Does a continuous and 1-1 function map Borel sets to Borel sets?

Question

Suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function which is 1-1, then does $f$ map Borel sets onto Borel sets?

Answer 1

The continuous image of a Borel set in a Polish space is analytic. So it is Lebesgue measurable. If the complement of that analytic set is analytic then it is Borel (a theorem by Suslin). Now we know that for injective functions we have $f(A \setminus B) = f(A) \setminus f(B)$ so the complement is analytic too, hence the continuous one-to-one image of a Borel set is Borel in a Polish space.

Maybe this is unnecessarly cluttered but it is the thing that pops up in my mind.

Edit:

Note that $f$ maps open sets to open sets. This is because injective functions $f:\mathbb{R} \to \mathbb{R}$ are strictly monotonic, so it maps open intervals to open intervals.

Define $D := \{Q \in P(\mathbb{R}) : f(Q) \textrm{ is Borel}\}$.

Now $\mathbb{R}$ is in $D$, it is closed under complements and unions (because it is injective), thus $D$ is a $\sigma$-algebra. So let $B$ be the Borel $\sigma$-algebra and $O$ the open sets of $\mathbb{R}$. Then $B \subset D$ if $O \subset D$. So we get that for all Borel sets the image is Borel.

Answer 2

Note, you also need the fact that $f(\mathbb{R})$ is borel. This fortunately is so, as $\mathbb{R}$ is $\sigma$-compact, and the continuous image of a compact set is compact, thus closed (since we are working in hausdorff spaces), thus borel.