# How to prove this function is L-continuous almost-everywhere?

Definition: Let $$(X, \mathcal{M}, \mu)$$ a measure space. Some property P (in this case, the continuity) is said to be satisfied almost everywhere in X if there exists a set $$N \in \mathcal{M}$$ such that $$\mu(N) = 0$$ and for all $$x \in X\setminus N$$, the property P holds.

Problem: Let $$f,g:\mathbb{R}\rightarrow\mathbb{R}$$ functions and g continuous in $$\mathbb{R}$$ with $$g\neq0$$. The function defined

$$f(x) = \begin{cases} g(x) & x \in \mathbb{R}\setminus\mathbb{Q} \\ 0 & x \in \mathbb{Q} \end{cases}$$ prove that $$f$$ is $$\mathcal{L}_{a.e}$$ continuous (that is, f is continuous almost-everywhere using the Lebesgue measure).

Is there an easy way to prove this?

• The definition of "$\mathcal{L}_{a,e}$ continuous" is unclear. What precisely does "for all $x\in X\setminus N$, the property $P$ holds" means when $P$ is continuity? (There are at least two reasonable interpretations, and one of them makes the statement you are trying to prove false.) Commented May 27, 2020 at 22:45
• @EricWofsey It means that there exists a set "N" where its Lebesgue measure is 0 and in the complement of N (in this case with respect to $R$) the function f is continuous. Therefore, f is continuous almost everywhere except in this "N". For example, the function f(x) = 0 if x < 0 and f(x) = 1 if x $\geq$ 0, is continuous almost everywhere because there exists N = {0} such that L(N) = 0 and f is continuous in $R\setminus \{0\}$. Commented May 27, 2020 at 23:19
• Ah, but what does "continuous in the complement of $N$" mean? It could mean either continuous at each point of the complement of $N$, or that the function is continuous when restricted to the complement of $N$. Those aren't the same thing! Commented May 27, 2020 at 23:22
• @EricWofsey the first thing you said! f is continuous for every $x \in \mathbb{R}\setminus N$ Commented May 27, 2020 at 23:29
• In that case the statement you are trying to prove is false. Commented May 27, 2020 at 23:39

If $$E=\mathbb Q$$ then $$E$$ has measure $$0$$ and the restriction of $$f$$ to $$\mathbb R \setminus E$$ is $$g$$ which is continuous.
You cannot prove the existence of a set $$E$$ of measure $$0$$ such that $$f$$ is continuous at each point of $$\mathbb R \setminus E$$. For example, if $$g(x)=1+|x|$$ then the corresponding $$f$$ is not continuous at any point!. If your null set exists then we get the conclusion that the real line has measure $$0$$!.