Measures: Sequential Continuity Disclaimer: This thread is meant as record and written in Q&A style.

Let $\Omega$ be a measure space.
It is well known that a measure is continuous from below as well as from above:
$$E_n\uparrow E\implies\mu(E_n)\uparrow\mu(E)$$
$$E_n\downarrow E\implies\mu(E_n)\downarrow\mu(E)\quad(\mu(E_n)<\infty)$$
On the other hand this fails in general if one replaces sequences by nets:
$$E_M\uparrow[0,1],\lambda(E_M)\equiv 0\quad(E_{M\subseteq[0,1]:\lambda(M)=0}:=M)$$
$$E_N\downarrow\varnothing,\lambda(E_N)\equiv 1\quad(E_{N\subseteq[0,1]:\lambda(N)=1}:=N)$$
with the ordering being given by inclusion respectively containement.
What about sequential continuity:
$$E_n\to E\implies \mu(E_n)\to\mu(E)$$
Here convergence of sets is meant as:
$$A_\lambda\to A:\iff\forall\omega\in\Omega\exists \lambda_\omega\in\Lambda\forall\lambda\geq\lambda_\omega:\omega\in A_\lambda\text{ if }\omega\in A\text{ and }\omega\notin A_\lambda\text{ if }\omega\notin A$$
 A: What you are asking/writing is not 100% correct.
First, observe that
$$
E_n \downarrow E \,\,\Rightarrow \,\, \mu(E_n) \downarrow \mu(E)
$$
does not hold in general. To see this, consider $E_n = [n,\infty)$ with $\mu = $ Lebesgue measure on $\Bbb{R}$.
The additional condition which is needed here is that $\mu(E_n) < \infty$ for some $n \in \Bbb{N}$. (For this, it is of course sufficient to have $\mu(X) <\infty$, where $X$ is the "surrounding" space.
Now note that your definition of $A_\lambda \to A$ is equivalent to
$$
\chi_{A_\lambda} \to \chi_{A} \text{ pointwise}.
$$
Hence, the question you are asking amounts to
$$
\text{Does }\,\, \int \chi_{A_n}  \,d\mu \to \int \chi_A \,d \mu \,\,\text{ if }\,\, \chi_{A_n} \to \chi_A \,\,\text{ pointwise?}
$$
This holds by dominated convergence as long as $\mu(\bigcup_{n=k}^\infty A_n) <\infty$ for some $k \in \Bbb{N}$, but fails in general (as you already observed).
A: Sequential continuity fails to hold in general as well:
$$E_n\to\varnothing,\# E_n\nrightarrow\#\varnothing\quad(E_n:=\{\frac{1}{n}\})$$
