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I am trying to understand the proof of the following Corollary 15.2 in the book Classical descriptive Set Theory by Kechris.

Corollary: Let $X$, $Y$ be standard Borel spaces and $f:X\rightarrow Y$ be Borel. If $A\subseteq X$ is Borel and $f|_A$ is injective, then $f(A)$ is Borel and $f$ is a Borel isomorphism of $A$ with $f(A)$.

So in other words, the Corollary says that $f(A)\in\mathcal{B}(Y)$ and that even $f:(A, \mathcal{B}(A))\rightarrow (f(A), \mathcal{B}(f(A)))$ is bijective, measurable and with measurable inverse.

It is a Corollary of the following Theorem:

Theorem: Let $X$, $Y$ be Polish spaces and $f:X\rightarrow Y$ be continuous. If $A\subseteq X$ is continuous and $f|_A$ is injective, then $f(A)$ is Borel.

Now I try to understand the proof of this corollary because I want to understand the following implication of this corollary: The corollary implies that if the function $f:X\rightarrow Y$ is injective then $f:(X, \mathcal{B}(X))\rightarrow (f(X), \mathcal{B}(f(X)))$ is bijective, measurable and with measurable inverse. And in particular it implies that for every $A\in\mathcal{B}(X)$ it follows that (because if $f$ is injective then every $f|_A$ is injective) $f(A)\in\mathcal{B}(f(X))$. And this is what I don't get because the first statement of the corollary actually only gives us $f(A)\in\mathcal{B}(Y)$.

So to trying to understand this I wanted to understand the proof, which is very short and simply says:

First we can clearly assume that $X$ and $Y$ are Polish. Then we can apply the (above) Theorem to the projection of $X\times Y$ onto $Y$ and the set $(A\times Y)\cap \operatorname{graph}(f)$.

So let this projection be $\operatorname{pr}:X\times Y\rightarrow Y, (x, y)\mapsto y$. Then applying the above theorem to the set $(A\times Y)\cap \text{graph}(f)=\text{graph}(f|_A)$ gives us $\operatorname{pr}\rvert_{\text{graph}(f|_A)}(\text{graph}(f|_A))=f(A)\in\mathcal{B}(Y)$. How does this prove the claim?

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    $\begingroup$ In the last paragraph, $(A\times Y)\cap \text{graph}(f)$ does not equal $A\times f(X)$. Try drawing a picture. Its projection does indeed equal $f(A)$, not $f(X)$. $\endgroup$ Commented Nov 11 at 14:24
  • $\begingroup$ @NateEldredge Thanks a lot for your correction! I edited the question. So is then $(A\times Y)\cap \text{graph}(f)=\text{graph}(f|_A)$? $\endgroup$ Commented Nov 11 at 14:35
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    $\begingroup$ Yes, that's correct. $\endgroup$ Commented Nov 11 at 14:37
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    $\begingroup$ There are still some steps after this to complete the proof that $f$ is a Borel isomorphism, but they should be straightforward verification, really just chasing definitions. Where are you getting stuck? $\endgroup$ Commented Nov 11 at 14:54
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    $\begingroup$ Keep in mind the following fact: if $Z$ is a standard Borel space and $W \subset Z$ is a Borel set, then a set $C \subset W$ is Borel as a subset of $W$ iff it is Borel as a subset of $Z$. If you are thinking of standard Borel spaces abstractly, then this is a definition; if you think of them as the Borel $\sigma$-algebras generated by topologies, with $W$ carrying the subspace topology, then this is an easy exercise (or possibly found elsewhere in Kechris). $\endgroup$ Commented Nov 11 at 15:05

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As mentioned in the comments, I think the key fact you're missing is:

Fact. If $C$ is a Borel subset of $X$, then a subset $D \subseteq C$ is Borel in $C$ if and only if it is Borel in $X$.

