Let $X\subset\mathbb{R}^d$ be compact. Given sequences of real-vauled random variables $f_n\to f$ and positive radon measures $\mu_n\to \mu$ both converging weakly for $n\to\infty$.

Under which further conditions can we deduce that

$$\lim_{n\to\infty}\int_X f_nd\mu_n =\int_X f d\mu ?$$

  • $\begingroup$ I think he means weak convergence $f_n$ to $f$ like convergence of random variables. There was a probability tag earlier. $\endgroup$ – Jakobian Aug 5 at 10:46
  • $\begingroup$ Yes sorry, that is what i meant! $\endgroup$ – freeba Aug 5 at 10:56

Claim: the conclusion holds for every sequence $(\mu_n)$ converging weakly to $\mu$ iff $f_n \to f$ uniformly.

Since $\mu$ is Radon and $X$ is compact, $\mu$ is a finite measure. Since $\mu_n(X) \to \mu(X)$ it follows that $sup_n \mu_n(X)<\infty$. So, if $f_n \to f$ uniformly then $\int f_n d\mu_n -\int fd\mu_n \to 0$ from which the conclusion follow easily.

Now suppose the conclusion holds for every sequence $(\mu_n)$ converging weakly to $\mu$. Let $x_n \to x$ and $\mu_n=\delta_{x_n}, \mu =\delta_x$. Then $\mu_n \to \mu$ weakly so $f(x_n)=\int f_n d\mu_n \to \int f d\mu=f(x)$. Since $X$ is compact the statement $f(x_n) \to f(x)$ whenever $x_n \to x$ is equivalent to uniform convergence of $f_n$ to $f$.

  • $\begingroup$ The OP is assuming $f_n\to f$ weakly. $\endgroup$ – uniquesolution Aug 5 at 10:37
  • $\begingroup$ That doesn't make sense to me. $\endgroup$ – Kabo Murphy Aug 5 at 10:42
  • $\begingroup$ Why not? I give you a weakly convergent sequence $f_n$ in the space $C(X)$ and a weakly convergent sequence of measures $\mu_n$. Now the questions is whether $\mu_n(f_n)$ converges or not. $\endgroup$ – uniquesolution Aug 5 at 10:43
  • $\begingroup$ He should say $f_n \to $ weakly in $C(X)$. The phrase 'real valued functions converging weakly' does not have a meaning. @uniquesolution $\endgroup$ – Kabo Murphy Aug 5 at 11:53

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