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I want to know the relationship between continuity and almost everywhere.

If a function f has only finitely many removable discontinuities, then there exists a continuous function $g$ such that $f = g$ a.e.

I know this is true. Also we can change ‘finitely many’ into ‘countably many’.

But if f has a discontinuity which is not removable, is there a continuous function g such that $f = g$ a.e?

For example, suppose $X=[0,1]$ and $f$ is a function on $X$ whose value is $0$ on $[0,1/2]$ and $1$ on $(1/2,1]$.

In this case, is there a continuous function $g$ such that $f = g$ a.e.?

Thank you.

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    $\begingroup$ Did you intend to say "$0$ on $[0,\frac12]$ and $1$ on $(\frac12,1]$? If so, then no, there is no such continuous $g$. $\endgroup$
    – Henry
    Commented Mar 20, 2020 at 10:10
  • $\begingroup$ can you tell me how to prove there is no such function? And is there any statement that generalizes this fact? $\endgroup$
    – anadad
    Commented Mar 20, 2020 at 10:12

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There is not. Suppose $g\colon X\rightarrow\mathbb{R}$ is such a function. $f^{-1}(0)$ has positive measure and $f=g$ a.e., so $g^{-1}(0)$ has positive measure; similarly, $g^{-1}(1)$ has positive measure. In particular, $0,1\in g(X)$, so $[0,1]\subseteq g(X)$ by the IVT. Now, the intuition is that the "jump" from $0$ to $1$ that $f$ does is something that a continuous $g$ could only do on a set with positive measure. Indeed, $(0,1)$ is an open set, so $g^{-1}((0,1))$ is open by continuity of $g$, non-empty by the IVT and hence has positive measure. However $f(x)\neq g(x)$ for all $x\in g^{-1}((0,1))$, because $f(X)=\{0,1\}$. Hence such a $g$ does not exist. I leave it to you to generalize this to arbitrary jump discontinuities.

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  • $\begingroup$ Every open set in a real line is a countable union of open intervals. And every open interval has a positive measure. Thus open set doesn’t have a measure zero? $\endgroup$
    – anadad
    Commented Mar 20, 2020 at 10:25
  • $\begingroup$ Yes, as long as it's non-empty. You don't need the structure result though, it follows directly by definition: If $U$ is open and non-empty, choose $x\in U$, by openness there is an $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\subseteq U$, hence $U$ has positive measure. $\endgroup$
    – Thorgott
    Commented Mar 20, 2020 at 10:27
  • $\begingroup$ Ohhh right..! Thank you!’ $\endgroup$
    – anadad
    Commented Mar 20, 2020 at 10:33
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Second part: Suppose there is a continuous function $g$ such that $f=g$ a.e... Thus implies that $g(x)=1$ for all $x >1/2$ by continuity. [Here we need the fact that the complement of a set of Lebesgue measure $0$ is dense which shows that $g=1$ on a dense subset of $(\frac 1 2, n)$]. Similarly $g(x)$ for all $x <\frac 12$. This contradicts continuity of $g$ at $\frac 1 2$.

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