An equivalence of $\sigma$-additivity

Let $(\Omega, \mathcal{A})$ be a measurable space and let $\nu: \mathcal{A}\to [0, \infty)$ be finitely additive with $\nu(\emptyset)=0$. Show the following:

$\nu$ is $\sigma$-additive if and only if the following holds:

If $A_j \in \mathcal{A}$ with $A_1 \supseteq A_2 \supseteq \ldots$ and $\mu(A_j)\geq \delta$ for all $j \geq 1$ for some $\delta >0$ then $\cap_{j=1}^\infty A_j\neq \emptyset$.

My proof for the forward direction is as follows: Suppose that $\nu$ is $\sigma$-additive. Since we also know that $\nu(\emptyset)=0$ and $\nu(A)\geq 0$ for all $A\in \mathcal{A}$ (since $\mu$ takes its values in $[0, \infty)$) then $\nu$ is a measure on $\mathcal{A}$. Let $\{A_j\}_{j=1}^\infty$ be a collection of sets in $\mathcal{A}$ such that $A_1\supseteq A_2 \supseteq \ldots$ and $\nu(A_j)\geq \delta$ for all $j\geq 1$ for some $\delta >0$. Suppose, to the contrary, that $\cap_{j=1}^\infty A_j=\emptyset$. Then $\nu(\cap_{j=1}^\infty A_j)=0$. However, by continuity from above we have $$0=\nu(\emptyset) = \nu(\cap_{j=1}^\infty A_j)=\lim_{n\to \infty} \nu(A_j)\geq \lim_{n\to \infty} \delta=\delta >0,$$ a contradiction. Thus, $\cap_{j=1}^\infty A_j\neq \emptyset$.

For the other direction I was going to consider an arbitrary collection $\{A_j\}_{j=1}^\infty$ of sets in $\mathcal{A}$ and define a new collection $\{E_n\}_{n=1}^\infty$ by $E_n=\cup_{j=1}^\infty A_j\setminus (\cup_{i=1}^n A_i)$ for each $n \geq 1$. Clearly, the $E_n$'s are decreasing, but I have no idea how to proceed from there.

Note that

If $A_j \in \mathcal{A}$ with $A_1 \supseteq A_2 \supseteq \ldots$ and $\nu(A_j)\geq \delta$ for all $j \geq 1$ for some $\delta >0$ then $\cap_{j=1}^\infty A_j\neq \emptyset$.

implies

If $A_j \in \mathcal{A}$ with $A_1 \supseteq A_2 \supseteq \ldots$ and $\bigcap_{j=1}^{\infty} A_j = \emptyset$, then $\inf_{j \geq 1} \nu(A_j)=0$, i.e. there does not exist $\delta>0$ such that $\nu(A_j) \geq \delta$ for all $j \geq 1$.

(The second statement is the contraposition of the first one.)

Hints:

1. Show that the sequence $(E_n)_{n \in \mathbb{N}}$ defined in your question is decreasing and satisfies $\bigcap_{n \geq 1}E_n = \emptyset$.
2. Conclude that $$0=\inf_{n \geq 1} \nu(E_n) = \lim_{n \to \infty} \nu(E_n).$$
3. Use the finite addivity of $\nu$ to show that $$\lim_{n \to \infty} \sum_{i=1}^n \nu(A_i) = \nu \left( \bigcup_{i=1}^{\infty} A_i \right).$$