If two functions are equal almost everywhere, the first is continuous a.e., is the second?

Question

If $f = g$ a.e. in $E \in \mathfrak{M}$ (the Lebesgue measurable sets) and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?

I think this is true.

My “proof”:

Let us denote $D_1 = \{ x \in E: f(x) \text{ discontinuous}\}$, $m(D_1) = 0$ and $D_2 = \{ x \in E: f(x) \neq g(x)\}$, $m(D_2) = 0$.

Define $D_3 = \{ x \in E: g(x) \text{ discontinuous}\}$.

If $f$ is identically $g$, then it is clear that the result follows as $D_3 = D_1$.

Otherwise, we have $D_3 \subseteq D_1 \cup D_2$, and so $m^*(D_3) \leq m^*(D_1 \cup D_2) \leq m^*(D_1) + m^*(D_2) = 0$.

So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.

Does this proof work?

Thanks!

Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.

Answer 1

The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.

(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)

Answer 2

No your proof does not work because the fact that $f(x)=g(x)$ on $E\setminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:

$D_2\cap (E \setminus D_1) \subseteq (E \setminus D_3)$

which means:

$D_3 \subseteq (E \setminus D_2) \cup D_1$