Is every borel set also an open set? I'm trying to prove this proposition:
Let $(X,M)$ be a measurable space and $(Y,\tau)$ be a topological space. If $f:X\to Y$ is measurable function then every borel set $E$ in $Y$ satisfies $f^{-1}(E)\in M$ 

Since $f$ is measurable, any $E$ in $\tau$ satisfies assertion above, so it is sufficent to prove that 
"A borel set in $Y$ is also an open set in $Y$"
But by definition of borel $\sigma$-algebra, it is the converse that is valid and soon recognized this approach is wrong idea. I couldn't proceed anymore.
What am i missing?
 A: Certainly not all Borel sets are open. I think the question is really asking you to prove the following claim:
Claim:  Let $(X, M)$ be a measurable space. Let $Y$ be a nonempty set and let $\tau$ be a collection of subsets of $Y$. If $f:X\rightarrow Y$ is a function that satisfies
$$\left(f^{-1}(E) \in M \quad \forall E \in \tau\right)$$
then $f$ also satisfies
$$\left(f^{-1}(E) \in M \quad \forall E \in \sigma(\tau)\right) \quad \Box$$


*

*In your case, $\tau$ is the collection of open sets in $Y$ and $\sigma(\tau)$ is the Borel sigma algebra generated by these open sets.


*The Claim is not trivial, but it can be proven in just a few lines by defining $A$ as the collection of all sets $E\subseteq Y$ such that $f^{-1}(E) \in M$. It follows that $\tau \subseteq A$, so $\sigma(\tau) \subseteq \sigma(A)$, and it remains to show $A$ is a sigma algebra (so $\sigma(A)=A$).

On ambiguity of definitions in the posted question:
As drhab comments, the standard definition of a "measurable function" $f:X\rightarrow Y$, where $(X,M)$ and $(Y, V)$ are two measurable spaces, is a function that satisfies
$$ f^{-1}(E) \in M \quad \forall E \in V$$
This requires a measurable space to be specified both for the domain and for the target set. If your posted question does not provide that, then we cannot go anywhere. On the other hand, if we are supposed to assume the two measurable spaces are $(X, M)$ and $(Y, \sigma(\tau))$, where $\sigma(\tau)$ is the Borel sigma algebra, then there is nothing to prove since the statement to be proven in the posted question is the very definition of a measurable function.
I think the ambiguity boils down to "what is your definition of a measurable function"? This is the same question asked in the first comment by Keen-amateur. It is possible that, in these cases when a topological space is specified instead of a full measurable space, you are using some alternative definition such as
"A measurable function $f:X\rightarrow Y$, where $(X, M)$ is a measurable space and $(Y,\tau)$ is a topological space, is a function that satisfies
$f^{-1}(E) \in M$ for all $E \in \tau$."
This is not a horrible definition of "measurable function" because the above claim immediately implies it is consistent with the standard definition. It is also the only definition I can think of that does not make your posted question either ambiguous or trivial.
