# A function with countable discontinuities is Borel measurable.

Let $f:[a,b] \to \mathbb{R}$ be bounded with countable discontinuities. Show that $f$ is Borel-measurable.

One solution uses the fact that A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere.

But is it possible to solve this problem similarly to finite discontinuity cases? In other words, $\{f>t\}$ can be decomposed something like $\bigcup_n \{f_n>t\}\cup\{\text{discontinuous points}\}$? If the discontinuites were finite, then just I can order them so it was possible. But I don't know similar method is possible in countable cases. (It will be good then we don't need the condition $f$ is bounded.)

• You wrote: One solution uses the fact that A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere. Maybe I'm missing something, but it seems to me that you're assuming every Riemann integrable function is Borel-measurable, which is somewhat like saying every real number is rational (since there are $c$ many Borel-measurable functions and $2^c$ many Riemann integrable functions). Apr 22 '13 at 17:14
• @DaveL.Renfro The solution start in there. $f$ is then riemann integrable, so I can define the partition to be increasing set of discontinuous points $P_k$ and lower and upper (Borel-)simple functions $g_{P_k},G_{P_k}$ and the limits of them $g,G$. And use the fact that for $x\in \cup P_k$, $f$ is continuous at $x$ iff $g(x)=G(x)$.
– Gobi
Apr 22 '13 at 23:19
• Actually I didn't proved the last statement, so there can be some errors.
– Gobi
Apr 22 '13 at 23:21

Let $f:[a,b]\rightarrow\mathbb{R}$ be a function with countable discontinuities, and let $c\in\mathbb{R}$. We must prove that $f^{-1}(-\infty,c)$ is measurable.

Let $A=\text{int}f^{-1}(-\infty,c)$ (the interior is taken in $[a,b]$) and $D=f^{-1}(-\infty,c)\setminus A$. Since $A$ is open, it is measurable, and since $f^{-1}(-\infty,c)=D\cup A$, it's only left for us to prove that $D$ is measurable. But notice that if $x\in D$, then $x$ is not an interior point of $f^{-1}(-\infty,c)$, that is, for every $r>0$, there exists $y$ such that $|y-x|<r$ but $f(y)\geq c$. Since $f(x)<c$, that means that $x$ is a discontinuity point of $f$. Therefore, $D$ is at most countable, hence measurable.

That argument can be used for any $f:E\subseteq\mathbb{R}^n\rightarrow\mathbb{R}$ which is Lebesgue-measurable, where $E$ is a Lebesgue-measurable subset of $\mathbb{R}^n$, and for which the set of discontinuities of $f$ has zero Lebesgue-measure.

(I'm using the following definition: a function $f:E\rightarrow\mathbb{R}$ is (Borel-)measurable iff for every $a\in\mathbb{R}$, $f^{-1}(-\infty,a)$ is a (Borel-)measurable set. It can be easily shown that any function which satisfies that condition also satisfies the following: for any Borel-measurable subset $A$ of $\mathbb{R}$, $f^{-1}(A)$ is a (Borel-)measurable subset of $E$)

• This is so great! Thank you.
– Gobi
Apr 22 '13 at 13:41
• I have a question. Your answer can be generallized to for any $f:E \to \mathbb{R}$ with the set of discontinuities measurable, then $f$ is measurable. But suppose that $E$ is not measurable. Your proof does not use any condition about $E$. Then $E=\cup \{ f<n \}$ should be measurable. What is the problem in my reasoning?
– Gobi
Apr 22 '13 at 15:33
• Or maybe the conclusion is that if $f$ is satisfying such conditions, then $E$ should be measurable?
– Gobi
Apr 22 '13 at 15:38
• The generalization should be that if all subsets of {discontinuies} are measurable.
– Gobi
Apr 22 '13 at 15:50
• You're right. Thanks. Oct 22 '13 at 0:27