If $X$ is a Borel set and $f$ has countably many discontinuities, prove that $f:X\rightarrow \mathbf{R}$ is Borel measurable.

Question

The Problem

Good evening! I am currently struggling with the following exercise.

Suppose $X$ is a Borel subset of $\mathbf{R}$ and $f:X\rightarrow \mathbf{R}$ is a function such that $\{x\in X: f \text{ is not continuous at } x\}$ is a countable set. Prove $f$ is a Borel measurable function.

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What I Know

Unfortunately I am pretty lost with this one. I was told that the following observation is an important item in my toolkit for handling this problem:

To show that $f$ is a Borel measurable function, it suffices to prove that $f^{-1}((a,\infty))=\{x\in X: f(x)>a\}$ is a Borel set for all $a\in\mathbf{R}$.

I have a sneaking suspicion that

1. $X$ is a Borel set
2. the set of discontinuities of $f$ is countable

are also equally important pieces of information here. But I don't know why.

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My Question

I think all I need here is a little push. I have all the pieces in front of me, I think, but I don't know how they fit together. In other words, I would really appreciate a tip on how you might approach this problem in an intuitive way. This is my first time working with this material, so you can safely assume that I am unfamiliar with the more "advanced" theorems that could be used here. The more basic, the better (in my case at least)!

Thank you all in advance!

Answer 1

In the question

A function with countable discontinuities is Borel measurable.

the user Luiz Cordeiro argues as follows:

We want to show that $f^{-1}((a,\infty))$ is a Borel set (a function $f$ satisfying this is the definition of a Borel measurable function). Let $a\in \mathbf{R}$ be arbitrary and consider the set $A = f^{-1}((a,\infty))$. We can write this set as a union of its interior and the complement of the interior. That is

$$A = \mathrm{int}(A)\cup\big[A\setminus \mathrm{int}(A)\big].$$

The interior is open and thus Borel measurable. If we can show that its complement in $A$ is Borel measurable, that is $A\setminus \mathrm{int}(A)$, then since any union of Borel measurable sets is Borel measurable it would follow that $A$ is Borel measurable. This follows by the property of the Borel sets being a $\sigma$-algebra.

Let $x\in A\setminus \mathrm{int}(A)$, then $x$ is not an interior point of $A$ meaning that for every $\delta>0$ we can find a point $y_\delta$ such that

$$|x-y_\delta|<\delta\text{ but }y_\delta \notin A.$$

What does this then mean? $A$ is the inverse of $(a,\infty)$ so where does $f(y_\delta)$ belong, certainly not to $(a,\infty)$? Is $f$ continuous at $x$?

Conclude that $A\setminus \mathrm{int}(A)$ is at most countable. How can we then argue that $A\setminus \mathrm{int}(A)$ is measurable?

Answer 2

Let $f_n(x)=f(\frac {[nx]} n)$. Then $f_n (x) \to f(x)$ at all points $x$ where $f$ is continuous. Each $f_n$ takes only countable number of values on unions of intervals of the type $[j/n, (j+1)/n)$ so they are all Borel measurable functions. Imitate the proof of the fact that point-wise limits if Borel measurable functions are Borel measurable to complete the proof.