Riemann integral proof

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 $\delta$.

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 $$\Big|\int f\,\Big| \le \int|f|$$
• Yiorgos, why do you have variables $[s,t], [c,d]$ and $[x,y]$? I don't know what they specify? – user104235 Jan 21 '14 at 4:46
• @user104235: The $[c,d]$ correspond to the subintervals of the partitions. – Yiorgos S. Smyrlis Jan 21 '14 at 5:14