$f:\Omega \rightarrow \mathbb{R}$ integrable $\Leftrightarrow$ $\exists g \ge 0$ so that $\int_{|f|\ge g}|f| \, \mathrm{d}\mu < \varepsilon$ $f:\Omega \rightarrow \mathbb{R} \cup\{-\infty\} \cup\{+\infty\}$ on the measure space $(\Omega,\mathscr{A},\mu)$ is integrable wrt $\mu$ $\Leftrightarrow$ there exists a integrable function $g \ge 0$ so that 
$$\int_{|f|\ge g}|f| \, \mathrm{d}\mu < \varepsilon, \; \forall \varepsilon>0$$
How can that be proven?

If $f$ is integrable, then $g=2|f|$ works, as $|f|\geq2|f|$ only when $f=0$ and $\int_{f=0}|f|=0$ anyways. Now, suppose there exists such a $g$. Then $\int_{\Omega} |f|=\int_{|f|<g}|f|+\int_{|f|\geq g}|f|\leq \int_{\Omega}|g|+\epsilon$, which gives that $f$ is integrable because $g$ is.
