Bounded integrable function

Question

Let $f : \mathbb{R} \to \overline{\mathbb{R}}$ be an integrable funtion. Given $\varepsilon > 0$ show that there is a bounded integrable function $g$ such that $\int |f - g| < \varepsilon$.

I was wondering if I could get a hint.

Answer 1

First, as $f$ is integrable, it takes infinite values on a negligible set, so we can assume that $f$ take its values on $\Bbb R$.

Writing $f=\max\{f,0\}+(f-\max\{f,0\})$, we can write $f$ as the difference of two measurable integrable non-negative functions. So we are reduced to the case $f\geqslant 0$ is integrable and measure.

To this aim, go back to the definition of Lebesgue integral, and recall that a simple function is bounded.