$P_n(A)\rightarrow P(A)$ implies $P$ is a probability measure? Here is the question: Let $\{P_n\}$ be a sequence of probability measures on $\sigma$-field (Also called $\sigma$-algebra) $\mathcal{F}$. Suppose that there exists some function $P$ on $\mathcal{F}$ satisfying that $P_n(A)\rightarrow P(A)$ for all $A\in\mathcal{F}$. Prove that $P$ is a probability measure.
What I try: The only problem is to verify countable additivity of $P$. But I can only prove it under the assumption that $P(A_n)\rightarrow 0$ (or $\sup_nP_n(A_k)\rightarrow0$ as $k\rightarrow0$) when $A_n\downarrow0$.
About $\sup_nP_n(A_k)\rightarrow0$: For any $k\in\mathbb{N}$ and any $\epsilon>0$, there exists $N=N(k,\epsilon)\in\mathbb{N}$, such that $|P_n(A_k)-P_N(A_k)|\leqslant\epsilon$ for all $n\geqslant N$. Since $\sup_nP_n(A_k)$ is decreasing, we have $$\lim_k\sup_nP_n(A_k)\leqslant\sup_nP_n(A_k)\leqslant\max_{n\leqslant N}P_n(A_k)+\epsilon.$$ Now I don't know what to do: we can't just let $k\rightarrow\infty$ 'cause $N$ is related to $k$.
I can prove the needed assumption: for any $\{A_n\}\subset\mathcal F$ with $A_n\downarrow\emptyset$, we have $P(A_n)\rightarrow0$. So this problem is now solved.
 A: Firstly, for any finite mutually disjoint sets $C_1,\cdots,C_N\in\mathcal F$, we have 
    \begin{eqnarray*}
 P(C_1+\cdots+C_N)&=&\lim_nP_n(C_1+\cdots+C_N)\\
 &=&\lim_n\big[P_n(C_1)+\cdots+P_n(C_N)\big]\\
 &=&\lim_nP_n(C_1)+\cdots+\lim_nP_n(C_N)\\
 &=&P(C_1)+\cdots+P(C_N). 
 \end{eqnarray*}
Now it remains to show $P(B_k)\rightarrow0$ for any sequence $\{B_k\}\subset\mathcal F$ with $B_k\downarrow\emptyset$.
Suppose we are given such a sequence $\{B_k\}$. Since $$P(B_k)=P((B_k-B_{k+1})+B_{k+1})=P(B_k-B_{k+1})+P(B_{k+1})\geqslant P(B_{k+1}),$$ $\{P(B_k)\}$ is a decreasing non-negative sequence and thus it converges to some non-negative limit. Assume this limit is srictly larger than $0$, next we show it leads to a contradiction.
(1) By assumption, suppose $\lim_kP(B_k)=7a>0$ for convenience. Since $\{P(B_k)\}$ is decreasing, there exists some $K_0\in\mathbb N$, such that $7a\leqslant P(B_k)\leqslant8a$ for all $k\geqslant K_0$.\
(2) For each fixed $k\geqslant K_0$, since $\lim_nP_n(B_k)=P(B_k)\in[7a,8a]$, there exists some $N_k\in\mathbb N$, such that $6a\leqslant P_n(B_k)\leqslant9a$ for all $n\geqslant N_k$.
(3) For each fixed $k\geqslant K_0$ and each fixed $n\geqslant N_k$, since $\lim_\ell P_n(B_\ell)=0$, there exists some $L_n\in\mathbb{N}$ such that $P_n(B_\ell)<a$ for all $\ell\geqslant L_n$.
By (2) and (3) we can conclude $$P_n(B_k-B_\ell)>6a-a=5a$$ for any $(n,k,\ell)\in\mathbb N^3$ that satisfies $k\geqslant K_0,n\geqslant N_k$ and $\ell\geqslant L_n$.
Then we can find two sequences of large integers $\{n_i\}\uparrow\infty,\{m_i\}\uparrow\infty$ such that for every $i\geqslant1$,  $m_i\geqslant K_0,n_i\geqslant N_{m_i}$ and $m_{i+1}\geqslant L_{n_i}$ hold and thus we have $$P_{n_i}(B_{m_i}-B_{m_{i+1}})>5a.$$
(Hint: For example, we can let $m_1=K_0$ and $n_1=N_{m_1}$ and define inductively $$m_i=1+\max\{L_{n_{i-1}}~,m_{i-1}\},~~ n_i=1+\max\{N_{m_i},n_{i-1}\}$$ for all $i\geqslant2$.)
Now let $$B=\bigcup_{i=1}^\infty(B_{m_{2i-1}}-B_{m_{2i}})=\sum_{i=1}^{\infty}(B_{m_{2i-1}}-B_{m_{2i}}).$$
Note that $B\in\mathcal F$ and $P_{n_{2t-1}}(B)\geqslant P_{n_{2t-1}}(B_{m_{2t-1}}-B_{m_{2t}})>5a$ for all $t\geqslant1$. It follows that $$P(B)=\lim_tP_{n_{2t-1}}(B)\geqslant5a.$$
Meanwhile, since $B\subset B_{m_1}$ and $B\cap(B_{m_{2i}}-B_{m_{2i+1}})=\emptyset$, we have $B\subset B_{m_1}\setminus(B_{m_{2i}}-B_{m_{2i+1}})$ and thus $P_{n_{2i}}(B)\leqslant P_{n_{2i}}(B_{m_1})-P_{n_{2i}}(B_{m_{2i}}-B_{m_{2i+1}})<P_{n_{2i}}(B_{m_1})-5a$.
Also note that $n_{2i}\geqslant n_1\geqslant N_{m_1}$, so by (2) we have $P_{n_{2i}}(B_{m_1})\leqslant 9a$. Thus $P_{n_{2i}}(B)<9a-5a=4a$. It follows that $$P(B)=\lim_tP_{n_{2t}}(B)\leqslant 4a.$$
It's a contradiction.
