Lemma 1: If $X$ is separable, then convergence in $d_P$ is equivalent to weak convergence.
Lemma 2: Let $\mu, \mu_1,\mu_2,\ldots \in \mathcal M$. Then $\mu_i \to \mu$ weakly if and only if $\int f \mathrm d \mu_i \to \int f \mathrm d \mu$ for all uniformly continuous and bounded functionals $f$.
Notice that $x \mapsto \delta_{x}$ is a homeomorphism from $X$ onto $\left\{\delta_{x} \mid x \in X\right\}$. If $\mathcal{M}$ is separable, then so is $\left\{\delta_{x} \mid x \in X\right\}$ and thus is $X$. Let's prove the other direction. Let $D$ be a countable dense subset of $X$. Let
$$
\mathcal D := \{\alpha_1 \delta_{a_1} + \cdots + \alpha_k \delta_k \mid a_1, \ldots, a_k \in D \text{ and }\alpha_1,\ldots, \alpha_k \in \mathbb Q_{\ge 0}\}
$$
Then $\mathcal D$ is countable. Let's prove that $\mathcal D$ is dense in $\mathcal{M}$. Fix $\mu \in \mathcal{M}$. For each $m\ge 1$, we pick $k_m$ such that
$$
\mu \left ( \bigcup_{j=1}^{k_m} B(a_j, 1/m) \right ) \ge \mu(X)-1/m.
$$
Let $A_{1}^{m} := B\left(a_{1}, 1 \right)$ and
$$
A_{j}^{m} := B\left(a_{j}, 1 / m\right) \setminus \bigcup_{i=1}^{j-1} B\left(a_{i}, 1 / m\right) \quad \forall j=2, \ldots, k_{m}.
$$
Then $(A_j^m)_{j=1}^{k_m}$ is disjoint, and their union is equal to $\bigcup_{i=1}^{k_m} B\left(a_{i}, 1 / m\right)$ for all $m\ge 1$. In particular,
$$
\mu(X) \ge\sum_{j=1}^{k_{m}} \mu\left(A_{j}^{m}\right) \ge \mu(X)-1/m \quad \forall m \ge 1.
$$
We approximate
$$
\mu\left(A_{1}^{m}\right) \delta_{a_{1}}+\cdots+\mu\left(A_{k_{m}}^{m}\right) \delta_{a_{k_{m}}} \quad \text{by} \quad
\mu_{m}:=\alpha_{1}^{m} \delta_{a_{1}}+\cdots+\alpha_{k_{m}}^{m} \delta_{a_{k_{m}}}
$$
such that $\alpha_{1}^{m}, \ldots, \alpha_{k_{m}}^{m} \in \mathbb Q_{\ge 0}$ and
$$
\sum_{j=1}^{k_{m}} \left|\mu\left(A_{j}^{m}\right)-\alpha_{j}^{m}\right|<2 / m.
$$
Let $g$ be a uniformly continuous and bounded functional on $X$. By Lemmas 1 and 2, we need to prove $\int g \mathrm d \mu_m \to \int g \mathrm d \mu$ as $m \to \infty$. In deed,
$$
\begin{align}
& \left |\int g \mathrm d \mu_m - \int g \mathrm d \mu \right | \\
= &\left | \sum_{j=1}^{k_{m}}\alpha_j^m g(a_j) - \int g \mathrm d \mu \right| \\
\le & \left | \sum_{j=1}^{k_{m}} \mu\left(A_{j}^{m}\right) g(a_j) - \int g \mathrm d \mu \right| + \frac{2}{m} \sup_j | g(a_j) | \\
\le& \left | \int \sum_{j=1}^{k_{m}} g(a_j) 1_{A_j^m} \mathrm d \mu - \int g \mathrm d \mu \right| + \frac{2}{m} \|g\|_\infty \\
=& \left | \int \sum_{j=1}^{k_{m}} [g(a_j) -g] 1_{A_j^m} \mathrm d \mu + \int g 1_{(\bigcup_{j=1}^{k_{m}} A_j^m)^c} \right| + \frac{2}{m} \|g\|_\infty\\
\le& \sum_{j=1}^{k_{m}} \int | g(a_j) -g| 1_{A_j^m} \mathrm d \mu + \|g\|_\infty \mu \left ( \left (\bigcup_{j=1}^{k_{m}} A_j^m \right )^c \right ) + \frac{2}{m} \|g\|_\infty\\
\le& \sum_{j=1}^{k_{m}} \sup_{x\in A_j^m} | g(a_j) -g(x)| \mu (A_j^m) + \frac{1}{m} \|g\|_\infty + \frac{2}{m} \|g\|_\infty.
\end{align}
$$
Each $A_{j}^{m}$ is contained in a ball with radius $1 / m$ around $a_{j}$. Since $g$ is uniformly continuous, for every $\varepsilon>0$ there is a $\delta>0$ such that $|g(y)-g(x)|<\varepsilon$ whenever $d(x,y)<\delta$, so $\left|g(a_{j}) - g(x)\right|<\varepsilon$ for all $j$ and $x \in A_{j}^{m}$. Then for $m$ such that $1/m < \min\{\varepsilon, \delta\}$, it follows from the above computation that
$$
\left|\int g \mathrm d \mu_{m}-\int g \mathrm d \mu\right| \leq \varepsilon \mu(X) + \frac{3}{m} \|g\|_{\infty}.
$$
This completes the proof.
