How do I prove there exists such a probability measure? Let $X$ be a metric space and $\{P_n\}$ be a sequence of probability (Borel) measures on $X$.
Assume that $\lim_{n\to \infty} \int_X g dP_n$ exists for each $g\in C_b(X,\mathbb{R})$, and denote this by $I(g)$.
Note that $I(g)\geq 0$ if $g\geq 0$. Hence $I:C_b(X,\mathbb{R})\rightarrow \mathbb{R}$ is a positive linear functional.
Also, assume that $I(g_k)\to 0$ for every monotonically decreasing sequence $g_{k+1}\geq g_k\geq \cdots \geq 0$ such that $g_k\to 0$ pointwise.
In this situation, how do I prove that there exists a probability (Borel) measure $P$ on $X$ such that $I(g)=\int_X gdP$ for each $g\in C_b(X,\mathbb{R})$?
I know Lusin’s theorem and Riesz-Kakutani representation theorem. I tried to use these two theorems to prove the existence of such $P$, but it did not work well. How do I prove this? Thank you in advance!
 A: Let $\tilde{X}$ denote the Stone-Cech compactification of $X$.  Then $\tilde{X}$ is a compact Hausdorff space, and, by density of $X$ in $\tilde{X}$, the dual of $C_{b}(X)$ is "naturally" isomorphic to the dual of $C_{b}(\tilde{X})$.  In particular, $P_{n} \rightharpoonup P$ with respect to $C_{b}(\tilde{X})$ for some probability measure $P$ on $\tilde{X}$.  
We claim that $P(X) = 1$.  (Note that $P(\tilde{X}) = 1$ since the constant functions are in $C_{b}(\tilde{X})$.)  To see this, it suffices to show that, for each $\epsilon > 0$, there is a $K \subseteq X$ such that $P(K) \geq 1 - \epsilon$. 
To do this, define a sequence $(g_{k})_{k \in \mathbb{N}} \subseteq C_{b}(X)$ such that 
\begin{equation*}
g_{k}(x) = \frac{\text{dist}(x,B(0,k))}{\text{dist}(x,B(0,k)) + \text{dist}(x,X \setminus B(0,k + \frac{1}{2}))}.
\end{equation*}
Observe that $g_{k}(x) = 0$ if $x \in B(0,k)$, $g_{k}(x) = 1$ if $x \in X \setminus B(0,k + \frac{1}{2})$, and $g_{k} \in [0,1]$ everywhere.  Moreover, $g_{k} \geq g_{k + 1}$ and $\lim_{k \to \infty} g_{k}(x) = 0$ if $x \in X$.
Observe, additionally, that $P(B(0,k + 1)) \geq \int_{\tilde{X}} (1 - g_{k}(x)) \, P(dx) = 1 - I(g_{k})$.  Thus, for each $\epsilon > 0$, there is a $k \in \mathbb{N}$ such that $P(B(0,k + 1)) \geq 1 - \epsilon$.  This completes the proof.      
All of these ideas can be found in these very nice lecture notes: https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf
