Let $f:[0,1]\rightarrow \mathbb{R}$ be an almost everywhere continuous function in the sense of Lebesgue measure. Prove that $f$ is Lebesgue measurable.

Here I tried to figure out inverse image of a Borel set under $f$ yet the set where f is not continuous might be very weird. I also tried to do with considering $f$ as a limit of continuous function, but I am not sure about an existence of such a sequence. Thanks in advance for any comment or ideas.


Use the fact that the Lebesgue measure is complete, i.e., every subset of a measurable set of measure zero is measurable and has measure zero (at worse, every measure has a completion). Now, let $I:=[0,1]$ , and let $S \subset [0,1]$ be the set of measure zero. Then $I= S \cup (S^c)$. We can use the classical definition that $f$ is measurable iff the inverse image of an open set is measurable, while breaking up $I:=[0,1]$ into the union of $S$ ; the measure-zero subset where $f$ is not continuous, and $I\S$ , where $f$ is continuous, and consider the inverse image intersected with each of the two sets (and then consider the union of the inverse images):

Now, let $U$ be open in $\mathbb R$ . Then , $f^{-1}(U)=[f^{-1}(U) \cap(S)] \cup [f^{-1}(U)\cap(S^c)] $. Now, $f$ is continuous in $S^c$, so that $f^{-1}(U)\cap S^c $ is open in $[0,1]$ , and $f^{-1}(U) \cap S $ is a subset of the measurable set $S$ of measure zero, so that $f^{-1}(U) \cap S:=V$ is measurable, ( with measure zero). Then $f^{-1}U)$ is the union of an open set --which is measurable , and a measurable set $V$ ( with measure zero, but we only care that it is measurable), so the inverse image of the open set $U$ of $\mathbb R$ is the union of two measurable sets, and so it follows, it is measurable, so that $f$ is measurable.

  • $\begingroup$ $f^{-1}(U)\cap S^c$ is open in $[0,1]$ or in $S^c$? $\endgroup$ – Anupam Jun 5 '14 at 9:36
  • $\begingroup$ Ah, yes, open in $S^c$, so measurable, union a set of measure zero, which is measurable. $\endgroup$ – user99680 Jun 5 '14 at 16:24
  • $\begingroup$ In another approach , $f^{-1}(U) \cap S^c$ is measurable in $[0,1]$, because $f^{-1}(U)$ is open in $[0,1]$, by assumed continuity of $f$, and $S^c$ is measurable in $S^c$ because it is the complement of a measurable set; $S$ is measurable in $[0,1]$(with measure zero) so that its complement $S^c$ is also measurable. $\endgroup$ – user99680 Jun 5 '14 at 16:37
  • $\begingroup$ If you can assume that $f^{-1}(U)$ is open, then it's measurable and it's proved. I don't think your last approach is correct. $\endgroup$ – Richard Clare Jan 16 at 21:53

If you show that the inverse image of any open interval of the form $(a,\infty)$ is a measurable set, you can conclude that the inverse image of any Borel set is a measurable set. This is because these intervals generates the Borel $\sigma-$algebra. I hope it can help you.

  • $\begingroup$ Thank you for your comment, but I dont know the inverse image of this interval under f (an a.e contionuous function). The set where f is not continous could be very strange I guess, that is what I was concerning about it. $\endgroup$ – seriously divergent Nov 25 '13 at 8:26

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