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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.

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.

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!

Thy Art is Math
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    Here is a similar question: https://math.stackexchange.com/questions/369268/a-function-with-countable-discontinuities-is-borel-measurable?rq=1 – OgvRubin Oct 18 '18 at 08:25
  • @OlofRubin Thank you. Unfortunately I do not understand the answer given in that post :( – Thy Art is Math Oct 18 '18 at 08:54

2 Answers2

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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?

OgvRubin
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    Thank you for clarifying that user's answer. This is very helpful. However, I have a dumb question for you: what do you mean by interior? A quick Google search indicates this is a notion from topology (could be wrong on this), which I don't have any experience in :( – Thy Art is Math Oct 18 '18 at 09:10
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    The interior is a topological term but we could argue solely in the metric of $\mathbf{R}$ when we define it. If $A$ is any subset of the real numbers then $\mathrm{int}(A)$ is the union of all open subsets contained $A$, this definition coincides with the union of all segments $(x-\delta,x+\delta)\subseteq A$ where $x$ and $\delta$ may vary. The interior is an open set, that is for every $x\in \mathrm{int}(A)$ we can find a $\delta$ such that $(x-\delta,x+\delta)\subseteq A$ – OgvRubin Oct 18 '18 at 09:11
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    As an example the interior of the set ${0}\cup [1,2]$ is the set $(1,2)$. – OgvRubin Oct 18 '18 at 09:14
  • Wow, that actually makes sense to me! Thank you so much for that! – Thy Art is Math Oct 18 '18 at 09:17
  • Now that I've had a chance to think about this, let me see if I am on the right track. Since $y_{\delta}$ is not in $(a,\infty)$, it must be in $(-\infty,a]$. So $f(y_{\delta})\leq a$. But how does this allow us to make any conclusions about $x$? We know $f(x)>a$, but how does $y_{\delta}$ help us? – Thy Art is Math Oct 18 '18 at 19:46
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    So this shows that any neighbourhood of the form $(x-\delta,x+\delta)$ contains a point $y_\delta$ such that $y_\delta\notin A$. As you correctly asserted $f(y_\delta)\leq a$. But since $x\in A$ we must have that $f(x)>a$. Now can $x$ be a point of continuity of $f$? That is for every $\varepsilon>0$ can we find a $\delta>0$ such that any $y$ satisfying $|x-y|<\delta$ also satisfies $|f(x)-f(y)|<\varepsilon$? To contradict this use that $|f(x)-f(y_\delta)|\geq |f(x)-a|>0$. – OgvRubin Oct 18 '18 at 19:59
  • Ah, that is the final piece, isn't? Beautiful. That is what I was missing! Thank you for your assistance, Olof. The time you took out of your day to assist me is immensely appreciated. To avoid "extended discussion" this will be my final comment here. Thanks again :) – Thy Art is Math Oct 18 '18 at 20:22
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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.

Kavi Rama Murthy
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  • Thank you for your answer -- it is clear, concise, and simple enough for the likes of me to understand. However, I am wondering: is there a way to approach this problem without being "creative," as you were? Because I would have never came up with that on my own. – Thy Art is Math Oct 18 '18 at 08:57
  • Actually, the construction is does not require much 'creativity'. You have seen how measurable functions are approximated by simple functions. Instead taking functions that are constant when $n \leq f(x) <n+1$ (as you do in that construction) I am taking functions that are constant on the intervals $[n,n+1)$. – Kavi Rama Murthy Oct 18 '18 at 09:02