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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.

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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.

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  • $\begingroup$ Thanks for the fast response! $\endgroup$ – thehardyreader Jan 6 at 10:12

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