# Convergence integral given weak convergence of measures and functions/random variables

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

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

• I think he means weak convergence $f_n$ to $f$ like convergence of random variables. There was a probability tag earlier. – Jakobian Aug 5 at 10:46
• Yes sorry, that is what i meant! – 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 $$f$$.
• The OP is assuming $f_n\to f$ weakly. – uniquesolution Aug 5 at 10:37
• 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. – uniquesolution Aug 5 at 10:43
• He should say $f_n \to$ weakly in $C(X)$. The phrase 'real valued functions converging weakly' does not have a meaning. @uniquesolution – Kavi Rama Murthy Aug 5 at 11:53