Given a function $f: \Omega \to \mathbb{R}$, let $\mathcal{A}$ be a $\sigma$-algebra on $\Omega$ and $\mathcal{B}$ be the Borel $\sigma$-algebra. Assume that $f$ is $\mathcal{A}-\mathcal{B}$ measurable, $\mu$ is a measure on $\mathcal{A}$, and $\mu (\Omega) < \infty$.
Given that $$\mu( \{ \omega: f(\omega) = \pm \infty \}) = 0,$$ which basically means $f$ is finite $\mu$-almost everywhere, I have to prove that $f$ is $\mu$-almost uniformly bounded: $$\forall \epsilon > 0, \exists E_{\epsilon} \in \mathcal{A} : \mu (E_{\epsilon} ^ c) < \epsilon$$ and $f$ is bounded on $E_{\epsilon}.$
My idea was that since the set of $\omega$ such that $f(\omega)$ is infinite has zero measure, then $\forall A \in \mathcal{A}$, if we let $A' = A \cap \{ \omega: f(\omega) = \pm \infty \}$, then $$\mu(A) = \mu (A').$$
Basically, if we remove the infinite elements, the measure remains the same, which means the measure of the complement also stays the same. If I could show that doing this process makes $f$ bounded on $A'$, then I could probably formulate a proof, but I don't think this notion is correct. $A'$ could still be an uncountable set, so I don't know how to show that $f$ would be bounded.
Is this approach correct? How should I do the proof?