Prove that there exists $B$ such that $\mu(f(B))>0$. Let $\mu$ be the Lebesgue measure. Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is measurable. $f(B)$ is measurable for every Borel set $B\subset \mathbb{R}$, and $\{y:f^{-1}(y) \text { is infinite}\}$ has measure $0$. Suppose there is some $A\subset\mathbb{R}$ such that $\mu(A)=0$ and $\mu (f(A))>0$. Prove there is a closed set $F$ such that $\mu(F)=0$ and $\mu(f(F))>0$. 
My try: 
$\mu(f(A))>0$ implies that there exists a subset $C$ of $A$ such that $f(C)$ is not measurable. I try to find a closed set $B$ such that $C\subset B$ and $\mu(B)=0$. Then we can get $\mu(f(B))>0$. However, I don’t know how to find such $B$.
 A: Partial answer:
Consider $C(A)=\mu(f(A))$. Three things can be checked:
1) $C(\emptyset)=0$.
2) If $A\subseteq B$, then $C(A)\leq C(B)$.
3) If $E_{1}\subseteq E_{2}\subseteq\cdots$, then $C\left(\displaystyle\bigcup_{n}E_{n}\right)=\lim_{n\rightarrow\infty}C(E_{n})$.
We wish to have the following extra condition:
4) If $K_{1}\supseteq K_{2}\supseteq\cdots$ are compact, then $C\left(\displaystyle\bigcap_{n}K_{n}\right)=\lim_{n\rightarrow\infty}C(K_{n})$.
If the fourth condition is also met, then Choquet Theorem shows that $C$ is capacitable in the sense that each Souslin set, in particularly, Borel set, is inner regular.
Now we can find a Borel set $B$ such that $A\subseteq B$ such that $\mu(B)=0$, then we also have $\mu(f(B))\geq\mu(f(A))>0$.
As $C$ is capacitable for $B$, so a compact set $K\subseteq B$ is such that $C(K)>(1/2)C(B)>0$, then of course $\mu(K)=0$ and $\mu(f(K))>0$.
The point is that, given any compact set, is there a way to show that $\mu(f(K))<\infty$? If so, then the fourth condition is met. Note that I haven't used the assumption that $\{y: f^{-1}(\{y\})~\text{is an infinite set}\}$ is measure zero. I think this may be crucial for the fourth condition to be met.
If the above proposal fails, naturally, one may ask, if this question holds for the measure $\mu$ such that $\mu(f(K))=\infty$ for some compact set $K$?
Edit:
Let $\mathbb{R}=\displaystyle\bigcup_{n=1}^{\infty}X_{n}$, where $(X_{n})$ is disjoint with positive measure. Consider 
\begin{align*}
\nu(S)=\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}\dfrac{\mu(S\cap X_{n})}{\mu(X_{n})}.
\end{align*}
Then $\nu$ is a finite measure and $\nu<<\mu$ together with $\mu<<\nu$.
We construct the capacity with respect to $\nu$, not $\mu$.
Now for every $y\in\displaystyle\bigcap_{n}f(K_{n})-N$, then $y=f(x_{n})$ for $x_{n}\in K_{n}$. The sequence $(x_{n})$ cannot have any subsequence with consists of mutually distinct elements, for then it will violate $y\notin N$ and hence $(x_{n})$ must be eventually constant. So $\displaystyle\bigcap_{n}f(K_{n})-f\left(\displaystyle\bigcap_{n}K_{n}\right)\subseteq N$.
With such, we have 
\begin{align*}
C\left(\bigcap_{n}K_{n}\right)&=\nu\left(f\left(\bigcap_{n}K_{n}\right)\right)\\
&=\nu\left(\displaystyle\bigcap_{n}f(K_{n})\right)-\nu\left(\displaystyle\bigcap_{n}f(K_{n})-f\left(\displaystyle\bigcap_{n}K_{n}\right)\right)\\
&=\nu\left(\displaystyle\bigcap_{n}f(K_{n})\right)\\
&=\lim_{n\rightarrow\infty}\nu(f(K_{n}))\\
&=\lim_{n\rightarrow\infty}C(K_{n}).
\end{align*}
Therefore the fourth condition is met. We can now apply Choquet Theorem.
Note that we at first have only that 
\begin{align*}
\nu(f(K))=\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}\dfrac{\mu(f(K)\cap X_{n})}{\mu(X_{n})}>0.
\end{align*}
But then we may choose an $N$ such that 
\begin{align*}
\sum_{n=1}^{N}\dfrac{1}{2^{n}}\dfrac{\mu(f(K)\cap X_{n})}{\mu(X_{n})}>0.
\end{align*}
We see that 
\begin{align*}
\mu(f(K))&\geq\mu\left(f(K)\cap\left(\bigcup_{n=1}^{N}X_{n}\right)\right)\\
&=\mu\left(\bigcup_{n=1}^{N}(f(K)\cap X_{n})\right)\\
&=\sum_{n=1}^{N}\mu(f(K)\cap X_{n})\\
&=\sum_{n=1}^{N}\dfrac{1}{2^{n}}\dfrac{\mu(f(K)\cap X_{n})}{\mu(X_{n})}2^{n}\mu(X_{n})\\
&\geq\gamma\sum_{n=1}^{N}\dfrac{1}{2^{n}}\dfrac{\mu(f(K)\cap X_{n})}{\mu(X_{n})}\\
&=\gamma\nu(f(K))\\
&>0,
\end{align*}
where $\gamma=\min\{2^{n}\mu(X_{n})\}_{n=1}^{N}>0$.
