Equality with continuous function on a large set

Question

By Lusin's Theorem, a measurable function $f$ on $[0,1]$ is continuous on a set $A$ of measure arbitrarily close to 1. Such $f$ need not be continuous anywhere. What about the property of being equal to some (everywhere) continuous function on a set? Our $f$ is equal to a continuous function on a compact set of measure arbitrarily close to 1 (use inner regularity and the Tietze Extension Theorem.) Must $f$ be equal to some continuous function on some dense set, even if it is only countable? An arbitrary real-valued function $f$ is equal to some continuous function on some infinite set. (For example, interpolate $f$ by a piecewise linear function at the midpoint of every interval in the complement of the Cantor set.) Again, is it possible in general to do this with a dense set?

Answer 1

A measurable function need not be equal to a continuous function on a dense set. To see this consider the following measurable function: $$f(x)=\begin{cases}0 & x<1/2\\ 1 & x\geq 1/2.\end{cases}$$ Suppose that there exists a continuous function $g:[0,1]\to \mathbb R$ and a dense set $A\subset[0,1]$ such that $f(x)=g(x)$ for each $x\in A$. Since $A$ is dense in $[0,1]$, $A\cap[0,1/2)$ is dense in $[0,1/2)$ and similarly $A\cap (1/2,1]$ is dense in $(1/2,1]$. Therefore, let $(x_n)$ and $(y_n)$ be a sequences in $A\setminus\{1/2\}$ converging to $1/2$ such that $(x_n)$ is monotonically increasing and $(y_n)$ is monotonically decreasing. Then $g(x_n)=f(x_n)=0$ for each $n$ and $g(y_n)=f(y_n)=1$ for each $n$. But since $g$ is continuous, we must have that $$0=\lim_{n\to\infty}g(x_n)=g(1/2)=\lim_{n\to\infty}g(y_n)=1,$$ which is a contradiction.

I am not sure about your second question.