Suppose $f \in \mathcal{L}^1(\Omega,\mathcal{A}, \mu).$ Prove that for each $\epsilon > 0$ there exists a bounded $\mathcal{A}$-measurable function $g$ such that $\int_{\Omega} |f-g| d\mu < \epsilon$.
My proof:
$$f \in \mathcal{L}^1(\Omega,\mathcal{A}, \mu) $$ $$\Rightarrow \int_{\Omega} |f| d\mu < \infty$$ $$\Rightarrow |f| \text{ is finite } \mu-ae$$ $$\Rightarrow \text{ Let } D = \{x: f(x) < \infty\}, \text{ then } \mu(D^c) = 0$$
Let $$g(x) = \begin{cases} f(x) & \text{if } x \in D \\ 0 & \text{otherwise } \end{cases}$$ then $g(x)$ is bounded and $\int_{\Omega} |f-g| d\mu = \int_D |f-g| d\mu = 0 < \epsilon$.
But this is a stronger result than stated as $g$ does not depend on $\epsilon$!
As mentioned in the comments, the function $f$ is not necessarily bounded on $D$ (that is, by a uniform constant), as the choice $g( x)= 1/\sqrt {x} \cdot 1_{(0,1)}(x)$.
However, we can define $E_n:=\{x\mid|f(x)|\lt n\}$ and $f_n(x)$ equal to $f(x)$ if $x$ belongs to $E_n$ and $0$ otherwise. Using the monotone convergence theorem, we can show that $\int_{\Omega}|f-f_n|\mathrm d\mu$ converges to $0$. Since each $f_n$ is bounded, we are done after a good choice of $n$ when $\varepsilon$ is given.