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I am trying to prove (or disprove) the following assertion:

Consider a probability triple $(X,\mathcal{B},\mu)$, $X$ separable Banach space (complete), $\mathcal{B}$ the Borel $\sigma-$algebra and $\mu$ a countably additive probability measure there on. Let $Y$ be a different separable Banach space.

Given a Baire 1 function $f:X\rightarrow Y$, there exists a $\mu-$almost everywhere continuous function $g:X\rightarrow Y$, such that $f(x)=g(x)$, $\mu-$almost everywhere.


I couldn't find a counterexample so far, even in $\mathbb{R}$, since all the major examples of Baire 1 functions that I know, seems to have this property: the characteristic function of the integers, the characteristic function of the Cantor set, etc.

It is interesting that it is true also for some Baire 2 functions. For example the Dirichlet function, which is equal to 0 $\lambda-$almost everywhere (where $\lambda$ is the Lebesgue measure).

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    $\begingroup$ Doesn't the characteristic function of a fat Cantor set (one of positive measure) provide a counterexample? $\endgroup$
    – Gro-Tsen
    Oct 7, 2021 at 10:21
  • $\begingroup$ What's a “$\mu$-almost continuous function”? I had read “$\mu$-almost everywhere continuous”, but now I realize that's not what you wrote (but maybe that's still what you mean?). $\endgroup$
    – Gro-Tsen
    Nov 10, 2021 at 17:22
  • $\begingroup$ Dear @Gro-Tsen thank you for your comment. I edited my post accordingly. The point remains: I don't know if it's a counterwxample, since there might exists an almost everywhere continuous function, almost equivalent to the indicator function of a fat Cantor set of positive measure. Am I wrong? $\endgroup$
    – Gioppa
    Nov 12, 2021 at 16:11

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Let $F$ be a closed subset of $[0,1]$ (say) with empty interior and $\mu(F)>0$ (where $\mu$ is Lebesgue measure), for example given by a fat Cantor set. Clearly, $1_F$ is Baire class $1$. I claim that $1_F$ cannot be almost everywhere equal to an almost everywhere continuous function.

Indeed, assume that $f = 1_F$ almost everywhere, say $f(x) = 0$ if $x \not\in F \cup N$ and $f(x) = 1$ if $x \in F \setminus N'$ with $N,N'$ having measure zero. I need to prove that $f$ is discontinuous on a set of positive measure. Note that $\mu(F \setminus N') > 0$ so I am done if I show that $f$ is discontinuous on $F \setminus N'$.

Assume toward a contradiction that $x \in F \setminus N'$ and that $f$ is continuous at $x$. Since $f(x) = 1$, there is an open neighborhood $V$ of $x$ such that $f(x) \neq 0$ on $V$, and in particular, $V \subseteq F \cup N$.

But since $F$ has empty interior, $x$ is in the closure of the complement of $F$, so $V$ meets this (open) complement, so there is a nonempty open interval $I \subseteq V$ which is disjoint from $F$. So we have $I \subseteq N$, which is a contradiction as $N$ has zero measure and $I$ has positive measure.

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  • $\begingroup$ +1. Baire-1 functions do have some good continuity properties, but not this good. See en.wikipedia.org/wiki/Baire_function#Baire_class_1 $\endgroup$ Nov 14, 2021 at 20:02
  • $\begingroup$ To me this is not the proof i was looking for. This proves that a Baire-1 function defined in the way you defined it does not have the property I was looking for. Am i wrong? $\endgroup$
    – Gioppa
    Feb 11, 2022 at 8:50
  • $\begingroup$ @Gioppa You asked to prove or disprove the assertion “a Baire class 1 function is equal a.e. to an a.e. continuous function”. I give you a counterexample that disproves the assertion. If this is not what you wanted, you should clarify or start a different question. $\endgroup$
    – Gro-Tsen
    Feb 11, 2022 at 9:33
  • $\begingroup$ Dear @Gro-Tsen, I apologize if my comment was rude in some way, it was not my intent. What I was saying is that you defined a function that is almost equivalent to the indicator function of F. It looks to me that this is not the only way of defining it. Am I wrong? $\endgroup$
    – Gioppa
    Feb 11, 2022 at 10:07
  • $\begingroup$ I'm sure there are plenty of other ways to construct a counterexample. I'm not offended, I just don't know what you're looking for: again, if you want a counterexample with different properties, or constructed in a different way, you can open another question (provided you can state precisely what you're looking for). On the other hand, if you're asking for a clarification of what I wrote, I can do that, but I don't understand what it is you're asking for. $\endgroup$
    – Gro-Tsen
    Feb 11, 2022 at 13:45

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