# Prove that $\lim\limits_{n\rightarrow \infty}P_n(A)=P(A)$ implies $\lim\limits_{n\rightarrow \infty} \int f ~dP_n = \int f ~dP$

Let $$P_n, n \in \mathbb{N}$$ and $$P$$ be probability measures on the measurable space $$(\Omega,\mathfrak{S})$$ and assume $$\forall A \in \mathfrak{S}: \lim\limits_{n\rightarrow \infty}P_n(A)=P(A)$$.

I want to prove now that for all bounded measurable functions $$f$$ the following holds

$$\lim\limits_{n\rightarrow \infty} \int f ~dP_n = \int f ~dP.$$

My approach was to first assume $$f\geq0$$ and approximate $$f$$ with step functions $$(u_k)\in \mathcal{T}$$ where $$u_k$$ converges uniformly to $$f$$. Let $$u_k = \sum\limits_{i} \alpha_i^{(k)} 1_{A_i^{(k)}}$$. Now I do the following: \begin{align} \lim\limits_{n\rightarrow \infty} \int f ~dP_n &= \lim\limits_{n\rightarrow \infty} \int \lim\limits_{k\rightarrow \infty} u_k ~dP_n \\ &= \lim\limits_{n\rightarrow \infty} \lim\limits_{k\rightarrow \infty} \int u_k ~dP_n \\ &=\lim\limits_{n\rightarrow \infty} \lim\limits_{k\rightarrow \infty} \sum\limits_{i} \alpha_i^{(k)} P_n(A_i^{(k)}) \\ &\stackrel{(*)}{=} \lim\limits_{k\rightarrow \infty} \lim\limits_{n\rightarrow \infty} \sum\limits_{i} \alpha_i^{(k)} P_n(A_i^{(k)}) \\ &= \lim\limits_{k\rightarrow \infty} \sum\limits_{i} \alpha_i^{(k)} P(A_i^{(k)}) \\ &= \lim\limits_{k\rightarrow \infty} \int u_k ~dP \\ &= \int \lim\limits_{k\rightarrow \infty} u_k ~dP \\ &= \int f ~dP. \end{align}

I can justify all steps, except for $$(*)$$. I now I would need uniform convergence of one of the two sequences with respect to the other but I don't see why this should be the case. On the other hand I don't have another approach for the proof. Does this break my argumentation? Could someone help me with this?

For the general case I would split $$f$$ into a positive and a negative part $$f=f^++f^-$$ and do basically the same argumentation.

My suggestion is to avoid all those double limits. For $$k$$ sufficiently large we have $$\int u_ kdP_n -\epsilon <\int fdP_n < \int u_ kdP_n +\epsilon$$ and $$\int u_ kdP -\epsilon <\int fdP < \int u_ kdP +\epsilon.$$ Fix one just $$k$$ and note that $$\int u_kdP_n \to \int u_k dP$$ as $$n \to \infty$$. You can easily finish the proof from these inequalities.