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So originally I was asked to prove that the set of all points of continuity for $f:\mathbb{R}\rightarrow\mathbb{R}$ is measurable (all of my analysis so far is in $\mathbb{R}^d$). I came up this solution:

For $f$ above, let $C=\{\,\, x \in \mathbb{R}\,\, |\,\, f \,\,\text{is continuous at} \,\,x \,\,\}$. For all $a\in C$, let $B_{1/n}(a)$ be the open ball centered at $a$ of radius $1/n$. Define $O_n = \bigcup_{a\in C}B_{1/n}(a)$. Since this is a union of open sets it is open (regardless of countability of the union). Furthermore, since it is open it is measurable. Then $C=\bigcap_{n=1}^\infty O_n$, and this is a $G_\delta$ set (countable intersection of open/measurable sets and hence measurable).

So then $C$ is measurable. But it was pointed out to me that based on what I have here, $C$ could be replaced with any set, including a non-measurable set. So I'm curious as to what goes wrong here.

Calvin Khor
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masjgomz
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2 Answers2

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Take $C$ to be the set of all irrational numbers in $\mathbb{R}$ and do what you propose. Then $O_n=\Bbb{R}$ for every $n$ so the intersection of all $O_n$ is not $C$. The same objection applies to your proof.

A correct proof can be found in an answer to this question. It is actually somewhat similar to what you are trying to do.

markvs
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The issue is the last equality $$C \overset{\Huge?}= \bigcap_{n=1}^\infty O_n$$ For instance, if $f:\mathbb R\to\mathbb R$ is defined by $f=1$ when $x>0$ and $0$ when $x\le 0$, then $C=\mathbb R\setminus \{0\}$, but the intersection over all $O_n$ gives $\mathbb R$.

Calvin Khor
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