Prove that $f$ is integrable on $[a,b]$ if and only its "positive part" $\max{(f,0)}$ and its "negative part" $\min{(f,0)}$ are integrable on $[a,b]$ 
Prove that $f$ is integrable on $[a,b]$ if and only its "positive part" $\max{(f,0)}$ and its "negative part" $\min{(f,0)}$ are integrable on $[a,b]$.

Proving the first direction, we know that $U(f,\mathcal{P})-L(f,\mathcal{P}) < \epsilon_1$ as we refine $\mathcal{P}$. Then we want to show that $U(\max{(f,0)},\mathcal{P})-L(\max{(f,0),\mathcal{P})} < \epsilon_2$. I want to use the fact that  $$U(|f|,\mathcal{P})-L(|f|,\mathcal{P}) \leq U(f,\mathcal{P})-L(f,\mathcal{P})<\epsilon_1,$$ but I am not sure how to incorporate that.
 A: Define the positive and negative parts as $f^+ = \max(f,0)$ and $f^- = \min(f,0) = \max(-f, 0)$. Then $f = f^+ - f^-$ and $|f| = f^+ + f^-$.
We then have
$$f^+ = \frac{1}{2}(|f| +f), \\ f^- = \frac{1}{2}(|f|- f).$$
As you surmised, if $f$ is integrable and we can show that $|f|$ is integrable as well, then it follows that $f^+$ and $f^-$ are integrable as the sum and difference, respectively, of integrable functions.
To show that integrability of $f$ implies integrability of $|f|,$ note that using the reverse triangle inequality. 
$$||f(x)| - |f(y)|| \leqslant |f(x) - f(y)|.$$
Let $P = (a=x_0,x_1, \ldots , x_n = b)$ be a partition of $[a,b]$.
Given any sub-interval $I_k = [x_{k-1},x_k],$ we have
$$\sup_{x \in I_k} |f(x)|- \inf_{x \in I_k}|f(x)| = \sup_{x,y \in I_k}||f(x)|- |f(y)|| \leqslant \sup_{x,y \in I_k}|f(x)- f(y)| = \sup_{x \in I_k} f(x)- \inf_{x \in I_k}f(x). $$
Hence, for any $\epsilon > 0$ there is a partition $P$ such that
$$0 \leqslant U(|f|,P) - L(|f|,P) \leqslant U(f,P) - L(f,P) < \epsilon,$$
and $|f|$ is integrable.
Conversely, if $f^+$ and $f^-$ are integrable then $f = f^+ - f^-$ is integrable as the difference of two integrable functions.
