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Let $(X, \mathcal{B}(X))$ and $(Y, \mathcal{B}(Y))$ be Polish spaces. Let $f:(X, \mathcal{B}(X))\rightarrow (Y, \mathcal{B}(Y))$ be $\mathcal{B}(X)$-$\mathcal{B}(Y)$-measurable and injective. According to a theorem then for every $A\in\mathcal{B}(X)$, it follows that $f(A)\in\mathcal{B}(Y)$. Now I want to show that we can say actually more, i.e., that the $\sigma$-algebra $\mathcal{B}(f(X))=\mathcal{B}(Y)|_{f(X)}:=\{f(X)\cap B:B\in\mathcal{B}(Y)\}$ is equal to the set $\{f(A):A\in \mathcal{B}(X)\}$. I am not sure if that holds without further assumptions. Maybe, I would have to assume that additionally that the inverse function of $f:(X, \mathcal{B}(X))\rightarrow (f(X), \mathcal{B}(f(X))$ (a bijection) is measurable. But I would like to try to show it without that additional assumption.

So my first try looks as follows:

$\mathcal{B}(f(X)) \supseteq \{f(A):A\in \mathcal{B}(X)\}$: Let $A\in\mathcal{B}(X)$. So for the inverse image of the inverse function $f^{-1}$ it holds that $(f^{-1})^{-1}[A]=f(A)$. Now due to the aforementioned theorem this is in $\mathcal{B}(Y)$. BUT: I don't know how I can deduce from that to deducing $(f^{-1})^{-1}[A]=f(A)$ is even in $\mathcal{B}(f(X))$, which is what I need. Can someone help here?

$\mathcal{B}(f(X)) \subseteq \{f(A):A\in \mathcal{B}(X)\}$: So consider any $B\in\mathcal{B}(Y)$ and the set $\{f(X)\cap B\}$.

Then $f^{-1}[f(X)\cap B]=f^{-1}[f(X)]\cap f^{-1}[B]=f^{-1}[B]\in\mathcal{B}(X)$. Hence, $f(f^{-1}[B])\in\{f(A):A\in\mathcal{B}(X)\}$.

Is this proof correct? Does this proof imply that $f:(X, \mathcal{B}(X))\rightarrow (f(X), \mathcal{B}(f(X))$ is bijective and that its inverse is also measurable? And lastly, is the statement that I am proving equivalent to saying that $A\in\mathcal{B}(X)$ if and only if $f(A)\in\mathcal{B}(f(X))$?

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  • $\begingroup$ According to which theorem $A\in\mathcal{B}(X)$, implies that $f(A)\in\mathcal{B}(Y)$ if $f$ is measurable and injective ? I don't think this is true for general measurable spaces. $\endgroup$ Commented Nov 10 at 22:43
  • $\begingroup$ Consider the identity map from $\mathbb R$ with power set to $\mathbb R$ with $\{\emptyset, \mathbb R\}$. This is injective and measurable, but does not map measurable sets to measurable sets. Such things have been pointed out to you many times in the past but you keep ignoring them. I am downvoting your question. $\endgroup$ Commented Nov 10 at 22:56
  • $\begingroup$ @KaviRamaMurthy Thanks for your comments. Yes I did indeed forgot to mention that the spaces need to be Polish. Sorry for this inconvenience $\endgroup$ Commented Nov 11 at 8:19

2 Answers 2

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Saying that $X$ and $Y$ are Polish spaces makes a significant difference. As you said, you already know the following theorem:

Theorem: Let $X$ and $Y$ are Polish spaces. Let $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ be their respective Borel $\sigma$-algebras. Let $f:(X, \mathcal{B}(X))\rightarrow (Y, \mathcal{B}(Y))$ be $\mathcal{B}(X)$-$\mathcal{B}(Y)$-measurable and injective. Then, for every $A\in\mathcal{B}(X)$, it follows that $f(A)\in\mathcal{B}(Y)$.

The conclusion of the Theorem can be rephrased as $ f(\mathcal{B}(X)) = \{f(A) : A\in\mathcal{B}(X) \} \subseteq \mathcal{B}(Y) $.

Now, let us define $\mathcal{B}(f(X))=\mathcal{B}(Y)|_{f(X)}:=\{f(X)\cap B:B\in\mathcal{B}(Y)\}$. The idea is to prove that, under the hypotheses of the theorem, we have $\mathcal{B}(f(X)) = f(\mathcal{B}(X)) $, that is, $$ \{f(X)\cap B:B\in\mathcal{B}(Y)\}=\{f(A) : A\in\mathcal{B}(X) \} $$ (the proof presented by the OP needs minor adjustments)

Proof:

Step 1: Let us prove that $\{f(X)\cap B:B\in\mathcal{B}(Y)\} \supseteq \{f(A) : A\in\mathcal{B}(X) \}$.

Given any $A\in\mathcal{B}(X)$, by the theorem we know that $f(A)\in\mathcal{B}(Y)$. Since $A \subseteq X$, we have that $f(A) \subseteq f(X)$ and so, $f(A) = f(X) \cap f(A) \in \{f(X)\cap B:B\in\mathcal{B}(Y)\}$.

