Portmanteau theorem for non-negative finite Borel measures

Question

Let

• $X$ be a metric space,
• $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
• $\mathcal M_+(X)$ the space of all finite nonnegative Borel measures on $X$,
• $\mathcal M_1(X)$ the space of all Borel probability measures on $X$, and
• $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$.

Then $\mathcal C_b(X)$ is a real Banach space with supremum norm $\|\cdot\|_\infty$. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$. Then $(\mathcal M(X), [\cdot])$ is a Banach space. For $\mu_n,\mu \in \mathcal M(X)$, we define the weak convergence $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X). $$

Then we have

Portmanteau theorem Let $\mu_n,\mu \in \mathcal M_1(X)$. Then the following statements are equivalent.

• (1) $\mu_n \rightharpoonup \mu$.
• (2) $\lim_n \int_X f \mathrm d \mu_n = \int_X f \mathrm d \mu$ for all bounded Lipschitz functions $f:X \to \mathbb R$.
• (3) $\liminf_n \int_X f \mathrm d \mu_n \ge \int_X f \mathrm d \mu$ for all lower semi-continuous functions $f:X \to \mathbb R$ that are bounded from below.
• (4) $\limsup_n \int_X f \mathrm d \mu_n \le \int_X f \mathrm d \mu$ for all upper semi-continuous functions $f:X \to \mathbb R$ that are bounded from above.
• (5) $\liminf_n \mu_n(O) \ge \mu(O)$ for all open subsets $O$ of $X$.
• (6) $\limsup_n \mu_n(C) \le \mu(C)$ for all closed subsets $C$ of $X$.
• (7) $\lim_n \mu_n(A) = \mu(A)$ for all Borel subsets $A$ of $X$ such that $\mu(\partial A) = 0$.

Now I would like to extend above result to $\mathcal M_+(X)$, i.e., we replace $\mu_n,\mu \in \mathcal M_1(X)$ by $\mu_n,\mu \in \mathcal M_+(X)$ such that $\mu_n(X) \to \mu(X)$. Could you have a check on my attempt?

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Proof: Let's prove that $(1) \implies (2) \implies \cdots \implies (7) \implies (1)$.

First, we consider $\mu(X)=0$.

• Assume (1).
• We have $\int_X |f| \mathrm d \mu_n \le \|f\|_\infty \mu_n(X)$. Then (2) holds.
• there is an increasing sequence $(g_m)$ of $m$-Lipschitz continuous bounded functions such that $g_m \nearrow f$ pointwise. Then $$ \int f \mathrm d \mu_n \ge \int g_m \mathrm d \mu_n \quad \forall m \in \mathbb N. $$ Then $$ \liminf_n \int f \mathrm d \mu_n \ge \liminf_n \int g_m \mathrm d \mu_n = \int g_m \mathrm d \mu \quad \forall m \in \mathbb N. $$ Then $$ \begin{align} \liminf_n \int f \mathrm d \mu_n &\ge \liminf_m \int g_m \mathrm d \mu \\ &\ge \int \liminf_m g_m \mathrm d \mu \quad \text{by Fatou Lemma} \\ &= \int f \mathrm d \mu. \end{align} $$ Then (3) holds.
• Clearly, (4) holds.
• (5) holds trivially because $\mu(O) \le \mu(X)=0$.
• We have $\limsup_n \mu_n(C) \le \limsup_n \mu_n(X) =\lim_n \mu_n(X)=0=\mu(C) \le \mu(X)=0$. Then (6) holds.
• We have $\lim_n \mu_n(A) \le \lim_n \mu_n(X) =0 =\mu(A) \le \mu(X)=0$. Then (7) holds.
• We have $\int_X |f| \mathrm d \mu_n \le \|f\|_\infty \mu_n(X)$. Then (1) holds.

Second, we consider $\mu(X)>0$. Because $\mu_n(X) \to \mu(X)$, we assume WLOG that $\mu_n(X)>0$ for all $n$. Let $\nu_n := \frac{\mu_n}{\mu_n(X)}$ and $\nu_n := \frac{\mu}{\mu(X)}$. Then $\nu, \nu_n \in \mathcal M_1(X)$. Because $\mu_n(X) \to \mu(X)$, we have $\mu_n \rightharpoonup \mu \iff \nu_n \rightharpoonup \nu$. We apply Portmanteau theorem for $\nu_n, \nu$ to get the equivalence. We multiply those equalities/inequalities for $\nu_n, \nu$ by $\mu_n(X)$ and $\mu(X)$ respectively. Then we get the desired result for $\mu_n, \mu$. This completes the proof.

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Remark: As shown here, $\mu_n \rightharpoonup \mu$ implies $\liminf_n [\mu_n] \ge [\mu]$, so we can actually replace replace "$\mu_n,\mu \in \mathcal M_1(X)$" by "$\mu_n,\mu \in \mathcal M_+(X)$ such that $\limsup_n [\mu_n] \le [\mu]$".