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Is it always possible to construct a measure $ \mu $ on a Hausdorff space Y such that the $ \mu $-measurable sets are exactly the Borel sets of Y?

By Theorem in 2.2.13 of Federer's book this question in answered negatively if we can answer positively to the following one:

Let X be a complete and separable metric space. Is there a continuous map $ f: X \rightarrow Y $ and a Borel set B of X such that $ f(B) $ is NOT a Borel set of Y?

  • Isn't there a problem with the null sets? Subsets of null sets should also be null sets and especially measurable but there could be non-measureable subsets of Borel null sets? – Dirk Mar 27 '14 at 15:23
  • I think the phrasing is slightly off. Do you mean to ask if it's possible to construct an outer measure $\mu$ such that the $\mu$-measurable sets are the Borel sets? – Ink Mar 27 '14 at 15:33
  • Yes in 'the language of outer measures' I am asking exactly this. For me measure means outer measure. –  Mar 27 '14 at 15:35

2 Answers2

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See Is projection of a measurable subset in product $\sigma$-algebra onto a component space measurable?.

I give partially answer on that question: There are complete and separable metric space(equivalently, Polish space) $X$, a Hausdorf space $Y$,a continuous map $f : X \to Y$ and a Borel set $B$ of $X$ such that $f(B)$ is not Borel measurable in $Y$.

Indeed,let $A$ be an analytic but non-Borel subset of a Polish space $X_1$. That means that there is a Polish space $Y_1$ and a Borel set $B_1 \subseteq X_1 \times Y_1$ such that $A$ is projection of $B_1$., that is $A=\{x \in X_1|(\exists y)(x,y) \in B_1\}$.

Now put $X =X_1 \times Y_1$, $Y=X_1$, $B=B_1$, and $f(x,y)=x$ for $(x,y)\in X$.

Then under $f$ an image of each Borel subset in $X$ is not Borel in $Y$.

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George
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  • In my sense, my answer must be correct. But there is a certain confusion. My answer, supported on Theorem 2.2.13 of Federer's book, implies that we can not define a Borel measure on $Y=X_1$ which is a Polish space. I know that there always exists a Borel probability measure in a Polish space. Where am I wrong ? – George Mar 28 '14 at 14:00
  • Federer's definition of a measure $\mu$ assumes it's completeness. Each analytic set(for example, $f(B)$ in my example) is universally measurable. Hence, it is $\mu$-measurable. Theorem 2.2.13 asserts that an arbitrary continuous map $f:X \to Y$ sends each Borel subset in to measurable (with respect to completion of an arbitrary Borel measure $\mu$) subset of $Y$. – George Mar 28 '14 at 14:46
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    Thank you for your answer. I am not so familiar with some of the concepts that you have introduced. By the way I have just written an answer to own my question. It is based on the theory of Federer's book in section 2.2 –  Apr 03 '14 at 16:53
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Under reasonably hypothesis on the space $ Y $ the answer to my question is NO. In fact referring to the theory of Federer's book in section 2.2 we can consider the following two theorems:

1) Let $ X $ be a complete separable metric space, Z Hausdorff space and $ f: X \rightarrow Z $ be a continuous map. Let $ \mu $ be a measure on $ X $ such that every closed set is $ \mu $-measurable. Then if $ B \subset X $ is a Borel set then $ f(B) $ is $ \mu $-measurable.

2) On every complete metric space without isolated points there exists a Suslin set that is not Borel.

Using 1) and 2) we can easily conclude:

3) Let $ X $ be a complete separable metric space without isolated points. Then every measure $ \mu $ on X such that every Borel set is $ \mu $-measurable admits a $ \mu $-measurable set that is not Borel.

Proof of 3): By contradiction. Let $ \mu $ be a measure that violates the statement 3). By 2) take $ S \subset X $ Suslin set that is not Borel. Since $ S = p(C) $ where p: $ \mathscr{N} \times X \rightarrow X $ is the projection and $ C $ is a closed set, we conclude by 1) that S is $ \mu $-measurable. Then we get the contradiction.