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I am studying Real analysis and I want to prove (or disprove) that (this is inspired by the fact that every continuous function is measurable)

Problem If $f:X\rightarrow \mathbb R$ is continuous a.e. then $f$ is measurable.

Applying the idea from this link I believe it is true only when $X$ is complete, where the space is complete here means that every subset of $D$ of measure zero in that measure space is measurable. Then $\mathbb R$ is complete since every subset of the set of Lebesgue measure zero $A$ must be measurable.

Q1 I am not sure if I understand this concept correctly. However, it seems false if $X$ is not complete and I use the example here (not sure if it is true please help me)

Let $Z\subset X$ be a measurable set of measure zero and $Z'\subset Z$ be a non-measurable subset of $Z$ (I don't know if there exists such space that this happens but I think it should exist for otherwise we may not need to use the "complete" term) and that $$f(x) = \cases{1 & , $x\in Z'$ \cr 0 & , $x\in(X\setminus Z)\cup (Z\setminus Z')$}$$ This function is continuous a.e. (on $X\setminus Z$) but $$\left\{x\in X; f(x)>\dfrac{1}{2}\right\}$$ is not measurable and $f$ is not a measurable function.

Q2 What about the proof when $X$ is complete here, I am trying to use the idea on that link so that I can understand myself. Let $D$ be the set of measure zero of $X$. Define $$f_1(x) = \cases{0 & , $x\in X\setminus D$ \cr f|_{D}(x) & , $x\in D$}$$ $$f_2(x) = \cases{0 & , $x\in D$ \cr f|_{X\setminus D}(x) & , $x\in X\setminus D$}$$ and each $f_1,f_2$ can be proven not hard that they are measurable functions (by using the fact that every subset of $D$ is measurable). Thus $f=f_1+f_2$ is measurable.

W. Wongcharoenbhorn
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1 Answers1

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Comment
Perhaps it will help to define "continuous a.e.". Let $f : \mathbb R \to \mathbb R$. Let $\lambda$ be Lebesgue measure. Let $\mathcal B$ be the sigma-algebra of Borel sets. Two possible definitions:

(a) Let $E = \{a \in \mathbb R : f \text{ is continuous at } a\}$. We say "$f$ is continuous a.e." if $E \in \mathcal B$ and $\lambda(\mathbb R \setminus E) = 0$.

(b) We say "$f$ is continuous a.e." if there exists a set $E \in \mathcal B$ such that $\lambda(\mathbb R \setminus E) = 0$ and the restriction $f\vert_{E}$ is a continuous function from $E$ to $\mathbb R$.

These two definitions are different. For example, if $E\subset \mathbb R$ is measure zero but dense in $\mathbb R$, then the indicator function $f = \mathbf1_E$ satisfies (b) but not (a). An example of such $E$ is $\mathbb Q$.

GEdgar
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