Let $f:\mathbb{R}^{d} \rightarrow [0,\infty) $ continuous function and denote by $P$ the space of probability measures on $\mathbb{R}^{d}$ and by $P_R \,=\, \{\nu \in P \, | \, \int f d \nu \leqslant R \, \} $ for $R>0$. We equip both spaces with the topology of weak convergence of probability measures.

I am trying to see if $P_R$ is a closed subset of $P$.

My attempt, which shows that it is indeed closed subset, is the following :

Let $\{\mu_n\} \subseteq P_R $ such that $\mu_n \rightarrow \mu \in P $. I will show that $\mu \in P_R$.

$\{\mu_n\} \subseteq P_R \,\,\Rightarrow \,\, \sup \limits_{n} \int f d \mu_n \, \leqslant R < \infty \,\, \Rightarrow \,\, \{\mu_n\}$ tight sequence in $P_R$ $\Rightarrow$ sequentially compact in $P_R$ by Prokhorov's theorem $\Rightarrow$ there exists subsequence $\{\mu_{k_n}\}_n \subseteq P_R$ which converges in $P_R$. So by uniqueness of the limit $\mu \in P_R$.

What do you think ? Is it correct ?

  • $\begingroup$ Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures) $\endgroup$ – Kavi Rama Murthy Dec 8 '18 at 0:03
  • $\begingroup$ No I meant less or equal in the definition of P_R.. sorry typo $\endgroup$ – vl.ath Dec 8 '18 at 9:38

Boundedness of the integrals for one particular $f$ does not give you tightness: take $f \equiv 0$, for example. Hence your argument is invalid. I suppose you meant $P_R=\{\nu: \int f d \nu \leq R\}$. Suppose $\nu_j \to \nu$ weakly. If $N$ is a positive integer and $f_N=\min \{f,N\}$ then $f_N$ is a bounded continuous function. Hence $\int f_N d \nu =\lim_{j \to \infty} \int f_N d\nu_j \leq \lim_{j \to \infty} \int f d\nu_j \leq R$. Now let $N \to \infty$ and apply Monotone Convefrgence Theorem.

  • $\begingroup$ I have one more question. Based on this math.stackexchange.com/questions/1142631/… , if I also assume that my fuction f goes to $\infty$ as $|x|\rightarrow \infty$, then is my attempt correct ? $\endgroup$ – vl.ath Dec 8 '18 at 15:09
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    $\begingroup$ @vl.ath In that case tightness is valid but still your argument is circular. You are already assuming that $P_R$ is closed when you say that the limiting measure $\mu \in P_R$. $\endgroup$ – Kavi Rama Murthy Dec 8 '18 at 23:17

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