Show that it exists $M\in \mathbb{R}$ such that $m(\{|f|>M\})<ε $ Let $f$ be a measurable function and almost everywhere finite function $f:[a,b]\to [-\infty,+\infty]$, show that for every $ε>0$ it exists a real number $M$  such that $m(\{|f|>M\})<ε $

"for every $ε>0$ it exists a real number $M$  such that ..." it reminds the definition of a continuous function, I don't know if I can use this somehow.
My approach was to say that:
Let $A$ be the set that $f$ is not finite, by hypothesis $m(A)=0$
$f^{-1}[-\infty,+\infty]=[a,b]$ thus it must exist a set $B$ s.t $f^{-1}(B)=A$
if $B$ contains an interval, then it exists $M\in \mathbb{R}$ such that $f^{-1}(-M,+M)=A' \subset A
\Rightarrow \{|f|>M\}=A' \Rightarrow m(\{|f|>M\})=0$
I am not sure if I can show that $B$ contains an interval, any hints ?
 A: As David C. Ullrich commented, a nice way of proving this proposition is doing the following:
First, since $f$ is measurable, it is also the case that $|f|$ is measurable (since we are talking about an extended real valued function, measurability means $(\mathcal{F}, \mathcal{B}(\mathbb{\overline{R}}))$ - measurability, where $\mathcal{F}$ is the $\sigma$ - algebra of our measure space). Now, we can create the following sequence of sets
\begin{equation*}
\forall \, n \in \mathbb{N}, \; B_n = \{ |f| > n \}.
\end{equation*}
Since $|f|$ is measurable, $B_n \in \mathcal{F}$ and $B_{n + 1} \subseteq B_n$, for all $n$. Further, $m(B_1) = m(\{ |f| > 1 \}) \leq m([a, b]) = b - a < \infty$, $\textit{i.e.}$, our sequence of sets has finite measure. These are all the ingredients we need to use monotone continuity from above and, consequently, we have that
\begin{equation*}
\lim_{n \rightarrow \infty} m(B_n) = m\left(\bigcap_{n = 1}^\infty B_n \right).
\end{equation*}
As David pointed out in the comments, it is clear that $\{ |f| = \infty \} = \cap_{n = 1}^\infty B_n$, which means that $\lim_{n \rightarrow \infty} m(B_n) = 0 \; (*)$. After this (really) long preamble we can finally prove the theorem.
Consider an arbitrary $\varepsilon > 0$. Since $(*)$, we can use the epsilon in the definition of limit to get that
\begin{equation*}
\exists \, n_0 \in \mathbb{N}, \, \text{s.t.} \, \forall \, n > n_0, \, | \mu(B_n) - 0| < \varepsilon.
\end{equation*}
Simply define $M = n_0 + 1$. Then
\begin{equation*}
m(\{ |f| > M \}) = m(\{ |f| > n_0 + 1 \}) = m(B_{n_0 + 1}) = |m(B_{n_0 + 1}) - 0| < \varepsilon.
\end{equation*}
which completes the proof. Hope it help!
