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We say a measurable function $f: \mathbb R^n \to \mathbb R$ is essentially continuous if the inverse image of any open set $O$ differs from an open set by a set of null measure, in the sense that there exists an open subset $U$ of $\mathbb R^n$ such that $\mu(f^{-1} (O) \, \Delta \, U) = 0$ where $\Delta$ denotes the symmetric difference and $\mu$ denotes the Lebesgue measure.

Question: Is it true that $f$ is essentially continuous if and only if $f$ agrees a.e. with a function that is continuous a.e.?

Remark: The chosen open set $U$ is allowed to be empty.

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    $\begingroup$ Let $O_1,\dots$ be a generator of the topology and for each $j$ let $U_j$ such that $N_j=f^{-1}(O_j)\Delta U_j$ is a nullset. Set $N=\bigcup_jN_j$, then on $\mathbb{R}^n\ -\ N$ the function should be continuous, right? $\endgroup$
    – user473423
    Commented Aug 24, 2023 at 12:35
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    $\begingroup$ Hmm I don’t see immediately that this is true (since N need not be closed)… might be though. $\endgroup$
    – Nate River
    Commented Aug 24, 2023 at 12:36
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    $\begingroup$ @Echo: In general, this cannot work without modifying $f$ on a null set, for example when $f=\chi_{\mathbb Q}$. In your argument, it seems unclear how one would redefine $f$ on $N$. $\endgroup$
    – Christian Remling
    Commented Aug 24, 2023 at 15:35
  • $\begingroup$ @Christian: the fuction $\chi_{\mathbb Q}$ is indeed continuous on the set $\mathbb{R} -\mathbb Q$. $\endgroup$
    – user473423
    Commented Aug 24, 2023 at 15:53
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    $\begingroup$ @Echo: Yes, the restriction $f\bigr|_{N^c}$ is continuous, but this is beside the point here: we need the unrestricted function to be continuous at all points $\notin N$, which is a stronger property and not satisfied by this $f$. $\endgroup$
    – Christian Remling
    Commented Aug 24, 2023 at 16:01

1 Answer 1

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I adapt the proof of a more general result: Theorem 4.12 in Chapter XI of K. Kuratowski and A. Mostowski’s book “Set Theory: with an introduction to descriptive set theory” (see page 408).

Claim. $f$ is essentially continuous iff there exists a null set $N$ such that $f|({\bf R}^n-N)$ is continuous.

Proof. ($\Rightarrow$) Let $U_1$, $U_2$, … be an open base of $\bf R$. By assumption, $$f^{-1}(U_n)=(O_n-A_n)\cup B_n,$$ where $O_n$ is open and $A_n$, $B_n$ are null sets. Set $N=\bigcup_nA_n\cup\bigcup_nB_n$, so that $N$ is a null set. We show that $g=f|({\bf R}^n-N)$ is continuous. Let $H\subset\bf R$ be open; we prove that $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$. Now $H=U_{k_1}\cup U_{k_2}\cup\cdots$, and so $$g^{-1}(H)=\bigcup_nf^{-1}(U_{k_n})-N=\bigcup_n\bigl((O_{k_n}-A_{k_n})\cup B_{k_n}\bigr)-N.$$ Since $A_{k_n}\cup B_{k_n}\subset N$, it follows that $$g^{-1}(H)=\bigcup_nO_{k_n}-N,$$ which completes the proof since $\bigcup_nO_{k_n}$ is open.

($\Leftarrow$) Let $N$ be a null set and suppose $g=f|({\bf R}^n-N)$ is continuous. Then, if $H\subset\bf R$ is open, the set $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$; that is, there exists open $O\subset{\bf R}^d$ such that $f^{-1}(H)-N=O-N$. Therefore, $$\begin{align*} f^{-1}(H) &= (f^{-1}(H)-N)\cup(f^{-1}(H)\cap N)\\ &=(O-N)\cup(f^{-1}(H)\cap N). \end{align*}$$ Since $N$ is a null set, $f^{-1}(H)\cap N$ is also a null set, and it follows that $f^{-1}(H)\triangle O$ is a null set. $\square$

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    $\begingroup$ This is not exactly what the OP is asking. The restriction can be continuous without $f$ being continuous at any point of $N^c$ (for example when $f=\chi_{\mathbb Q}$, $N=\mathbb Q$). $\endgroup$
    – Christian Remling
    Commented Aug 24, 2023 at 16:04
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    $\begingroup$ @ChristianRemling: That is a valid point. But I think that if $f$ is continuous when restricted to a full-measure set, then there is an a.e. continuous function $g$ that is a.e. equal to $f$. (See mathoverflow.net/questions/145957/…) Combined with this post, I think that answers the original question. $\endgroup$
    – Will Brian
    Commented Aug 24, 2023 at 18:29
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    $\begingroup$ @Will Brian Yes, these two conditions are indeed equivalent! So we have three equivalent definitions which is pretty satisfying. When I’m free later, given permission from the OP I will edit the answer to include this and show that the original question is also answered. $\endgroup$
    – Nate River
    Commented Aug 25, 2023 at 9:22

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