Suppose we have $X_n \to X$ convergence in $P_k$-probability for every $k$. And we are given a probability measure $R= \sum_{k\ge 1} \lambda_k P_k$, where $\sum \lambda_k = 1, \lambda_k \ge 0$.
Then how do we show that $X_n \to X$ converges in $R$-probability?
I can see that this easily holds if the number of positive $\lambda_k$ is finite, but I cannot come up with an argument in the infinite case. I would greatly appreciate any help.
You can change your representation of convergence of probability to see the hidden pattern.
Please note that the notation of convergence in probability of $X_n \rightarrow X$ ( in $\mathbb{R}$) is equivalent to say that:
$$ \mathbb{E} \left( \min( |X_n-X|, 1) \right) \xrightarrow{n \rightarrow +\infty} 0$$
Besides, under your new measure, we have:
$$\mathbb{E^{R}} \left( \min( |X_n-X|, 1) \right) =\sum_{k \ge 0} \lambda_k \mathbb{E}^k \left( \min( |X_n-X|, 1) \right)$$
So you see why.
Fix $\varepsilon, \delta > 0$. One has $$R(|X_n - X| \geq \varepsilon) = \sum \lambda_k P_k(|X_n - X| \geq \varepsilon) $$ Since $\sum \lambda_k = 1$ and $\lambda_k \geq 0$ there is $K$ such that for every $n$ $$\sum_{k \geq K} \lambda_k P_k(|X_n - X| \geq \varepsilon) \leq \sum_{k \geq K} \lambda_k \leq \frac{\delta}{2}$$ Then since $X_n \to X$ in $P_k$-probability for $k < K$ we can pick an $N$ such that for each $k < K$, $$P_k(|X_n - X| \geq \varepsilon) \leq \frac{\delta}{2K}$$ for $n > N$. Hence for $n > N$, $$R(|X_n - X| \geq \varepsilon) \leq \frac{\delta}{2} + K \frac{\delta}{2K} = \delta$$ as desired.
