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I'm trying to understand the following proof of the statement : "Almost sure convergence implies convegence in distribution"

The definition of convergence in distribution is given as follows :

$X_n$ converges in distribution to $X$ if and only if for all bounded real function $f$ we have :

$$\lim_{n \rightarrow +\infty }E\left[f(X_n)\right]=E\left[f(X)\right]$$

The proof goes like this :

If $X_n$ converges almost surely to $X$ then $f(X_n)$ converges almost surely to $f(X)$. Now using the dominated convergence theorem which is :

$$\lim_{n \rightarrow +\infty } \int f_n d\mu = \int f d\mu $$

we get :

$$\lim_{n \rightarrow +\infty }E[f(X_n)]=E[f(X)].$$

My question is this : How using the dominated convergence theorem gets us from : $$f(X_n)\rightarrow^{a.s.} f(X)$$ to $$\lim_{n \rightarrow +\infty }E[f(X_n)]=E[f(X)]$$

given that the almost sure convergence is given by : $P\left[\lim_{n \rightarrow + \infty} X_n = X\right] = 1$?

One of my attempts is to write the convergence in distribution in the form of integrals like this :

$$\lim_{n \rightarrow +\infty }E[f(X_n)]=E[f(X)]$$

is equivalent to :

$$\lim_{n \rightarrow +\infty }\int f(y) \phi_n(y) dy=\int f(y) \phi(y) dy$$

with $\phi$ the density of $X$ and $\phi_n$ the density of $X_n$. But this is a little different from the monotone convergence theorem result. In the dominated convergence theorem we have the same measure, $\mu$, but writing the expectations gives us two different measures, which are $\phi_n dy$ and $\phi dy$, respectively the cumulative distributions for $X_n$ and $X$

Any help please? I appreciate if you can tell me why my attempt is not leading anywhere and at the same time give your own proof. Thank you!

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    $\begingroup$ I think $f$ should be continuous also. $\endgroup$ – Danielsen Jul 7 '16 at 1:22
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In order to apply dominated convergence, write the expectations as integrals over the probability space $(\Omega,{\cal F},P)$, not the real line: $$E(f(X_n))=\int_\Omega f(X_n(\omega))\,P(d\omega)\to\int_\Omega f(X(\omega))\,P(d\omega)=E(f(X)).$$

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  • $\begingroup$ Why do you use the same $P(dw)$ for both? $X_n$ and $X$ don't necessairly have the same distribution $\endgroup$ – Dave ddd Jul 8 '16 at 12:53
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    $\begingroup$ The fact that $X_n$ and $X$ have different distributions is irrelevant. This is the correct way to do it ;) $\endgroup$ – user940 Jul 8 '16 at 12:55
  • $\begingroup$ Ok, now I get it. When writing the expectation over the probability space $(\Omega, F, P)$ we only care about $w$ and so the distribution of $X$ is irrelevant. Thank you very much $\endgroup$ – Dave ddd Jul 8 '16 at 13:13
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As the comment by Danielsen points out, the most common definition of the convergence in distribution is $$\lim_{n \rightarrow +\infty }\mathbb E\left[f\left(X_n\right)\right]=\mathbb E\left[f(X)\right]$$ for each continuous and bounded function $f$.

If $f$ is such a function and $X_n\to X$ almost surely, then $X_n(\omega)\to X(\omega)$ for each $\omega\in\Omega\setminus N$, where $N$ is a measurable set of probability zero. Since $f$ is continuous, we have $\left(f(X_n)\right)(\omega)\to\left(f(X)\right)(\omega)$ for each $\omega\in \Omega\setminus N$, hence $f(X_n)\to f(X)$ almost surely. Since the function $f$ is supposed to be bounded, the sequence $\left(f\left(X_n\right)\right)_{n\geqslant 1}$ is dominated by a constant hence integrable function.

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  • $\begingroup$ Thank you this is very clear for the first part of the proof. What about the second part? I mean applying the dominated convergence theorem to get the expectations? As you can see, in the dominated convergence theorem we have the same measure, $\mu$, but writing the expectations gives us two different measures, which are $\phi_n dy$ and $\phi dy$, respectively the cumulative distributions for $X_n$ and $X$ $\endgroup$ – Dave ddd Jul 8 '16 at 12:24
  • $\begingroup$ It seems that you are not in the good road for the second part for the following reasons. 1. At best, you will only solve the case where all the random variables have a density. 2. The connection between the convergence of densities and almost sure convergence is not clear (the latter involves the distribution of the whole sequence and not only of marginals). $\endgroup$ – Davide Giraudo Jul 8 '16 at 13:03
  • $\begingroup$ What do you mean by "Almost sure convergence involves the distribution of the sequence and not only of marginals"? $\endgroup$ – Dave ddd Jul 8 '16 at 13:07
  • $\begingroup$ Almost sure convergence is a property of the distribution of $(X_n)_n$. $\endgroup$ – Davide Giraudo Jul 8 '16 at 13:08

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