# Borel Functions That Are Continuous

I am self-studying measure theory. I know that not all Borel functions are continuous. But I would like to explore certain conditions that make a Borel function continuous.

In this post, we saw that boundedness does not make a Borel function continuous. So I was wondering what if the domain of a Borel function $$f$$ is a compact subset of $$\mathbb{R}^d$$, or (to make it more restrictive) a closed interval $$[a,b]$$ ($$a,b\in\mathbb{R}$$)?

Basically, I was wondering if any of the following four statements are true:

Statement 1$$\quad$$ Every Borel function on a compact subset of $$\mathbb{R}^d$$ is continuous.

Statement 2$$\quad$$ Every bounded Borel function on a compact subset of $$\mathbb{R}^d$$ is continuous.

Statement 3$$\quad$$ Every Borel function on a closed interval $$[a,b]$$ is continuous.

Statement 4$$\quad$$ Every bounded Borel function on a closed interval $$[a,b]$$ is continuous.

Evidently, if Statement 1 were correct, then the rest are all correct. If Statement 3 were correct, then Statement 4 would be correct.

I tried to prove them, but I got stuck. For example, when proving Statement 4, I couldn't incorporate the definition of a bounded function that is measurable with respect to $$\mathscr{B}(\mathbb{R})$$ with the $$\epsilon-\delta$$ definition of conitnuity. Could someone please help me out? Thank you very much!

Here is the definition of a Borel function that I learned:

Definition$$\quad$$ Let $$(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$$ be a measurable space, and let $$A$$ be a subset of $$\mathbb{R}^d$$ that belongs to $$\mathscr{B}(\mathbb{R}^d)$$. A function $$f:A\to[-\infty,+\infty]$$ is measurable with respect to $$\mathscr{B}(\mathbb{R}^d)$$ if it satisfies for each real number $$t$$ the set $$\{x\in A:f(x) < t\}$$ belongs to $$\mathscr{B}(\mathbb{R}^d)$$. In this case, we say the function $$f$$ is a Borel function.

• There are Borel level $2$ functions $f:[0,1] \rightarrow \mathbb R$ (about as low as you can get in the Borel/Baire hierarchy) whose graphs are dense subsets of $[0,1] \times \mathbb R$ (about as far from being continuous as you can get), and bounded Borel level $2$ functions $f:[0,1] \rightarrow \mathbb R$ whose graphs are dense subsets of $[0,1] \times [\inf f, \, \sup f].$ Apr 19 at 17:03

You can't prove any of your four statements because they are all false. For a counterexample to each, take the indicator function of any proper Borel subset of your domain.

The Borel functions that are continuous are precisely those that satisfy the $$\epsilon-\delta$$ definition of continuity.

• Thank you very much! This is neat: The Borel functions that are continuous are precisely those that satisfy the $\mathbf{\epsilon-\delta}$ definition of continuity. Is there a proof of it? Apr 19 at 17:09
• @Beerus You misunderstand. All continuous functions are Borel. If you want to look at the Borel functions which are continuous, you just need to use the definition of continuity. Apr 19 at 17:19
• Thanks a lot! Now I get your point. Apr 19 at 17:20
• Yes, thank you very much! May 6 at 22:38

Define $$f:[0,1]\to\mathbb{R}$$ given by $$f(x)=0$$ if $$x=0$$ and $$f(x)=1$$ otherwise. Then $$f$$ is Borel and bounded but not continuous, so $$4$$ is false and thus $$1,2,3$$ are too.

• Thanks a lot! One more quick question: The book I am reading (Measure Theory by Donald Cohn) wrote the following in Example 2.3.7: "Every bounded Borel function, and hence every continuous, function, on $[a,b]$ is Lebesgue integrable." How should I understand this "hence" part? Apr 19 at 17:03
• @Beerus If $f$ is continuous in a bounded and closed interval $[a,b]$ then $f$ is bounded, and every continuous function is Borel. Thus, the fact that all bounded Borel functions on $[a,b]$ are Lebesgue integrable implies all such continuous functions are integrable too. Apr 19 at 17:05
• Oh, I see! I misunderstand that. Thanks again! Apr 19 at 17:08