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?


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 $\Bbb R^d$.


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. $f^{-1}(B) \in B(\Bbb R^p)$.

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  • $\begingroup$ @NoahSchweber thanks! $\endgroup$ – user384138 Nov 21 '16 at 21:44
  • $\begingroup$ I just don't quite understand what would this lead us to. Say, I managed to prove that. We can take a countable union, we can take a complement show that under $f^{-1}$ they will be in this sigma algebra. But what's next? $\endgroup$ – BoBoB Nov 21 '16 at 22:13
  • $\begingroup$ @Ilia let's call this set $A$. If $\tau$ denotes the usual topology, we have $\tau \subset A \therefore \sigma(\tau) \subset A$ i.e. $B(\Bbb R)\subset A$. On the other hand clearly $A \subset B(\Bbb R)$. $\endgroup$ – user384138 Nov 22 '16 at 5:13
  • $\begingroup$ @OpenBall I think I got it now. I proved that the collection $\mathcal{A}$ of all open sets whose inverse $f^{-1}$ is a Borel set is a $\sigma$-algebra (I think it would suffice to prove that it's a sigma ring). Borel $\sigma$-ring is the smallest $\sigma$-ring (or $\sigma$-algebra) that is contained in $\mathcal{A}$. Hence if we take any Borel set, then it's inside our set $\mathcal{A}$ and has a Borel set as its inverse. $\endgroup$ – BoBoB Nov 22 '16 at 17:35
  • $\begingroup$ Not open, sorry, just all sets for which preimage is Borel $\endgroup$ – BoBoB Nov 22 '16 at 19:09

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.

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  • $\begingroup$ Note that it's crucial that we're working with preimage, not image. In general, the intersection of the images contains the image of the intersection (take $f:\mathbb{R}\rightarrow\mathbb{R}$ constant and let $A, B$ be disjoint subsets of $\mathbb{R}$). $\endgroup$ – Noah Schweber Nov 21 '16 at 21:38
  • $\begingroup$ I'm curious, why the downvote? $\endgroup$ – Noah Schweber Nov 22 '16 at 4:16

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.

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  • $\begingroup$ What does this have to do with the question? $\endgroup$ – Noah Schweber Nov 21 '16 at 21:38
  • $\begingroup$ @NoahSchweber You were too quick to undervote. Unfair I would say! I was fighting with the poor editing capabilities of my tablet. $\endgroup$ – mathcounterexamples.net Nov 21 '16 at 21:50

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