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

Solution 1:

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 $0 \le U(|f|,P) -L(|f|,P)$.)

Addendum:

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