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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.

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    $\begingroup$ How do you justify $D_3 \subseteq D_1 \cup D_2$ ? $\endgroup$ – Yves Daoust Dec 12 '18 at 20:46
  • $\begingroup$ @YvesDaoust My reasoning was that if $f(x) = g(x)$ on $E\setminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well? $\endgroup$ – TuringTester69 Dec 12 '18 at 20:47
  • $\begingroup$ Continuity of $f$ in $E\setminus D_1$ does not imply continuity of $g$, because of $D_2$. $\endgroup$ – Yves Daoust Dec 12 '18 at 20:51
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    $\begingroup$ Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals. $\endgroup$ – Andrés E. Caicedo Dec 12 '18 at 20:52
  • $\begingroup$ @Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer. $\endgroup$ – James Yang Dec 12 '18 at 20:54
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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.)

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$f$ continuous on $E\setminus D_1$ implies $f$ continuous on $E\setminus(D_1\cup D_2)$. And of course, $g$ continuous on $E\setminus(D_1\cup D_2)$. Then the measure of $D_3$ is at most that of $D_1\cup D_2$.

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    $\begingroup$ Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$. $\endgroup$ – Andrés E. Caicedo Dec 12 '18 at 20:57
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    $\begingroup$ @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ?? $\endgroup$ – Yves Daoust Dec 12 '18 at 20:59
  • $\begingroup$ The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point. $\endgroup$ – Andrés E. Caicedo Dec 12 '18 at 21:02
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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:

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

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