A simple proposition in theory of stable convergence of random sequence I'm reading some materials about "stable convergence" and have been stuck with proof for the following proposition.
Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space, on which a sequence of random variables $\{X_n\}$ is defined. Let $(\Omega_1, \mathscr{F}_1, \mathbb{P}_1)$ be another probability space, and define the "extended probability space" $(\tilde{\Omega}, \tilde{\mathscr{F}}, \tilde{\mathbb{P}})$ of $(\Omega, \mathscr{F}, \mathbb{P})$ by setting $(\tilde{\Omega}, \tilde{\mathscr{F}}, \tilde{\mathbb{P}}) = (\Omega, \mathscr{F}, \mathbb{P}) \times (\Omega_1, \mathscr{F}_1, \mathbb{P}_1)$. Let $X$ be a random variable defined on $(\tilde{\Omega}, \tilde{\mathscr{F}}, \tilde{\mathbb{P}})$. Now suppose given any $k \geq 1$, for all Borel set $A \in \mathscr{B}(\mathbb{R})$ and $B \in \sigma(X_1, X_2, \cdots, X_k)$, it holds
$$
\mathbb{P}((X_n \in A) \cap B) \longrightarrow \tilde{\mathbb{P}}(X \in A) \mathbb{P}(B)
$$
Assertion: we have that for all Borel set $A \in \mathscr{B}(\mathbb{R})$ and $B \in \sigma(X_k, k \geq 1)$,
$$
\mathbb{P}((X_n \in A) \cap B) \longrightarrow \tilde{\mathbb{P}}(X \in A) \mathbb{P}(B)
$$
I attempted by dint of monotone class theorem: set 
$$\mathscr{H}=\{B \in \sigma(X_k, k \geq 1): \mathbb{P}((X_n \in A) \cap B) \longrightarrow \tilde{\mathbb{P}}(X \in A) \mathbb{P}(B)
\}
$$
and try to prove it's a $\sigma$-algebra, but didn't manage to achieve that. Actually I doubt that this proposition is not true... Any hint will be greatly appreciated! 
 A: Define $\mathcal A :=\bigcup_{n\geqslant 1}\sigma\left(X_i,1\leqslant i\leqslant n\right)$. Then $\mathcal A$ is an algebra which generates $\sigma\left(X_i,i\geqslant 1\right)$. Let us fix a positive $\varepsilon$ and $B\in \sigma\left(X_i,i\geqslant 1\right)$. Then we can find $B'\in \mathcal A$ such that $\mathbb P\left(B\Delta B'\right)\lt\varepsilon$.
We thus have 
$$\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B\right)- \widetilde P\left(\left\{X\in A\right\} \right)\mathbb P(B) \right|\leqslant\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B\right)-\mathbb P\left(\left\{X_n\in A\right\}\cap B'\right) \right|\\+
\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B'\right)-\widetilde P\left(\left\{X\in A\right\} \right)\mathbb P\left(B' \right) \right|+
\widetilde P\left(\left\{X\in A\right\} \right)\left|\mathbb P\left(B \right)-\mathbb P\left(B' \right)\right|.$$
Now, using the fact that for any sets $S_1, S_2$ and $S_3$, $(S_1\cap S_2)\Delta (S_1\cap S_3)=S_1\cap\left(S_2\Delta S_3\right)$ and $\left|\mathbb P(S_1)-\mathbb P(S_2)\right|\leqslant\mathbb P\left(S_1\Delta S_2\right)$, we derive that 
$$\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B\right)- \widetilde P\left(\left\{X\in A\right\} \right)\mathbb P(B) \right|\leqslant\mathbb P\left(\left\{X_n\in A\right\}\right)\mathbb P\left( B\Delta B'\right)\\+
\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B'\right)-\widetilde P\left(\left\{X\in A\right\} \right)\mathbb P\left(B' \right) \right|+
\widetilde P\left(\left\{X\in A\right\} \right)\mathbb P\left(B\Delta B' \right)\\ 
\leqslant 2\varepsilon+\left|\mathbb P\left(\left\{X_n\in A\right\}\cap B'\right)-\widetilde P\left(\left\{X\in A\right\} \right)\mathbb P\left(B' \right) \right|.$$
We derive by the assumption that for any positive $\varepsilon$, 
$$\limsup_{n\to +\infty} \left|\mathbb P\left(\left\{X_n\in A\right\}\cap B\right)- \widetilde P\left(\left\{X\in A\right\} \right)\mathbb P(B) \right|\leqslant 2\varepsilon,$$
which is what we wanted. 
