# Approximating an a.e. finite function by a bounded measurable function

Question: Let $$(\Omega,\mathcal{F}, \mu)$$ be a measure space with $$\mu\geq 0$$, $$f\geq 0$$, $$\mu(\Omega)<\infty$$ and let there be a $$A\in \mathcal{F}$$ s.t. $$\mu(A)=0$$ and $$f|_{\Omega \setminus A} <\infty$$.

Show that $$\forall \epsilon>0, \exists$$ a bounded measurable function $$g$$ s.t. $$\mu(\{w\in \Omega: f(w)\neq g(w)\})<\epsilon.$$

It sounds believable because $$f$$ is finite except for a measure zero set but am not sure where to start. Appreciate a hint.

$$\{x\in \Omega \setminus A: f(x) >N\}$$ decreases to the empty set as $$N$$ increases to $$\infty$$, so its measure tends to $$0$$. Take $$g=fI_{x: f(x) \leq N}$$ where $$N$$ is so large that $$\mu (\{x\in \Omega \setminus A: f(x) >N\}) <\epsilon$$.