$\int_{U}f(x)dx$ converge iff $\int_{U}|f(x)|dx$ converge

Question

Problem: Let be $U\subset\mathbb{R}^m$ a open set. Show that $\int_{U}f(x)dx$ converge iff $\int_{U}|f(x)|dx$ converge.

It seems to be a simple matter, but I'm not having ideas to solve it. The settings are as follows:

Definition of integral improper: Given a continuous function (limited or not) it says that the improper integral $\int_{U}f(x)dx$ is convergent when, for all exhaustion $U=\cup K_i$, there is the limit $\lim_{i \rightarrow \infty}\int_{K_i}f(x)dx$ and its value is independent of the exhaustion taken.

Definition of Exhaustion: An exhaustion of $U$ is a sequence of J-measurable $K_i\subset U$ compact such that $U=\cup K_i$ and $K_i\subset int(K_{i+1}) \forall i\in\mathbb{N}$.

Answer 1

Is this solution?

Proof. Define $f_+,f_-$ parts positve and negative of $f$. Then, we have that $f=f_++f_-$ and $|f|=f_+-f_-$. Now, suppose that $\int_Uf(x)dx$ converge. Hence $\lim_{i\rightarrow \infty}\int_{K_i}f(x)dx=\lim_{i\rightarrow\infty}[\int_{K_i} f_+ + \int_{K_i}f_-]$ exist. Then, how $\int_{K_i} f_+,\int_{K_i}f_-$ converge, implies that $\int_{K_i}|f|$ converge too.

Is the reciprocal similar?