Let $\mu_n$ be a sequence of Radon measures weak-$*$ converging to a measure $\mu$. Moreover assume that $\{f_n\}_{n\in\mathbb{N}}$ is a sequence of measurable (continuous, whatever regulartiy you wish) functions converging in some sense to $f$.

My question is: in which cases can we deduce that:

$$ \int f_nd\mu_n\to\int f d\mu$$

as $n\to\infty$, or at least for a subsequence?

It is really well known that if $f_n$ are continuous, and for istance $\mu_n$ are finite and uniformily bounded ($\lVert\mu_n\rVert\leq C$ for any $n$), or $f_n$ have compact support uniformily contained in a bounded open set, and converges uniformily to $f$ the claim follows easily:

$$\left\lvert\int f_nd\mu_n-\int fd\mu\right\rvert\leq\left\lvert\int f_nd\mu_n-\int fd\mu_n\right\rvert+\left\lvert\int fd\mu_n-\int fd\mu\right\rvert\\ \leq\int\lVert f_n-f\rVert d\mu_n+\left\lvert\int fd\mu_n-\int fd\mu\right\rvert\leq C\lVert f_n-f\rVert+\left\lvert\int fd\mu_n-\int fd\mu\right\rvert.$$

On the other hand, if we assume some more regularity on the measures the claim follows as well, for istance if we assume that $\mu_n(B_r(x))=h(r)$ for any $n$ and any $x\in supp(\mu)$ and $f_n$ is radially simmetric, uniformily integrable for $\mu_n$ and converges pointwise to $f$ then the claim holds for a subsequence. I remark that this sentence may be inaccurate, but if you want to figure out what I am talking about, you can find references in Mattila's book, or De Lellis notes on rectifiability (since uniformily distributed measures naturally arise as blowups of measures with finite and non zero density).

Are there any other important cases (general enough) in which this holds?

  • $\begingroup$ If $\{f_n\} \subset C_0(X)$ with $f_n \to f$ in $C_0(X)$ and $\mu_n \stackrel*\rightharpoonup \mu$ in $C_0(X)^*$, then $\int_X f_n \mathrm{d}\mu_n \to \int_X f\mathrm{d}\mu$. $\endgroup$ – gerw Jan 4 '18 at 7:40

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