Image of separable complete metric space is measurable I'm wondering about the following problem;
suppose that $ \mu $ is a finite Borel measure on a separable metric space $ X $. Now, suppose we have a continuous map $ f: F \rightarrow X $ from a separable, complete metric space $ F $ to $ X $. My aim is to prove that $ f(F) $ is $ \mu $-measurable.
I was trying to work around the following theorem:
Suppose we have a complete, separable metric space $ X $ with Borel measure $ \mu $. Then for every subset $ E \subset X$ of finite measure and $ \varepsilon > 0 $ there exists a compact $ K \subset E $ such that $ \mu(E \setminus K)  < \varepsilon$.
I'm not sure that this can be applied directly, but somehow we might be able to use the construction of $ K $. We pick closed $ F \subset E $ with $ \mu(E \setminus F) < \varepsilon/2 $ and inductively construct a sequence of closed $ F_i $ such that $ \mu(F_i \setminus F_{i+1}) \leq \varepsilon/2^i $ and every $ F_i $ can be covered by a finite number of balls of diameter $ \leq 1/i $. The intersection then is the $ K $ we're looking for.
I'm not quite sure how to run this sort of argument, using also continuity of $ f $. I would appreciate some hints
Edit I made an error; the space $ X $ is not necessarily complete, but it is separable
 A: $f[F]$ is a so-called analytic set, which is universally measurable.
But there might be a more elementary approach.
A: I've added a few more details to Henno's excellent hint. 
The identity map $X\overset{\text{id}}{\to} X$ lifts to 
an isometry $X\overset{i}{\to} \bar X$, from $X$ into the completion $\bar X$ of $X$. 
If $C$ is a countable dense subset of $X$, then since $i(X)$ is dense in $\bar X$,  the countable set $i(C)$ is dense in $\bar X$. That is, $\bar X$ is a Polish space. 
Now the composition $F \overset{f}{\to}X\overset{i}{\to}\bar X$ is a continuous
 mapping of $F$ into $\bar X$ so that the image $i(f(F))$ is universally measurable in $\bar X$ (as mentioned in Henno's answer).
Since $i$ is one to one, we have $i^{-1}(i(f(F)))=f(F).$ Also since $i$ is continuous, it is also a Borel measurable map $(X,{\cal{B}})\overset{i}{\to} (\bar X,\bar{\cal{B}})$ and therefore 
$(X,{\cal{B}}^u)\overset{i}{\to} (\bar X,\bar{\cal{B}}^u)$ is also measurable
 for the universal completions. It follows that $f(F)\in{\cal B}^u,$ that is, 
$f(F)$ is universally measurable in $X.$
