If $f$ is integrable then $|f|$ is also integrable Show that if $f$ is integrable on $[a, b]$ then $\lvert f \rvert$ is also integrable. Hint: Show that
$$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$
I have:
$$U (P , \lvert f \rvert) \ge U (P , f )$$ and $$ L(P , \lvert f \rvert) \ge L(P,f)$$
Thus,
$$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$
 A: Consider a partition $$P=\{x_{0},x_{1},x_{2},\dots,x_{n}\} $$ of $[a, b] $ and for each $i=1,2,\dots,n$ let us use the notation $$M_{i}(f) =\sup\, \{f(x) \mid x\in[x_{i-1},x_{i}]\},\, m_{i}(f)=\inf\,\{f(x) \mid x\in[x_{i-1},x_{i}]\} $$ Then we have by definition $$U(P, f) =\sum_{i=1}^{n}M_{i}(f)(x_{i}-x_{i-1}),\,L(P,f)=\sum_{i=1}^{n}m_{i}(f)(x_{i}-x_{i-1})$$ and hence $$U(P, f) - L(P, f) =\sum_{i=1}^{n}(M_{i}(f)-m_{i}(f))(x_{i}-x_{i-1})$$ It is easy to prove that $$M_{i} (f) - m_{i} (f) =\sup\, \{|f(x) - f(y) | : x, y\in[x_{i-1},x_{i}]\} $$ Now by triangle inequality we know that $$||f(x) |-|f(y) ||\leq |f(x) - f(y) |$$ Therefore it follows that $$M_{i} (|f|) - m_{i} (|f|) \leq M_{i} (f) - m_{i}(f) $$ which leads to the desired inequality $$U(P, |f|) - L(P, |f|) \leq U(P, f) - L(P, f) $$ From this inequality and criterion for Riemann integrability it is almost obvious that if $f$ is Riemann integrable on $[a, b] $ then $|f|$ is also Riemann integrable on $[a, b] $. 
A: This answer shows that if $f$ and $g$ are Riemann integrable so is $h=\max\{f,g\}$. Use this with $g=-f$, then $h=|f|$.
Nota: In your post you "deduce" from $U (P , \lvert f \rvert) \geqslant U (P , f )$ and $ L(P , \lvert f \rvert) \geqslant L(P,f)$ that $U (P , \lvert f \rvert) − L(P , \lvert f \rvert) \leqslant U (P , f ) − L(P , f )$. This does not hold, in fact one can deduce no upper bound of $x-y$ from the hypothesis that $x\leqslant x'$ and $y\leqslant y'$. Instead one would need bounds such as $x\leqslant x'$ and $y\geqslant y'$, which indeed imply $x-y\leqslant x'-y'$.
