Suppose $\mu_n$ is a sequence of measures on $(X, \mathcal{A})$ such that $\mu_n(X) = 1$ for all $n$ and $\mu_n(A)$ converges as $n \to \infty$ for each $A \in \mathcal{A}$. Cal the limit $\mu(A)$. Does it follow that $\mu$ is a measure?
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This is overkill since I use the non trivial result that if the sequences $ x_n, x$ are in $l_1$, then $x_n \to x$ in norm iff $x_n(k) \to x(k)$ for all $k$.
It is straightforward to verify that $\mu \emptyset = 0$ and $\mu A \ge 0$ for any $A \in \cal A$.
Suppose $A_k \in \cal A$ are disjoint, let $x_n(k) = \mu_n A_k$, $x(k) = \mu A_k$, then we have $x_n(k) \to x(k)$ for each $k$ and hence $x_n \to x$ (in $l_1$) and hence $\|x_n\|_1 \to \|x\|_1$.
In particular, we have $\mu( \cup_k A_k) = \lim_n \mu_n ( \cup_k A_k) = \lim_n \sum_k \mu_n A_k = \lim_n \|x_n\|_1 = \|x\|_1 = \sum_k \mu A_k$.
Hence $\mu$ is a measure and $\mu X = 1$.
copper.hat
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