Convergence of a sequence of measures. On a measure space $(E,\mathcal{E},\mu)$, let $(\mathcal{F}_n)$ a filtration on $\mathcal{E}$, with $\mathcal{F}_n \uparrow \mathcal{E}$, and let $(\mu_n)$ be a sequence of finite measures defined on $\mathcal{E}$ such that
$$ \mu_n(A) = \mu(A) \qquad \text{for all $A \in \mathcal{F}_n$}. $$
Is it true that the sequence $(\mu_n)$ converges towards $\mu$? Which kind of convergence is expected? Is $(\mu_n)$ a Cauchy sequence?
 A: I have found a proof using the Monotone Class Theorem which states that:

If a class $\mathscr{I}$ of subsets of $E$ is closed under intersection then the monotone class generated by $\mathscr{I}$, namely $\mathscr{M}(\mathscr{I})$ is included in the sigma-field generated by $\mathscr{I}$, namely $\sigma(\mathscr{I})$.

Let us also remind that a monotone class is a class $\mathscr{M}$ such that

*

*$E$ lies in $\mathscr{M}$

*$\forall A,B\in\mathscr{M}, A\subseteq B \Rightarrow B \setminus A \in \mathscr{M}$

*$\mathscr{M} \ni A_n \uparrow A \Rightarrow A \in \mathscr{M}$
Answer to the Question.
Let $\mathscr{F} = \bigcup_n \mathscr{F}_n$. Clearly, for all $A \in \mathscr{F}$, $\mu_n(A) \to \mu(A)$, because, if $A \in \mathscr{F}$, there exists an index $n$ such that $A \in \mathscr{F}_n$ and, by hypothesis, $\mu_m(A) = \mu(A)$ for all $m > n$. Notice that $\mathscr{F}$ is closed under intersection. Now, let us denote by $\mathscr{H}$ the class of all subsets $A$ of $E$ such that $A \in \mathscr{E}$ and $\mu_n(A) \to \mu(A)$:
$$ \mathscr{H} = \big\{ A \in \mathscr{E} : \mu_n(A) \to \mu(A)\big\} $$
The class $\mathscr{F}$ is included in $\mathscr{H}$ so, in particular, $E \in \mathscr{F} \subseteq \mathscr{H}$ implies $E \in \mathscr{H}$. Also, if $A$ and $B$ belong to $\mathscr{H}$ and $A \subseteq B$, then $\mu_n(A) \to \mu(A)$ and $\mu_n(B) \to \mu(B)$, so
$$\mu_n(B\setminus A) = \mu_n(B) - \mu_n(A) \to \mu(B) - \mu(A) = \mu(B\setminus A)$$
thus $B\setminus A \in \mathscr{H}$. Finally, if $(A_j)$ is an increasing sequence of elements of $\mathscr{H}$, then $\mu_n(A_j) \to \mu(A_j)$ and $\lim_j \mu(A_j) = \mu(A)$ because $\mu$ is a measure on $\mathscr{E}$. Thus $A \in \mathscr{H}$.
It is therefore proved that $\mathscr{H}$ is a monotone class containing $\mathscr{F}$. Now, by definition $\sigma(\mathscr{F}) = \mathscr{E}$, and by monotone class theorem: $$\mathscr{E} = \sigma(\mathscr{F}) = \mathscr{M}(\mathscr{F}) \subseteq \mathscr{M}(\mathscr{H}) = \mathscr{H}. $$
Since $\mathscr{H} \subseteq \mathscr{E}$, it follows that $\mathscr{H} = \mathscr{E}$, i.e. that $(\mu_n)$ converges strongly towards $\mu$.
