# Discontinuity and almost everywhere

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.

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

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.
• 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. Commented Mar 20, 2020 at 10:27
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 $$\frac 1 2$$.