Convergence in probability for each $P_k$ implies convergence in probability in $\sum_k \lambda_k P_k$

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.

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.