Prove that a sequence of measures is uniformly controled by a finite measure Suppose $\mu$ is a finite measure on a $\sigma$-algebra $\mathcal{F}$ of a set $X$, $\{\nu_n\}_{n\in\mathbb{N}}$ is a sequence of finite measures on $\mathcal{F}$ s.t. $\nu_n$ is absolutely comtinuous with respect to $\mu$ for all $n\in\mathbb{N}$ and $\lim_{n\to\infty}\nu_n(E)=:\nu(E)<\infty$ for all $E\in\mathcal{F}$. Show that
(1)
$$\lim_{\mu(E)\to 0}\sup_{n\in\mathbb{N}}\nu_n(E)=0$$
This means it suffices to prove: $\forall\epsilon>0$, $\exists \delta>0, N\in\mathbb{N}$ s.t.
$$\nu_n(E)<\epsilon,\quad\forall n\geq N, \mu(E)<\delta$$
(2) $\nu$ is a finite measure.
Once we have (1), we can get (2) quite easily. But I am having difficulty on (1) and can't figure out a way to approach.
 A: Let $E_1\Delta E_2$ denotes the symmetric difference of two sets. Define the following equivalence relation:
$$E_1\sim E_2\quad\Longleftrightarrow \quad \mu(E_1\Delta E_2)=0$$
Define
$$d(E_1,E_2)=\mu(E_1\Delta E_2)$$
then under this equivalence relation $(\mathcal{F},d)$ is a complete metric space.
As $\nu_n$ is absolutely continuous with respect to $\mu$, and each $\nu_n$ is finite, thus $\forall\epsilon>0$, $\exists\delta_n>0$ s.t.
$$\nu_n(E)<\epsilon\text{ whenever }\mu(E)<\delta$$
So if $\mu(E_1\Delta E_2)<\delta_n$, we have
$$|\nu_n(E_1)-\nu_n(E_2)|\leq \nu_n(E_1\Delta E_2)<\epsilon$$
This means each $\nu_n$ is a continuous function on $(\mathcal{F},d)$.
On the other hand, as $\lim_{n\to\infty}\nu_n(E)=\nu_E$ for all $E\in\mathcal{F}$, then $\forall \epsilon>0, E\in\mathcal{F}$, $\exists N$ s.t.
$$|\nu_n(E)-\nu_m(E)|<\epsilon,\quad\forall n,m>N$$
Set
$$\begin{aligned}F_N(\epsilon)&=\{E\in\mathcal{F}: |\nu_n(E)-\nu_m(E)|<\epsilon,\forall n,m\geq N\}\\
&=\bigcap_{n=N}^\infty\bigcap_{m=N}^\infty\{E\in\mathcal{F}: |\nu_n(E)-\nu_m(E)|<\epsilon\}
\end{aligned}$$
It is easy to see that $F_N(\epsilon)$ is closed for all $N$ and $\mathcal{F}=\bigcup_{N\in\mathbb{N}}F_N(\epsilon)$.
Thus by Baire's category theorem, there is $N_0$ s.t. $F_{N_0)}(\epsilon)$ contains an interior point, i.e. $\exists E_0\in F_{N_0}(\epsilon)$ and $\delta>0$ s.t. $E\in F_{N_0}(\epsilon)$ whenever $\mu(E\Delta E_0)<\delta$.
Now choose $\delta>0$ sufficiently small s.t.
$$\mu(A)<\delta\quad\Rightarrow\quad\nu_n(A)<\epsilon,\forall n=1,\dots,N_0$$
Assume $A\in\mathcal{F}$ with $\mu(A)<\delta$, then set 
$$E_1=E_0\setminus A,\quad E_2=E_0\cup A=E_1\cup A$$
then $\mu(E_0\Delta E_1)<\delta,\mu(E_0\Delta E_2)<\delta$, so $E_1,E_2\in\mathcal{F}_{N_0}(\epsilon)$, so for all $n\geq N_0$:
$$\begin{aligned}\nu(A)&\leq |\nu_{N_0}(A)|+|\nu_n(A)-\nu_{N_0}(A)|\leq\epsilon+|\nu_n(A)-\nu_{N_0}(A)|\\
&=\epsilon+|\nu_n(A)+\nu_n(E_1)-\nu_n(E_1)-\nu_{N_0}(A)-\nu_{N_0}(E_1)+\nu_{N_0}(E_1)|\\
&\leq\epsilon+|\nu_n(A\cup E_1)-\nu_{N_0}(A\cup E_1)|+|\nu_n(E_1)-\nu_{N_0}(E_1)|\\
&=\epsilon+|\nu_n(E_2)-\nu_{N_0}(E_2)|+|\nu_n(E_1)-\nu_{N_0}(E_1)|\\
&\leq 3\epsilon
\end{aligned}$$
