Suppose (X, $\mathcal{M}$, $\mu$) is a $\sigma$-finite measure space, and let $g\in L^\infty(\mu)$. Can we always consider that for sets of the form ($\epsilon>0$ sufficiently small)
$$ A_{\epsilon} = \{x \in X: |g(x)|\geq ||g||_{\infty} - \epsilon > 0 \} $$
there exists a measurable subset $B\subset A_{\epsilon}$ such that $0<\mu(B)<+\infty$? Why or why not? I would need this result in order to prove some results for my measure theory class. Given the definition of the L infinity norm, it is clear to me that $\mu(A_{\epsilon})>0$, but taking for instance a constant function in an infinite measure space, I think the measure of said set could also be infinite. In this scenario, I doesn't look trivial to me that we can assume a positive and finite measure subset exists in an arbitrary measure space.
Edit: Changed (X, $\mathcal{M}$, $\mu$) from an arbitrary space to a $\sigma$-finite measure space to avoid obvious counterexamples.
I will prove a more general result, namely for any $A$ with $\mu(A) > 0$, there exists $B \subseteq A$ with $0 < \mu(B) < +\infty$, assuming $\mu$ is $\sigma$-finite. Indeed, as $\mu$ is $\sigma$-finite, there exists an increasing sequence of measurable sets $X_i$ s.t. $\cup_{i = 1}^\infty X_i = X$ and $\mu(X_i) < +\infty$ for all $i$. But then $A_i = A \cap X_i$ is an increasing sequence of measurable sets with $\cup_{i = 1}^\infty A_i = A$, so $\mu(A_i) \rightarrow \mu(A)$. In particular, as $\mu(A) > 0$, there exists $i$ s.t. $\mu(A_i) > 0$. We also have $\mu(A_i) \leq \mu(X_i) < +\infty$, so setting $B = A_i$ proves the claim.