Is there a way to obtain the Prohorov's theorem from the case of probabilities? Definition: we define a family $\mathcal{F}$ of finite positive measure over a metric space $X$ to be uniformly tight if for all $\varepsilon>0$ there exists a compact set $K$ such that $\mu (X\smallsetminus K) \leq \varepsilon$ for all $\mu \in \mathcal{F}$
I know the following theorem due to Prohorov:
Theorem: suppose $(X,d)$ is a Polish space (i.e. complete and separable). Let $\mathcal{A}=\mathcal{B}(X)$ the $\sigma$-algebra generated by the open sets. Let $\mathcal{F}$ be a family of probability over $(X,\mathcal{A})$. Then $\mathcal{F}$ is relatively sequentially weakly compact if and only if it is uniformly tight.
I would like to know if we can deduce the following from the previous one:
Theorem: suppose $(X,d)$ is a Polish space (i.e. complete and separable). Let $\mathcal{A}=\mathcal{B}(X)$ the $\sigma$-algebra generated by the open sets. Let $\mathcal{F}$ be a bounded family of finite positive measure over $(X,\mathcal{A})$. Then $\mathcal{F}$ is relatively sequentially weakly compact if and only if it is uniformly tight.
My attempt: I tried normalizing the bounded measures in this way $\mu'=\frac{1}{\mu(X)}\mu$ but I do not know if this will work. I am looking for a canonical answer. Any help will be appreciated.
 A: We shall denote with $M$ the constant controlling the mass of all the measures in the family. Let us focus on the first implication: if the family $\mathcal{F}$ is w.s.c. (weakly sequentially compact), then so is $\tilde{\mathcal{F}} = \{\mu+\delta_{x}\}_{\mu\in\mathcal{F}}$ for some $x\in X$. Define $\mathcal{F}^\prime=\{\nu\}=\biggl\{{\frac{\mu+\delta_x}{\mu(X)+1}}\biggr\}_{\mu\in\mathcal{F}}$: this family is clearly w.s.c. and by Prohorov for probabilities it is uniformly tight, but then
$\sup_{\nu\in\mathcal{F}^\prime}\nu(X\setminus K_\varepsilon)\leq\varepsilon$ implies
$$\sup_{\mu\in\mathcal{F}}\mu(X\setminus K_\varepsilon)\leq\varepsilon(1+M),$$ which is the uniform tightness of $\mathcal{F}$.
For the other implication we assume $\mathcal{F}$ to be uniformly tight, then $\tilde{\mathcal{F}}$ is uniformly tight and so is $\mathcal{F}^\prime$ (trivial inequalities involved plus the fact that if $K$ is compact then so is $K\cup\{x\}$). This means that $\mathcal{F}^\prime$ is w.s.c. and so is $\tilde{\mathcal{F}}$, which in turn implies that $\mathcal{F}$ is w.s.c. and this concludes the proof.
If you want a proof from scratch you can read Theorem 8.6.2 in Bogachev's book Measure Theory.
