# A closed (?) subset of the set of probability measures. Is my reasoning correct?

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 ?

• Obviously false if you have strict inequality in the definition of $P_R$. (Take degenerate measures) – Kavi Rama Murthy Dec 8 '18 at 0:03
• No I meant less or equal in the definition of P_R.. sorry typo – 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.
• 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 ? – vl.ath Dec 8 '18 at 15:09
• @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 $\mu \in P_R$. – Kavi Rama Murthy Dec 8 '18 at 23:17