# Baire 1 function equivalence in measure

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).

• Doesn't the characteristic function of a fat Cantor set (one of positive measure) provide a counterexample? Oct 7, 2021 at 10:21
• 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?). Nov 10, 2021 at 17:22
• 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? Nov 12, 2021 at 16:11

## 1 Answer

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.

• +1. Baire-1 functions do have some good continuity properties, but not this good. See en.wikipedia.org/wiki/Baire_function#Baire_class_1 Nov 14, 2021 at 20:02
• 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? Feb 11, 2022 at 8:50
• @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. Feb 11, 2022 at 9:33
• 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? Feb 11, 2022 at 10:07
• 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. Feb 11, 2022 at 13:45