15:23