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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.

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  • $\begingroup$ 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 $[0,1] \times [\inf f, \, \sup f].$ $\endgroup$ Apr 19 at 17:03

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Beerus
    Apr 19 at 17:09
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    $\begingroup$ @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. $\endgroup$ Apr 19 at 17:19
  • $\begingroup$ Thanks a lot! Now I get your point. $\endgroup$
    – Beerus
    Apr 19 at 17:20
  • $\begingroup$ Yes, thank you very much! $\endgroup$
    – Beerus
    May 6 at 22:38
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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.

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  • $\begingroup$ 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? $\endgroup$
    – Beerus
    Apr 19 at 17:03
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    $\begingroup$ @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. $\endgroup$ Apr 19 at 17:05
  • $\begingroup$ Oh, I see! I misunderstand that. Thanks again! $\endgroup$
    – Beerus
    Apr 19 at 17:08

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