Challenge Yet another counterexample challenge

[micromass]

Here's the proof that the set of continuity points is $G_\delta$ (meaning a countable intersection of open sets), which is of course equivalent that the set of continuity points is $F_\sigma$ by complementation.

So let $f:\mathbb{R}\rightarrow \mathbb{R}$ be any function. For $n\in \mathbb{N}$, we let
A_n = \left\{x\in \mathbb{R}~\vert~\exists r_{n,x} >0:~\forall x', x''\in B(x, r_{n,x}):~|f(x'') - f(x')| < 1/n\right\}
where $B(x,r) = \{a\in \mathbb{R}~\vert~|x-a|<r\}$.

1) First we show that each $A_n$ is open.
If $x\in A_n$, then $B(x, r_{n,x})\subseteq A_n$. And thus $A_n = \bigcup_{x\in A_n} B(x,r_{n,x})$ is the union of open sets and thus open.

2) We show that if $f$ is continuous at $x$, then $x\in A_n$ for each $n$
Indeed, if $f$ is continuous at $x$, then there is some $r_{n,x}>0$ such that for all $x' \in B(x,r_{n,x})$ holds that $|f(x) - f(x')| < \frac{1}{2n}$. And then the triangle inequality implies that for $x',x''\in B(x,r_{n,x})$ holds that
|f(x&#039;) - f(x&#039;&#039;)| \leq |f(x&#039;) - f(x)| + |f(x) + f(x&#039;&#039;)| &lt; \frac{1}{2n} + \frac{1}{2n} = \frac{1}{n}

And thus $x\in A_n$.

3) We show that if $x\in A_n$ for each $n$, then $f$ is continuous at $x$.
Indeed, let $\varepsilon>0$ be arbitrary and take $n$ such that $\frac{1}{n}<\varepsilon$. Since $x\in A_n$, it follows that there is some $r_{n,x}>0$ such that for all $x' \in B(x,r_{n,x})$ holds that $|f(x) - f(x')| <\frac{1}{n}<\varepsilon$. So $f$ is continuous at $x$.

So it holds that the set of continuity points of $f$ is equal to $\bigcap_{n\in \mathbb{N}} A_n$; which is the intersection of open sets, and thus a $G_\delta$.

 

[micromass]

Here is another example of a set of measure $0$ that is not $F_\sigma$ and thus not the discontinuity set of any function.

Let $A_n$ be a Cantor set in $[0,1]$ of measure $\frac{n-1}{n}$. Let $A = \bigcup_n A_n$, then since each $A_n$ is nowhere dense, we have that $A$ is of the first category. On the other hand $\lambda(A) = \lim_{n\rightarrow +\infty}\lambda(A_n) = 1$.

Then $A' = [0,1]\setminus A$ is of the second category and of measure $0$. If it was the countable union of closed sets $F_m$, then each $F_m$ would be closed and of measure $0$, it would then be nowhere dense. This would mean that $A'$ is of the first category, which is not true.

 

[micromass]

Thank you everybody for participating to this thread! If anybody is interested in more counterexamples, get the book by Gelbaum and Olmsted "Counterexamples in analysis"

https://www.amazon.com/dp/0486428753/?tag=pfamazon01-20