TITLE: Borel set preserved by continuous map
QUESTION [7 upvotes]: Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a continuous map. Show that if $A$ is a Borel subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is a Borel subset of $\mathbb{R}^m$.

I know that for $A$ open subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is open. Likewise for closed. A Borel subset is a countable unions/intersections of these sets, done repeatedly. From those facts, combined with $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, the statement should be intuitively true.
But to do it rigorously, I want to write $A$ as open and closed sets. But since the countable union/intersection can be done repeatedly, I don't know how I can write it that way.

REPLY [4 votes]: I suggest a proof of your theorem in more general case.
Suppose $X$ and $Y$ are measurable and topological spaces respectively.
It is easy to prove fallowing lemma:
If $f$ map $X$ to $Y$, The Collection of all sets $E\subset Y$ such that $f^{-1}(E)$ is measurable set of $X$ is a $\sigma-$$Algebra$ of $Y$.
Now consider $\sigma-$$Algebra$ of $X$ is $Borel$ $\sigma-$$Algebra$ (ie suppose $X$ was a topological space in this case). Under this assumptions any continuous function $f$ is measurable.
Now Consider $Y$ equipped with It's $Borel$ $\sigma-$$Algebra$. According the above lemma and Continuity of $f$ such a $\sigma-$$Algebra$ of $Y$ (that defined in lemma) must contains $Borel$ $\sigma-$$Algebra$ of $Y$ (because of deffinition of $Borel$ $\sigma-$$Algebra$). So inverse image of any $Borel$ $subset$ of $Y$ is a $Borel$ $subset$ of $X$.