Continuous function and Borel sets If $f:\mathbb{R}^p \rightarrow \mathbb{R}$ is continuous and $B\subset \mathbb{R}$ is Borel set, how to show that $f^{-1}(B)$ is also Borel set?
I was trying to construct a $\sigma$-ring of sets whose preimage is Borel, but they're not necessarily all borel in that sigma ring...
What would be the main idea for the proof here?
 A: Hint: prove that $\mathcal A := \{B \in B(\Bbb R): f^{-1}(B) \in B(\Bbb R^p)\}$ is a sigma algebra containing the open sets, where $B(\Bbb R^d)$ is the Borel sigma algebra on $\Bbb R^d$.
Added:
This way, $\tau \subset \mathcal A$ (where $\tau$ is the usual topology of $\Bbb R$), "applying $\sigma$ both sides" we get $\sigma(\tau)\subset \cal A$, therefore $B(\Bbb R)\subset \cal A$. Which means that each $B \in B(\Bbb R)$ is in $\cal A$, i.e. $f^{-1}(B) \in B(\Bbb R^p)$.
A: Induction on Borel rank.
The base cases are if $B$ is $\Sigma^0_1$ (open) or $\Pi^0_1$ (closed). Well, $f$ is continuous, so by definition the preimage of an open set is open - that is, the $\Sigma^0_1$ case is handled. Similarly, since the preimage of a complement is the complement of the preimage, the $\Pi^0_1$ case is handled.
Now there are two kinds of induction step:


*

*Successor: $B$ is $\Sigma^0_{\alpha+1}$ or $\Pi^0_{\alpha+1}$.

*Limit: $B$ is $\Sigma^0_\lambda$ or $\Pi^0_\lambda$ for $\lambda$ limit.
The point is, in either case $B$ is a countable union (or countable intersection) of Borel sets of lower rank, and since unions and intersections commute with taking preimages, the preimage of $B$ is a countable union or countable intersection of (by the induction hypothesis) Borel sets, and hence Borel.
A: A Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement.
As the inverse images of open sets under a continuous function are open sets and inverse images of a countable union is the countable union of the inverse images, same for countable intersection and relative complement. We get the the inverse image of a Borel set under a continuous function is a Borel set.
