# are almost everywhere bounded functions measurable?

We know that if $$f : [a,b] → R$$ is measurable, then $$f$$ is “almost bounded” on $$[a,b]$$ in the sense that, for every $$\epsilon > 0$$, there is an open set $$G$$ with $$\ell(G) < \epsilon$$ such that $$f$$ is bounded on the closed set $$[a,b]\setminus G$$.

Indeed:

for each integer $$n$$, let $$E_n = \{x ∈ [a,b] : | f (x)| > n\}.$$ These sets all have finite Lebesgue measures, they form a decreasing sequence of sets and $$\bigcap_{n=1}^\infty E_n=\emptyset.$$ Consequently $$\ell(E_n)\rightarrow 0$$. Now just take an integer $$N$$ so that $$\ell(E_N)<\epsilon/2$$ and use $$E = E_N$$. Observe that $$| f (x)| \le N$$ for all $$x \in [a,b]\setminus E$$. Since $$l(E) < \epsilon/2$$ we can also find an open set $$G\supseteq E$$ for which $$\ell(G) < \epsilon$$ and for which the statement of the theorem must hold.

Is the contrary true? That is: are almost everywhere bounded functions measurable?

If $$E$$ is any non-measurable set and $$f=\chi_E$$ (i.e. $$f(x)=1$$ for $$x \in E$$, $$0$$ for $$x \notin E$$ ) then $$f$$ satisfies your hypothesis (for any open set $$G$$!) but $$f$$ is not measurable.