Borel sets and measurability 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?
 A: 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 $Y$. 
A: 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.
