# Composition of Borel-measurable function and continuous function

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $h: \mathbb{R} \rightarrow \mathbb{R}$ be Borel measurable. Is $(h \circ g): \mathbb{R} \rightarrow \mathbb{R}$ necessarily Borel measurable?

For any $a \in \mathbb{R}$: $(h \circ g)^{-1}(a, \infty) = g^{-1}(h^{-1}(a, \infty))$, and $h^{-1}(a, \infty)$ is Borel since $h$ is Borel, so let's write $B = h^{-1}(a, \infty)$. The question is now whether or not $g^{-1}(B)$ is Borel.

I see no reason for $g^{-1}(B)$ to be Borel. I understand that continuity of $g$ implies that $g$ is also Borel measurable, but $B$ is not necessarily open. I am failing to see why $g^{-1}(B)$ would be Borel.

If the statement that I'm trying to prove is false, I would really appreciate a counter-example. Thanks in advance.

• The composition of Borel measurable functions on $\mathbb{R}$ is Borel measurable. The key fact is that any open set in $\mathbb{R}$ is a countable union of disjoint open intervals. Since preimages behave well with respect to set operations and the Borel sets are generated by open sets, the claim follows. – Ethan Alwaise Apr 27 '17 at 2:12

## 1 Answer

Note that by definition, $h^{-1}$ maps $\mathcal{B}_\mathbb{R}$ into $\mathcal{B}_\mathbb{R}$, so we need only prove that the same is true of $g^{-1}$, i.e. that all continuous functions $g:\mathbb{R}\to\mathbb{R}$ are Borel-measurable. More generally, let $g:X\to Y$ be a map of topological spaces, and I claim that $g^{-1}$ takes $\mathcal{B}_Y$ into $\mathcal{B}_X$.

Why is this true? Let $$\mathcal{M} = \{E\subseteq Y\mid g^{-1}(E)\in\mathcal{B}_X\}$$ then note that $\mathcal{M}$ is easily seen to be a $\sigma$-algebra, and we must show that $\mathcal{B}_Y\subseteq\mathcal{M}$. But to do this, we need only show that $V\in\mathcal{M}$ for all opens $V\subseteq Y$, and this follows since $g^{-1}(V)\subseteq X$ is open, by continuity, and therefore, $g^{-1}(V)\in\mathcal{B}_X$. If this is too general for you, simply replace $X$ and $Y$ with $\mathbb{R}$ and everything remains true.

Therefore, we have proved that continuous maps are Borel-measurable, and the statement is proved.