Let $E$ be a Hausdorff topological space and $X:=C_b (E)$ the space of $\mathbb R$-valued continuous bounded functions on $E$. We endow $X$ with the supremum norm $\| \cdot \|_\infty$. Then $(X, \| \cdot \|_\infty)$ is a Banach space. I'm trying to prove below convergence result.
Theorem: Let $\mu, \mu_n$ be non-negative finite Borel measures such that $\mu_n \overset{\ast}{\rightharpoonup} \mu$. Let $f,f_n \in X$ such that $f_n \to f$ in $\| \cdot \|_\infty$. Then $\int_E f_n \mathrm d \mu_n \to \int_E f \mathrm d \mu$.
Could you have a check on my attempt?
My attempt: Let $X^*$ be the continuous dual of $X$. We define $L_n, L: X \to \mathbb R$ by $$ \langle L_n, g \rangle := \int_E g \mathrm d \mu_n \quad \text{and} \quad \langle L, g \rangle := \int_E g \mathrm d \mu_n \quad \forall g \in X. $$
Then $L_n, L \in X^*$ with $\|L_n\|_{E^*} = \mu_n(E)$ and $\|L\|_{E^*} = \mu(E)$. By definition of weak convergence of measures, $\mu_n \overset{\ast}{\rightharpoonup} \mu$ if and only if $\langle L_n, g \rangle \to \langle L, g \rangle$ for all $g \in X$. This implies $\mu_n \overset{\ast}{\rightharpoonup} \mu$ if and only if $L_n \to L$ in the weak$^*$ topology $\sigma (X^*, X)$. We also have $f_n \to f$ in $\| \cdot \|_\infty$.
The claim then follows from below proposition taken from Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations.
Proposition 3.13. Let $E$ be a normed space and $\left(f_{n}\right)$ a sequence in $E^{\star}$. Then
- $\left[f_{n} \stackrel{\star}{\rightarrow} f\right.$ in $\left.\sigma\left(E^{\star}, E\right)\right] \Leftrightarrow\left[\left\langle f_{n}, x\right\rangle \rightarrow\langle f, x\rangle, \forall x \in E\right]$.
- If $f_{n} \rightarrow f$ strongly, then $f_{n} \rightarrow f$ in $\sigma\left(E^{\star}, E^{\star \star}\right)$. If $f_{n} \rightarrow f$ in $\sigma\left(E^{\star}, E^{\star \star}\right)$, then $f_{n} \stackrel{\star}{\rightarrow} f$ in $\sigma\left(E^{\star}, E\right)$.
- If $f_{n} \stackrel{\star}{\rightarrow} f$ in $\sigma\left(E^{\star}, E\right)$ then $\left(\left\|f_{n}\right\|\right)$ is bounded and $\|f\| \leq \lim \inf \left\|f_{n}\right\|$.
- If $f_{n} \stackrel{\star}{\rightarrow} f$ in $\sigma\left(E^{\star}, E\right)$ and if $x_{n} \rightarrow x$ strongly in $E$, then $\left\langle f_{n}, x_{n}\right\rangle \rightarrow\langle f, x\rangle$.
