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

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up vote 11 down vote accepted

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.

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Good answer. Another way to put things (but with less details) is that $f$ is a homeomorphism (because as you say it is an open map), i.e., its inverse is continuous, hence its inverse is Borel measurable, hence $f$ maps Borels to Borels. –  Jonas Meyer Oct 7 '10 at 15:40
    
But $f$ is not surjective? So how can we talk about an "inverse"? –  Jonas Teuwen Oct 7 '10 at 15:42
    
The inverse is defined on its image, which is an open interval. –  Jonas Meyer Oct 7 '10 at 15:43
    
Ah, right, I didn't think about that long enough, I thought you would get problems with the domain... Nice. –  Jonas Teuwen Oct 7 '10 at 15:49
    
You meant saying that continuous and injective. Otherwise you can cook some example by "switching" two intervals off the identity map. –  Asaf Karagila Oct 7 '10 at 17:03
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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.

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