Since $v$ is Continuous:$$\int v(x) \mu(dx): \mu \mapsto \int v(x) \mu(dx)$$ is lower semicontinuous.

This is because we can write $v$ as the pointwise limit of nondecreasing family $(v_{n})_{n\in \mathbb{N}}$ of continuous real-valued functions by

Then using this we can prove the following lemma

https://math.stackexchange.com/questions/3859243/characterization-of-weak-convergence-with-lower-semicontinuity \begin{proof} We first know that by the definition of lower semicontinuity that for l.s.c. $f$, the set $$\{ x\in X:f(x)>\alpha \},\,\alpha \in \mathbb{R}$$is Open. Hence, we can then apply Portmanteau’s Theorem to get $$\mu(\{ x\in X:f(x)>\alpha \})\le\liminf_{ k \to \infty }\mu_{k}(\{ x\in X:f(x) >\alpha \}) .$$Now, integrating over the positive reals and applying Fatou’s Lemma we get $$\begin{align*} \int\limits _{0}^{\infty}\mu(\{ x\in X:f(x)>\alpha \}) \, d\alpha &\le \int\limits _{0}^{\infty}\liminf_{ k \to \infty } \mu_{k}(\{ x\in X:f(x)>\alpha \}) \, d\alpha\\ &\le \liminf_{ k \to \infty } \int\limits _{0}^{\infty}\mu_{k}(\{ x\in X:f(x)>\alpha \}) \, d\alpha \end{align*}$$where the first inequality is simply by monotonicity of the integral and the second is by Fatou’s. Hence, $$\int\limits _{\mathbb{X}}f \, d\mu \le\liminf_{ n \to \infty }\int\limits f \, d\mu_{n}$$ \end{proof}

\begin{proof} By Monotone Convergence Theorem we have $$\begin{align*} \int\limits v(x) \, \mu(dx) &= \lim_{ n \to \infty } \int\limits v_{n}(x) \, \mu(dx) &\text{(MCT)}\\ &= \lim_{ n \to \infty } \left( \lim_{ k \to \infty } \int\limits v_{n}(x) \, \mu_{k}(dx) \right)&\text{(Portmanteau)}\\ &\le \lim_{ n \to \infty } \left( \liminf_{ k \to \infty } \int\limits v_{n}(d) \, \mu_{k}(dx) \right) &\text{(w.c of }\mu)\\ &\le \liminf_{ k \to \infty } \int\limits v(x) \, \mu_{k}(dx) &\text{(MCT)} \end{align*}$$so we first apply MCT to get an integral in terms of a continuous and bounded function. This allows us to apply Portmanteau’s Theorem to then write the integral as a limit of the measure, we then get the inequality from the fact that $v_{n}\in C_{b}(\mathbb{X})$ meaning the integral itself is continuous and bounded by the Feller Property property of $\mu_{k}$ and hence is lower semicontinuous. Then we apply MCT again to take the limit of the function.

\end{proof}

Therefore, if $\mu_n$ is a sequence in $F_m$ converging to $\mu$; we have that$$m \geq \liminf_{n \to \infty} \int \mu_n(dx) v(x) \geq \int \mu(dx) v(x)$$ and therefore  $\int \mu(dx) v(x) \leq m$. This shows that $\mu \in F_{m}$, and therefore the set $F_m$ is Closed.