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).



Baire 1 function equivalence in measure
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.


