Measure weak convergence and the one-point compactification Let $X$ be a locally compact space.
Let $\tilde{X} = X \cup \{\infty\}$ be its one-point compactification.
Suppose $\tilde{X}$ is metrizable and let $A_n \subset X$ be a sequence of finite subsets of $X$ with $n$ elements each.
Now, consider the sequence of (borel) probabilities in $\tilde{X}$ given by
$$
  \mu_n
  =
  \frac{1}{n}
  \sum_{a \in A_n} \delta_a,
$$
where $\delta_a$ is de Dirac Measure supported in $\{a\}$.
That is,
$$
  \delta_a(B)
  =
  \left\{
    \begin{array}{ll}
      1,& a \in B
      \\
      0,& a \not \in B
    \end{array}
  \right..
$$
Suppose that in the weak-* topology
$$
  \mu_n
  \rightarrow
  \alpha \mu
  +
  (1 - \alpha) \delta_\infty,
$$
where $\mu$ is a probability over $\tilde{X}$ such that $\mu(\{\infty\}) = 0$.
I want to know how can I choose $B_n \subset A_n$ such that
$$
  \nu_n
  =
  \frac{1}{n}
  \sum_{a \in A_n \setminus B_n} \delta_a
$$
converges to $\alpha \mu$. Is it always possible?

PS: By the weak-* convergence $\gamma_n \rightarrow \gamma$,
I mean that for every continuous function
$f: \tilde{X} \rightarrow \mathbb{R}$,
$$
  \int f\, \mathrm{d}\gamma_n
  \rightarrow
  \int f\, \mathrm{d}\gamma.
$$
What I want can be translated to
$$
  \int f\, \mathrm{d}\nu_n
  \rightarrow
  \alpha \int f\, \mathrm{d}\mu.
$$
In particular, using $f = 1$, this means that
$$
  \frac{|A_n \setminus B_n|}{|A_n|}
  =
  \nu_n(X)
  \rightarrow
  \alpha.
$$
That is,
$$
  \frac{|B_n|}{|A_n|}
  =
  1 - \alpha.
$$
Since $\tilde{X}$ is metrizable,
I believe $B_n$ can be specified in terms of
$B_{\frac{1}{n}}(\infty)$,
the ball of radius $\frac{1}{n}$ and center $\infty$.
 A: The talk about one-point compactification isn't necessary. We can forget $X$ and consider $\tilde X$ just as a compact metric space  with a chosen point $\infty$. Let $N_\epsilon$ be the $\epsilon$-neighborhood of $\infty$.  The goal is   to remove subsets $B_n\subset A_n$ such that $$\frac{1}{n}\sum_{a\in B_n}\delta_a\to (1-\alpha)\delta_\infty\tag1$$ 
Testing weak* convergence of $\mu_n$ with the function $f(x)=(1-\epsilon^{-1}d(x,\infty))^+$ we find that for large enough $n$, $\int f\,d\mu_n$ is between  $(1-\alpha)\pm \delta$, where $\delta$  is small when $\epsilon$ is small; specifically, $\delta$ depends on $\mu(N_\epsilon\setminus\{\infty\} )$. Therefore, at least $\lfloor (1-\alpha-\delta)n\rfloor$ of the points in $A_n$ are in $N_\epsilon$.
Proceed as follows. Take a sequence $\epsilon_j\to 0$. For each $j$, find $\delta_j$ as above, so that $\delta_j\to 0$. Also find $M_j$ so that at least $\lfloor (1-\alpha-\delta_j)n\rfloor$ of the points in $A_n$ are in  $N_{\epsilon_j}$ when $n\ge M_j$. 
For $M_j\le n<M_{j+1}$, let $B_n$ consist of $\lfloor (1-\alpha-\delta_j)n\rfloor$ points arbitrarily chosen from $A_n\cap N_{\epsilon_j}$. The property (1) follows from  $\epsilon_j\to 0$ and $\delta_j\to 0$.
