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detail prokhorov's thm
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RemyDegenne committed Jan 1, 2024
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Expand Up @@ -59,13 +59,23 @@ \section{Prokhorov's theorem}
\begin{proof}\uses{lem:exists_finite_union_inter_lt}
\end{proof}

\begin{lemma}\label{ProbabilityMeasure_compact}
If $E$ is a compact metric space, then $\mathcal P(X)$ with the Prokhorov metric is a compact metric space.
\end{lemma}
\begin{theorem}[Riesz-Markov-Kakutani representation theorem, for compact spaces]\label{thm:riesz_markov_kakutani}
Let $E$ be a compact T2 space and let $\psi$ be a positive linear functional on $C(E, \mathbb{C})$. There exists a regular measure $\mu$ finite on compacts such that
\begin{align*}
\forall f \in C(E, \mathbb{C}), \ \psi(f) = \int_{x \in E} f(x) \partial\mu(x) \: .
\end{align*}
\end{theorem}

\begin{proof}
\end{proof}

\begin{lemma}\label{lem:ProbabilityMeasure_compact}
If $E$ is a compact metric space, then $\mathcal P(E)$ with the Prokhorov metric is a compact metric space.
\end{lemma}

\begin{proof}\uses{thm:riesz_markov_kakutani}
\end{proof}

\begin{theorem}[Prokhorov's theorem]\label{thm:prokhorov}
Let $E$ be a complete separable metric space and let $S \subseteq \mathcal P(E)$. Then the following are equivalent:
\begin{enumerate}
Expand All @@ -75,7 +85,7 @@ \section{Prokhorov's theorem}
\end{enumerate}
\end{theorem}

\begin{proof}\uses{lem:prokhorov_aux1}
\begin{proof}\uses{lem:prokhorov_aux1, lem:ProbabilityMeasure_compact}
Lemma~\ref{lem:prokhorov_aux1} provides one direction.

TODO: proof of the other one.
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