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I've seen Prohorov's theorem in the following formulation (I'll cite only one direction). $\mathcal{M}_{\leq 1}(E)$ is the set of subprobability measures on $(E, \mathcal{B}(E))$:

Let $(E, d)$ be a metric space and $\mathcal{F} \subset \mathcal{M}_{\leq 1}(E)$. Then:

$\mathcal{F}$ is tight $\Longrightarrow$ $\mathcal{F}$ is weakly relatively sequentially compact.

Question 1:

Now what does "weakly relatively sequentially compact" mean? Only explanation I've found: "relatively sequentially compact regarding the weak topology". I don't know if I understand this. Or can somebody explain it to me?

Couldn't one just write it with weak limits?

Would the following formulation be equivalent ($\text{w-lim}$ denotes the weak limit):

Let $(E, d)$ be a metric space and $\mathcal{F} \subset \mathcal{M}_{\leq 1}(E)$. Then:

$\mathcal{F}$ is tight $\Longrightarrow$ Every sequence $(\mu_n)_{n\in\mathbb{N}} \in \mathcal{F}$ has a subsequence $(\mu_{n_k})_{k\in\mathbb{N}}$, so that a $\mu\in\mathcal{M}_{\leq 1}(E)$ exists with $$\underset{k\rightarrow \infty}{\text{w-lim}}\; \mu_{n_k} = \mu \, .$$

Question 2:

In this formulation of the theorem subprobability measures were used, I think that's a bit strange, makes everything complicated and when are we really ever interested in them?

But the theorem would surely still hold if we would replace $\mathcal{M}_{\leq 1}(E)$ with $\mathcal{M}_1(E)$, which is the set of probability measures on $(E, \mathcal{B})$. Right?

Because for probability mesures $\mu_k$ we have that their weak limit must be a probability measure, too: $$\underset{k\rightarrow \infty}{\text{w-lim}}\; \mu_{k} = \mu \Longrightarrow 1 = \int 1 \, d\mu_n \overset{n\rightarrow\infty}{\longrightarrow} \int 1 \, d\mu = 1\Rightarrow \mu(E) = 1 \, .$$

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Question 1. Your formulation is indeed equivalent. Note that the " regarding the weak topology" should be replace by "topology of weak convergence of sub-probability measures" as this corresponds rather to a weak-$*$ topology.

Question 2.

But the theorem would surely still hold if we would replace $\mathcal{M}_{\leq 1}(E)$ with $\mathcal{M}_1(E)$

This is indeed true (if you do the replacement everywhere).

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