At the beginning of Chapter 5 of Ambrosio's Gradient Flows book, he introduces the idea of a narrowly convergent sequence of measures as...
"A sequence $(\mu_n) \subset \mathcal{P}(x)$ is narrowly convergent to $\mu \in \mathcal{P}(x)$ as $n\to\infty$ if
$$ \lim_{n\to\infty}\int_{X}f(x)d\mu_n(x) = \int_X f(x)d\mu(x) $$ for every function $f \in C_b^0(X)$.
Ambrosio then claims that it is sufficient to check this on a subset of $C_b^0(X)$ such that $span(\mathcal(F))$ is a uniformly dense subset of $C_0^b(X)$. I have attempted a proof, but I honestly dont know if it is correct, or how I would be intended to show it in the first place. The attempt follows.Suppose there exists $\mathcal{F}\subset C_0^b(X)$ such that $span(\mathcal{F})$ is uniformly dense in $C_0^b(X)$. Let $g \in span(\mathcal{F})$. Then
$$ \lim_{n\to\infty}\int g\mu_n = \lim_{n\to\infty}\int \sum_{i=1}^{\infty}c_ig_i d\mu_n$$ However, as each $g_i \in \mathcal{F}$ and we have narrow convergence on $\mathcal{F}$,
$$\lim_{n\to\infty}\int \sum_{i=1}^{\infty}c_ig_i d\mu_n = \int \sum_{i=1}^{\infty}c_ig_i d\mu = \int g d\mu$$.
We have now shown the claim holds on $span(F)$ itself. Now let $f \in C_0^b(X)$. Fix $\varepsilon>0$ and by uniform density we see that there must exist $g \in span(\mathcal(F))$ for which.
$$ \| f - g\|_{\infty} < \varepsilon$$
Therefore, $$|\lim_{n\to\infty}\int f-g d\mu_n| < \varepsilon$$. Splitting the integral we see that $$ | \lim_{n\to\infty}\int f d\mu_n - \int g d\mu | < \varepsilon.$$ Because the choice of $\varepsilon$ was arbitrary, we can show that $$ |\lim_{n\to\infty} \int f d\mu_n - \int f d\mu| < \varepsilon $$ and therefore $$\lim_{n\to\infty} \int f d\mu_n \to \int f d\mu.$$
I feel most like I have handwaved the later half. My largest concern is the span definition -- surely with an infinite set the number of coefficients I would need could blow up -- and the sum inside the integral could be infinite, no?
Everything with the span is fine. Only ordinary finite linear combinations occur in the span, and the integral is linear. For the latter half, note that
$$\bigg|\int f~\mathrm d\mu-\int f~\mathrm d\mu_n\bigg|=\bigg|\int f~\mathrm d\mu-\int g~\mathrm d\mu+\int g~\mathrm d\mu-\int g~\mathrm d\mu_n+\int g~\mathrm d\mu_n-\int f~\mathrm d\mu_n\bigg|$$ $$\leq \bigg|\int f~\mathrm d\mu-\int g~\mathrm d\mu\bigg|+\bigg|\int g~\mathrm d\mu-\int g~\mathrm d\mu_n\bigg|+\bigg|\int g~\mathrm d\mu_n-\int f~\mathrm d\mu_n\bigg|$$ $$\leq \int |f-g|~\mathrm d\mu+\bigg|\int g~\mathrm d\mu-\int g~\mathrm d\mu_n\bigg|+\int |g-f|~\mathrm d\mu_n$$ $$\leq\epsilon+\bigg|\int g~\mathrm d\mu-\int g~\mathrm d\mu_n\bigg|+\epsilon.$$ By narrow convergence, the last part is at most $3\epsilon$ for $n$ large enough.