If $f$ is integrable, then $|f|$ is integrable

Question

In order to prove that if $f(x)$ is integrable, then $|f(x)|$ is, I'm using the definition that $f$ is integrable $\iff \forall \epsilon>0$ there exists a partition $P$ such that

$$S(f,P)-s(f,P)<\epsilon$$

but

$$S(f, P) = \sum M_i(t_{i}-t_{i-1})$$ $$s(f, P) = \sum m_i(t_{i}-t_{i-1})$$

What I need to prove is that

$$S(f,P)-s(f,P)<\epsilon \implies S(|f|,P)-s(|f|,P)<\epsilon$$

I cannot find an obvious relation in order to prove it. Could somebody help me?

Whqat happens to $M_i$, $m_i$ when replacing $f$ with $|f|$? — Hagen von Eitzen, Oct 02 '16 at 19:46
In fact if $g$ is any function continuous on a closed interval containing the range of $f,$ then $g \circ f$ is Riemann integrable. $g(x)=|x|$ is a special case of this. — zhw., Oct 02 '16 at 19:51

score 2 · Accepted Answer · answered Oct 02 '16 at 19:48

2

hint: $||M_i|-|m_i|| \le |M_i-m_i|$

answered Oct 02 '16 at 19:48

DeepSea

77,651

And I need to use that $M_i-m_i\ge 0$ right? – Guerlando OCs Oct 02 '16 at 19:56
Yes. $M_i \ge m_i$ for each subinterval. – DeepSea Oct 02 '16 at 20:26

If $f$ is integrable, then $|f|$ is integrable

1 Answers1