In connection to my question here, let $P_{n}(n=1,2, \ldots)$ on $([0,1], \mathcal{B}([0,1]))$, where $\mathcal{B}([0,1])$ is the Borel $\sigma$-algebra on $[0,1]$, be a sequence of measures which converges weakly to the finite measure $P$ (i.e. $P_{n} \Rightarrow P$).
Claim: For all $x \in supp \; P$, and $\forall\;\epsilon>0$, $\exists\;K \equiv K_{x,\epsilon} \in \mathbb{N}$ s.t. $\forall\;k\geq K$, $supp\;P_k \cap (x-\epsilon, x+\epsilon)\neq \emptyset$.
Proof: By $a.e.$ continuity of the distribution function associated with $P$, $F(v) \equiv P[0,v]\;\forall\;v\in[0,1]$, there exist $x_1 \in (x-\epsilon,x)$ and $x_2 \in (x,x+\epsilon)$ such that $x_1, x_2 \in C(F)$. $x \in supp\;P,\;\therefore P((x_1,x_2))=P((x_1,x_2])>0$, where the equality follows from $x_1, x_2 \in C(F)$. $(x_1,x_2]$ is a continuity set of $P$, $\therefore\;\exists\;K$ s.t. $\forall\;k \geq K, P_k((x_1,x_2])>0$. $\therefore \;supp\;P_k \cap (x_1,x_2] \neq \emptyset, \therefore\;\;supp\;P_k \cap (x – \epsilon, x+ \epsilon) \supseteq \;supp\;P_k \cap (x_1,x_2] \neq \emptyset$.
Please let me know if this is correct and also if this is the correct (tightest?) way to make the notion of "convergence in measure implies supports getting close" precise?
Best Answer
Your proof is correct.
But next examples show that the statement "weak convergence of measures implies supports are getting closer" implies that we should have "convergence" of supports in the next cases:
a) $\sum_{i=1}^n \frac{1}n \delta_{\frac{i}n}$ converges to $U[0,1]$.
b) $N(0, \frac{1}n) \to \delta_0$.
c) $N(1, \frac{1}n) \to \delta_1$.
If we will say that weak convergence of measure implies supports getting closer then we have to say that
a) $\{ \frac{1}n, \frac2{n}, \ldots, 1\}$ are getting closer to $[0,1]$
b)$\mathbb{R}$ is getting closer to $\{0\}$ and at the same time c)$\mathbb{R}$ is getting closer to $\{1\}$.
So it's an illustration of corollaries of such a thesis, which are not good enough.