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 ?
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.