Increasing Sequence of Step Functions Converging to Characteristic Function of Fat Cantor Set

Question

This is related to this question.

I want to find an open set $G\subseteq [a,b]$ such that there does not exist an increasing sequence of step functions on $[a,b]$ which converges almost everywhere to $\chi_{G^c}$.

By that question, if $G^c$ has interior, I can find an increasing sequence $s_n$ of step functions which converges to the characteristic function of the interior. If the noninterior points form a set of measure zero, the sequence will still converge almost everywhere.

So there are two options. A nowhere dense closed set which has positive measure, or a closed set with nonempty interior but which boundary has not zero measure.

For the first case, I found this, a Fat Cantor Set, and from a little answer here for this case I should be able to get the example.

So we can propose this

Proposition. Let $F\subseteq [0,1]$ be the fat Cantor Set. Then there does not exist an increasing sequence $s_n$ of step functions such that $s_n\nearrow \chi_F$ almost everywhere in $[0,1]$.

Is there any idea of how can I prove that inexistence?

The second case seems to be around something similar.

Answer 1

I was all the day thinking and then the problem suddenly seemed easier.

Let $s$ be a step function such that $s\leq f$, where $f=\chi_F$. Let $\{x_0,\dots,x_n\}$ be a partition of $[0,1]$. Note that for any $x\in [0,1]$, then, since the interval $(x-\epsilon,x+\epsilon)\nsubseteq F$ for any $\epsilon>0$, there exists some $y\in(x-\epsilon,x+\epsilon)$ such that $f(y)=0$.

Let's suppose $x\in(x_{k-1},x_k)$, and let $\epsilon>0$ be such that $(x-\epsilon,x+\epsilon)\subseteq (x_{k-1},x_k)$. Then for all $y\in (x_{k-1},x_k)$, we have $s(y)\leq f(y)=0$. If we add the restriction for $s$ to be greater than 0 (Since $f\geq 0$ it doesn't matter), then $s(y)=0$ for $y\in[0,1]$, except perhaps at $x_0,\dots,x_n$.

Then if there existed such sequence (Of increasing and nonnegative step functions converging to $f$ almost everywhere), $s_n=0$ almost everywhere on $[0,1]$, thus $\lim_{n\to\infty}s_n(x)=0$ for almost any $x\in[0,1]$, so $f$ would be $0$ almost everywhere, which is a contradiction, since $F$ does not have zero measure.