This problem is from Royden's Real Analysis, (4-th e.d.), Chapter 18, Problem 66 (pp. 393).
Problem Statement
Let $(X, \mathcal{M})$ be a measurable space and $\{ \nu_n \}$ a sequence of finite measures on $\mathcal{M}$ that converges setwise on $\mathcal{M}$ to $\nu$. Let $\{ E_k \}$ be a descending sequence of measurable sets with empty intersection. Show that for each $\epsilon > 0$, thre is a natural number $K$ for which $\nu _n (E_K) < \epsilon$ for all $n$.
In other words, to show the convergence in $k$ is uniform.
My attempt
Since $\nu_n$'s are bounded and $\bigcap _{k = 1} ^\infty E_k = \emptyset$, the continuity of measure implies that $$ \lim _{k \to \infty} \nu_n(E_k) = \nu_n \left( \bigcap _{k = 1} ^\infty E_k \right) = \nu _n (\emptyset) = 0 \quad \text{for all } n. $$ On the other hand, setwise-convergence of $\nu_n$ implies that $$ \lim _{n \to \infty} \nu_n (E_k) = \nu (E_k) \quad \text{for all } k. $$ So we have $$ \begin{matrix} \nu_1(E_1) & \nu_2(E_1) & \nu_3(E_1) & \cdots & \to \nu(E_1) \\ \nu_1(E_2) & \nu_2(E_2) & \nu_3(E_2) & \cdots & \to \nu(E_2) \\ \nu_1(E_3) & \nu_2(E_3) & \nu_3(E_3) & \cdots & \to \nu(E_3) \\ \vdots & \vdots & \vdots & & & \\ \downarrow & \downarrow & \downarrow & & & \\ \nu_1(\emptyset) & \nu_2(\emptyset) & \nu_3(\emptyset) & = 0 \end{matrix} $$ Moreover, because $\{ E_k \}$ is descending, $\nu_n (E_k) \geq \nu_n(E_{k+1})$, so $\{ \nu(E_k) \}$ is a finite decreasing sequence bounded below by $0$.
This is the most information I can infer from the setting. I tried showing that for every $\epsilon > 0$, there is $K$ and $N$ such that $$ n > N \implies \nu_n(E_K) < \epsilon , $$ which I do by contradiction. Suppose otherwise, then there is an $\epsilon_0 > 0$, and for each $k \in \mathbb{N}$, a strictly increasing subsequence $\{ n(i, k) \} _{i = 1} ^\infty$ which $\nu_{n(i, k)}(E_k) \geq \epsilon _0$ for all $i$. $\color{red}{\textrm{If one can show that $\nu(E_k) \to 0$}}$, then both $\nu(E_K)$ and $|\nu_{n(i, K)}(E_K) - \nu(E_K)|$ can be bounded by $\epsilon_0 / 2$, arriving at a contradiction and finishing the proof. However, although $\{ \nu(E_k) \}$ is a boudned decreasing sequence, I cannot show and is unsure whether $\nu(E_k) \to 0$.
Please let me know if there is any idea on this problem. Any help is appreciated.
Editted
It turns out if $\color{red}{\text{$\nu$ is not finite}}$, the conclusion does not hold. See @copper.hat 's comment below for a counter-example.
Despite this, if we assumed $\nu$ is also finite, it there a way to prove the statement?