I want to show that a collection of probability distributions $\{\mu_1,…,\mu_n \}$ on $\mathbb{R}^d$ is tight. I have managed to prove that every single measure is tight, and now I want to use it to prove that the collection is tight. I have done the following:
I have shown that every probability measure on $\mathbb{R}^d$ is tight. i.e. $\mu(K) > 1-\epsilon$.(For $K$ I used the closure of $\cap_{k\in \mathbb{N}} \cup_{i \leq n_k} B_{1/k} (x_i)$, but I could've chosen the closed balls of radius k which cover all of $\mathbb{R}^d$) Now I want to show that I can have that $\mu_n(K) > 1-\epsilon$ for all $n$. Let $\epsilon>0$. As every $\mu_n$ is tight on $\mathbb{R}^d$ such that $\mu_n(K_n) \geq 1 – \frac{\epsilon}{2}$. Let $H_n = \cup_{j=1}^n K_j$ and $H = \cup_n K_n$. Then, $H_n$ is compact for every $n$ and $H = H_k \cup (H \backslash H_k)$ for all $k$. Since $H\backslash H_k \searrow \phi$, I get by uniform $\sigma$-additivity that $\mu_n(H \backslash H_{k_0}) < \frac{\epsilon}{2}$ for some $k_0$ and all $n$. Now, I have that $\mu_n (H) = \mu_n (H_{k_0}) + \mu_n (H\backslash H_{k_0}) < \mu_n (H_{k_0}) + \frac{\epsilon}{2}$.
Then, I have that $1-\frac{\epsilon}{2} < \mu_n (K_n) \leq \mu_n (H) < \mu_n (H_{k_0}) + \frac{\epsilon}{2}$ for all $n$.
So, I deduce that $H_{k_0}$ is compact and $\mu_n (H_{k_0}) > \epsilon$ for all $n$, which proves the tightness of the sequence $\{\mu_n \}_{n=1,..,n}$
(1) I want a simpler way to do this, more specifically by considering that $K$ are closed balls of radius $k$ which their union covers all of $\mathbb{R}^d$.
(2) I want to know how I can use the converse of Prokhorov's theorem to show this. So, I'd have to show that the collection is relatively compact which means its closure is compact. I don't know how this can be done.
(3) In my proof, where I used the uniform $\sigma$-additivity, is it correct to use this or not.
Best Answer
Let $K_i=\{x \in \mathbb R^{d}: \|x\|\leq i\}$. Then $K_i$ is compact and $K_i$'s increase to $\mathbb R^{d}$. Hence $\mu_j(K_i) \to \mu_j (\mathbb R^{d})=1$ for each $j$. It follows that there exists $i$ such that $\mu_j(K_i) >1-\epsilon$ for all $j \in \{1,2,...,n\}$. This proves tightness.
Any finite set in a metric space (or even a topological space) is compact. Hence $\{\mu_1,\mu_2,...,\mu_n\}$ is compact and Prohorov's Theorem gives tightness.
Uniform countable additivity is fine since you are dealing with a finite collection of measures.