Once $A$ and $f(A)$ are both Borel sets, then we no longer have to identify the ambient space when talking about whether a subset of either one is Borel, and it is simple to conclude that $f|_A$ is a Borel isomorphism. If we do want to be more precise about it, it's just chasing sets around:

  • To show $f|_A : A \to f(A)$ is Borel: If $B$ is a Borel subset of $f(A)$, it is a Borel subset of $Y$, so $f|_A^{-1}(B) = f^{-1}(B) \cap A$ which is Borel in $X$ (as $f : X \to Y$ is Borel), hence Borel in $A$.

  • To show $f|_A^{-1} : f(A) \to A$ is Borel: Let $B$ be a Borel subset of $A$, hence a Borel subset of $X$. By the same argument already given by Kechris, $f(B)$ is Borel in $Y$, hence Borel in $f(A)$. Since $f(B) = (f|_A^{-1})^{-1}(B)$, we are done.

The Fact can be interpreted in a couple different ways. In the category of measurable spaces, where "Borel set" just means "member of the designated $\sigma$-algebra", it is almost true by definition. Indeed, when $(X, \mathcal{S})$ is a measurable space and $C \subseteq X$, then when we talk about $C$ as a measurable space in its own right, the convention is that we equip it with the relative $\sigma$-algebra $\mathcal{S}|C = \{B \cap C : B \in \mathcal{S}\}$. See 10.B of Kechris. Now when we moreover have $C \in \mathcal{S}$, then it is easy to see that $\mathcal{S}|C = \{B \in \mathcal{S} : B \subseteq C\}$ as the fact asserts.

In the category of topological spaces, where "Borel set" means "member of the Borel $\sigma$-algebra generated by the open sets", and where $C$ is equipped with the subspace topology, it's true as well. Kechris says in 10.B: "Notice that if $\mathcal{S} = \sigma(\mathcal{E})$, then $\mathcal{S}|C = \sigma(\mathcal{E}|C)$." If we take $\mathcal{E}$ to be the open sets of $X$, then $\mathcal{E}|C$ is the open sets in the subspace topology of $C$, and so we have $\mathcal{S} = \mathcal{B}(X)$ the Borel $\sigma$-algebra of $X$, and $\mathcal{S}|C = \mathcal{B}(C)$ the Borel $\sigma$-algebra of the subspace topology on $C$. We can now apply the previous paragraph about measurable spaces.

To see Kechris's "Notice that": to show $ \sigma(\mathcal{E}|C) \subseteq \mathcal{S}|C$, verify that $\mathcal{S}|C$ is a $\sigma$-algebra containing $\mathcal{E}|C$. For the reverse direction, consider the collection $\mathcal{F} = \{ F \in \mathcal{S} : F \cap C \in \sigma(\mathcal{E}|C)\}$, and verify that $\mathcal{F}$ is a $\sigma$-algebra containing $\mathcal{E}$. It follows that $\mathcal{S} \subseteq \mathcal{F}$, which is to say that for every $F \in \mathcal{S}$ we have $F \cap C \in \sigma(\mathcal{E}|C)$; that is, $\mathcal{S}|C \subseteq \sigma(\mathcal{E}|C)$.

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  • $\begingroup$ Thanks for your very helpful answer! Just out of curiosity, is the fact that you are mentioning well known in general? I.e., can it be found in many textbooks? Because I havent come across it yet ... Also couldnt find it in Kechris' book so far $\endgroup$ Commented Nov 13 at 8:19
  • $\begingroup$ @guest1: I think it is certainly well known, and gets used implicitly all the time, but I can't recall having seen it explicitly stated in a textbook. $\endgroup$ Commented Nov 13 at 16:21
  • $\begingroup$ It is not stated implicitely in Kechris's book, but follows easily from stuff in the book. "Note also that if $Y$ is a subspace of $X$, then $(Y,\mathcal B(Y))$ is a subspace of $(X,\mathcal B(X))$ (i.e. $\mathcal B(Y)=\mathcal B(X)\mid Y$)" is stated at the beginning of 11.A in Kechris's book. Now if $Y$ is Borel in $X$ and $A$ is Borel in $Y$, then $A=Y\cap B$ for some $B\in\mathcal B(X)$, by Kechris's remark, so that $A$ is also Borel in $X$, being the intersection of two Borel sets ($Y$ and $B$). $\endgroup$ Commented Nov 14 at 1:03

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