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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?

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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.

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  • $\begingroup$ So the matter is even more radical, that is, there are even bounded functions that are not measurable. Thank you so much $\endgroup$
    – Jim Croce
    Aug 11, 2022 at 10:00

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