# Measurable function is bounded almost everywhere

Let $f : [a, b] \to \mathbb{R}$ be a measurable function. Given $\varepsilon > 0$ show that there is some $M > 0$ such that $|f(x)| \leq M$ for all $x \in [a, b]$ except on a set of finite measure less than $\varepsilon$.

Could I get a hint?

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Hint: $[a,b]$ is the increasing union of $f^{-1}([-M,M])$ over $M=1$, $2$, $3$, $\dots$.

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So clearly $f^{-1}([-M, M])$ is measurable and increasing, but I'm not sure exactly how to show that $$m^*\left([a, b] \setminus \bigcup_{n \geq 1}^M f^{-1}([-n, n])\right) < \varepsilon$$ –  icaruss Apr 26 '13 at 1:03
Use countable additivity of the measure to show that $\lim_{M\to\infty} \mu(f^{-1}([-M,M]))=\mu([a,b]).$ –  David Moews Apr 26 '13 at 18:58