Step 2: Let us prove that $\{f(X)\cap B:B\in\mathcal{B}(Y)\} \subseteq \{f(A) : A\in\mathcal{B}(X) \}$.

Given any $B\in\mathcal{B}(Y)$. Note that $$ f^{-1}(f(X)\cap B)=f^{-1}(f(X))\cap f^{-1}(B)= X \cap f^{-1}(B) = f^{-1}(B) \in\mathcal{B}(X) $$ So, $f^{-1}(f(X)\cap B) \in \mathcal{B}(X)$, and $f(X)\cap B =f(f^{-1}(f(X)\cap B)) \in \{f(A) : A\in\mathcal{B}(X) \}$.

$\square$

Now, note that $\mathcal{B}(f(X))=\{f(X)\cap B:B\in\mathcal{B}(Y)\}$ is the Borel $\sigma$-algebra generated by the topology induced on $f(X)$ by the topology of $Y$. Clearly, $f:(X, \mathcal{B}(X))\rightarrow (f(X), \mathcal{B}(f(X))$ is bijective. Moreover, for any $A \in \mathcal{B}(X)$, $(f^{-1})^{-1}(A) = f(A) \in \mathcal{B}(f(X))$, so $f^{-1}: (f(X), \mathcal{B}(f(X)) \rightarrow (X, \mathcal{B}(X))$ is measurable.

Remark 1: Under the hypotheses of the Theorem, it is immediate that $A\in\mathcal{B}(X)$ if and only if $f(A)\in\mathcal{B}(Y)$. In fact, from the theorem we already know that, if $A\in\mathcal{B}(X)$, then $f(A)\in\mathcal{B}(Y)$. On the other hand, since $f$ is $\mathcal{B}(X)$-$\mathcal{B}(Y)$-measurable and injective, we have that, if $f(A)\in\mathcal{B}(Y)$, then $A = f^{-1}(f(A)) \in \mathcal{B}(X)$ (where $f^{-1}$ indicate the pre-image by $f$).

Remark 2: The OP may also be interested taking a look at https://mathoverflow.net/questions/503735/understanding-proof-that-any-injective-borel-measurable-function-between-polish

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  • $\begingroup$ thank you for your answer. I think there is one thing that may not be correct: In the last line of Step 2, you are saying that $f(X)\cap B=f(f^{-1}(f(X)\cap))$. However, wikipedia says that in general $f(f^{-1}(B))\subseteq B$ with equality if $B\subseteq f(X)$. See the table in en.wikipedia.org/wiki/Image_(mathematics)#General $\endgroup$ Commented Nov 12 at 11:04
  • $\begingroup$ But okay it says that $f(f^{-1}(B))=B\cap f(X)$ then it is actually ok $\endgroup$ Commented Nov 12 at 11:19
  • $\begingroup$ @guest1 , For any set $C \subseteq Y$ , $f(f^{-1}(C)) = C \cap f(X)$. Taking $C = f(X) \cap B$, we have $f(f^{-1}(f(X) \cap B)) = (f(X) \cap B) \cap f(X)= f(X) \cap B$. So, we have $f(X) \cap B = f(f^{-1}(f(X) \cap B))$. This is what I used in the last line of Step 2. $\endgroup$ Commented Nov 12 at 12:48
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You wrote "Let $f:(X, \mathcal{B}(X))\rightarrow (Y, \mathcal{B}(Y))$ be $\mathcal{B}(X)$-$\mathcal{B}(Y)$-measurable and injective. According to a theorem then for every $A\in\mathcal{B}(X)$, it follows that $f(A)\in\mathcal{B}(Y)$. "

However, this is false. Here is a simple counter-example: let $X$ be $[0,1]$ with the discrete topology and let $Y$ be $[0,1]$ with the usual topology. It follows immediately that $\mathcal{B}(X) = 2^{[0,1]}$ and $\mathcal{B}(Y)$ is the usual Borel $\sigma$-algebra on $[0,1]$. Let $f: [0,1] \rightarrow [0,1]$ be defined by $f(x)=x$.

Clearly, for any $B \in \mathcal{B}(Y)$, $f^{-1}(B) \in \mathcal{B}(X) = 2^{[0,1]}$. So, $f$ is $\mathcal{B}(X)$-$\mathcal{B}(Y)$-measurable. Moreover, $f$ is injetive (in fact, bijective).

However, $\mathcal{B}(Y) \varsubsetneq \mathcal{B}(X)$, so let $A \in \mathcal{B}(X) \setminus \mathcal{B}(Y)$. Clearly, $A\in\mathcal{B}(X)$ and $f(A)=A \notin\mathcal{B}(Y)$.

Moreover, $\mathcal{B}(Y)|_{f(X)} = \{[0,1]\cap B:B\in\mathcal{B}(Y)\}=\mathcal{B}(Y)$ and $ \{f(A):A\in \mathcal{B}(X)\} = \mathcal{B}(X)$. So, clearly, $\mathcal{B}(Y)|_{f(X)} \varsubsetneq \{f(A):A\in \mathcal{B}(X)\}$.

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  • $\begingroup$ Thank you for your answer! I did indeed forget to mention that the spaces involved are Polish spaces. Thank you for pointing this out $\endgroup$ Commented Nov 11 at 8:22

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