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I would like to show that if $X_n$ has a density $f_n (x)$ and if for all $x$, $f_n (x) \to f(x)$ as $n \to \infty$, then for all bounded measurable functions $g$, $Eg(X_n) \to Eg(X)$. The way I was thinking to show this is by passing the limit under the integral as in this case we would have

\begin{equation} \lim Eg(X_n) = \lim \int_{-\infty}^{\infty} g(x) f_n(x) \mathrm{dx} = \int_{-\infty}^{\infty} \lim g(x) f_n(x) \mathrm{dx} = \int_{-\infty}^{\infty} g(x) f(x) \mathrm{dx} = Eg(X) \end{equation}

And in order to justify this I thought I could use the fact that

\begin{equation} g(x) f_n(x) \leq M f_n (x) \end{equation}because of the boundednness assumption. Since this is clearly inegrable for all $n$, the DCT holds. Could you please tell me if this is a valid argument?

Thank you.

JohnK
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    No, it doesn't work this way.. in order to apply the dominated convergence theorem, you need an integrable dominating function which does not depend on $n$, i.e. you have to find $h \in L^1$ such that $|g(x) f_n(x)| \leq h(x)$ for all $n$ and all $x$. – saz Oct 17 '16 at 17:39
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    Which version of the DCT? Not this, but the generalised DCT cf. here does it. – Daniel Fischer Oct 17 '16 at 17:40
  • @DanielFischer Where can I read about the generalised DCT? Thank you. – JohnK Oct 17 '16 at 17:41
  • Just edited in a link. – Daniel Fischer Oct 17 '16 at 17:42
  • @DanielFischer Truth be told I wasn't going for it but it certainly makes my incorrect proof a correct one! Glad to know that it exists. – JohnK Oct 17 '16 at 17:45
  • @saz Thank for pointing that out, my omission. – JohnK Oct 17 '16 at 17:45
  • @DanielFischer, Do we know $\int f_n \rightarrow \int f$? Or this can be avoided? – Ranc Oct 17 '16 at 17:53
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    @Ranc The $f_n$ are all explicitly probability densities. It's not explicitly stated of $f$, but from the context, $f$ must be the probability density of $X$. So $\int f_n = 1 = \int f$ for all $n$. – Daniel Fischer Oct 17 '16 at 17:59
  • Daniel @DanielFischer that was a good hint so if you want post that as an answer with a little more detail so that I can mark it as the correct one. – JohnK Oct 17 '16 at 19:38

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The standard form of the DCT isn't sufficient to (directly) yield the conclusion, since it requires that all functions are dominated by the same function. However, there is a more general form of the dominated convergence theorem that immediately yields the conclusion here, since the $f_n$ and $f$ are all probability density functions, thus the condition that the integrals of the dominating functions ($M\cdot f_n$ here) converge to the integral of their pointwise limit ($M\cdot f$) is satisfied.

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