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Prove that if the real valued function $f$ on the interval $[a,b]$ is Riemann integrable on $[a,b]$ then so is $|f|$, and $|\int^b_a f(x)dx|\le \int^b_a |f(x)|dx$.

The definition I have for Riemann integral is: Let $a,b\in \mathbb{R}$ $a<b$ and let $f$ be a real valued function on $[a,b]$. We say that $f$ is Riemann integrable on $[a,b]$ if there exists a number $A\in \mathbb{R}$ such that for any $\epsilon>0$ there exists $\delta>0$ such that $|S-A|<\epsilon$ whenever $S$ is Riemann sum for $f$ corresponding to any partition of $[a,b]$ of width less that $\delta$.

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Hint. To show that $|f|$ is integrable Riemann, it suffices to show that for every partition $P$ of $[a,b]$ $$ U(|f|,P)-L(|f|,P)\le U(f,P)-L(f,P), $$ which in order to prove it suffices to show that, for every $[s,t]\subset [a,b]$ $$ \sup_{x,y\in[c,d]} |f(x)|-|f(y)|= \sup_{x\in[c,d]}|f(x)|- \inf_{x\in[c,d]}|f(x)|\le \sup_{x\in[c,d]}f(x)- \inf_{x\in[c,d]}f(x)=\sup_{x,y\in[c,d]} f(x)-f(y), $$ which is a consequence of the fact that $$ \big| |f(x)|-|f(y)|\big|\le |f(x)-f(y)|. $$

For the second part of the question, just use the fact that $$ -|f(x)|\le f(x)\le |f(x)| $$ and hence $$ \int |f| \le \int f \le \int |f| $$ and finally $$ \Big|\int f\,\Big| \le \int|f| $$

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  • $\begingroup$ Yiorgos, why do you have variables $[s,t], [c,d]$ and $[x,y]$? I don't know what they specify? $\endgroup$ – user104235 Jan 21 '14 at 4:46
  • $\begingroup$ @user104235: The $[c,d]$ correspond to the subintervals of the partitions. $\endgroup$ – Yiorgos S. Smyrlis Jan 21 '14 at 5:14
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Hint of part 1: Use the triangle inequality: //f(x)/ - /f(y)// < /f(x) - f(y)/ for x, y in [x(i), x(i+1)] which are two partition points of [a, b].

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