- Consider the map
$$
g(x) := \min\{d(x, U^c), 1\} \quad \forall x \in X.
$$
Notice that $\min\{a,c\}- \min\{b,c\} \le a-b$ for all $a\ge b$, so $g$ is $1$-Lipschitz. We define
$$
g_n := \sqrt[n]{g} \quad \forall n \ge 1.
$$
Then $0 \le g_n \nearrow 1_U$. By monotone convergence theorem, there is $N$ such that
$$
\int (1_U-g_N) \mathrm d \mu < \varepsilon/2.
$$
Notice that the map $[0, \infty) \to \mathbb R,x \mapsto x^{\alpha}$ is unifomly continuous for all $\alpha \in [0,1]$. This implies $g_N$ is uniformly continuous. On the other hand, $\operatorname{Lip}_{b}(X)$ is dense (w.r.t. $\| \cdot \|_\infty$) in the space of uniformly continuous bounded maps. So there is $f \in \operatorname{Lip}_{b}(X)$ such that
$$
f\le g_N \quad \text{and} \quad\|f-g_N\|_\infty < \varepsilon/(2 \mu(X)).
$$
It follows that
$$
\begin{align}
\int (1_U-f) \mathrm d \mu &= \int (1_U-g_N) \mathrm d \mu + \int (g_N-f) \mathrm d \mu \\
&\le \varepsilon/2 + \varepsilon/2 = \varepsilon.
\end{align}
$$
- Every finite Borel measure on a metric space is outer regular, so there is $U$ open in $X$ such that $\mu(U\setminus A) < \varepsilon/2$. By 1., there is $f \in \operatorname{Lip}_{b}(X)$ such that $\int |1_U - f| \mathrm d \mu<\varepsilon/2$. The claim the follows.
Best Answer
Let $\mathcal N$ be the collection of all negative sets. Notice that $\mathcal N$ is closed under countable union and countable intersection. Let $\alpha := \inf \{\mu(B) \mid B \in \mathcal N\}$. Let $(N_n)$ be a sequence in $\mathcal N$ such that $\mu(N_n) \searrow \alpha$. Let $N := \bigcup_n N_n$. Clearly, $N$ is negative and thus $N \in \mathcal N$. We have $\alpha \le \mu(N) \le \mu(N_n)$, so $\mu(N) = \alpha$.
Let $P := X \setminus N$. We will prove that $P$ is a positive. Assume the contrary that $P$ is not positive. Then $\exists B_0 \in \mathcal B(X)$ s.t. $B_0 \subset P$ and $\mu (B_0)<0$. Then $B_0$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0$ and $\mu (B)>0$.
We pick the smallest $k_1 \in \mathbb N$ such that $\exists B_1 \in \mathcal B(X)$ s.t. $B_1 \subset B_0$ and $\mu (B_1) \ge 1/k_1$. Then $\mu(B_0\setminus B_1)<0$. Then $B_0\setminus B_1$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0 \setminus B_1$ and $\mu (B)>0$.
We pick the smallest $k_2 \in \mathbb N$ such that $\exists B_2 \in \mathcal B(X)$ s.t. $B_2 \subset B_0 \setminus B_1$ and $\mu (B_2) \ge 1/k_2$. Then $\mu(B_0\setminus B_1\setminus B_2)<0$. Then $B_0\setminus B_1 \setminus B_2$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0 \setminus B_1 \setminus B_2$ and $\mu (B)>0$.
Inductively, $\exists B_{n+1} \in \mathcal B(X)$ s.t. $B_{n+1} \subset B_0 \setminus B_1 \setminus \cdots \setminus B_n$ and $\mu(B_{n+1})>1/k_{n+1}$. Clearly, $k_n \to \infty$. If not, $\mu(\bigcup_{n=1}^\infty B_n) = \infty$.
Let $C := \bigcup_{n \ge 1} B_n \subset B_0$. Then $\mu(C) >0$. We claim that $B_0 \setminus C$ is negative. If not, $\exists D \in \mathcal B(X)$ s.t. $D \subset B_0 \setminus C$ and $\mu (D)>0$. Take $M \in \mathbb N$ such that $\mu(D) > 1/M$. There is $N \in \mathbb N$ such that $$ \frac{1}{k_N-1} \le \frac{1}{k_{N}} + \frac{1}{M} \le \mu(B_N \cup D). $$
This contradicts the minimality of $k_N$, so $B_0\setminus C$ is negative. This contradicts the minimality of $\alpha$. Hence $P$ is positive.
Let $(N', P')$ be another Hahn decomposition of $\mu$. Let $B \in \mathcal B(X)$. Then $$ \mu(B\cap N) = \mu(B\cap N\cap N') + \mu(B\cap N \cap P') = \mu(B\cap N\cap N'). $$ By symmetry, $\mu(B\cap N') = \mu(B\cap N\cap N')$. So $\mu(B\cap N) = \mu(B\cap N')$. Similarly, $\mu(B\cap P) = \mu(B\cap P')$.