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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 for finite non-negative Borel measures, i.e.,

Portmanteau theorem Let $\mu_n,\mu \in \mathcal M_+(X)$ such that $\mu_n(X) \to \mu(X)$. 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-continuous 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 $\mu(\partial A) = 0$.

I would like to ask if above theorem extends to $\mathcal M(X)$, i.e., we replace "$\mu_n,\mu \in \mathcal M_+(X)$ s.t. $\mu_n(X) \to \mu(X)$" by "$\mu_n,\mu \in \mathcal M(X)$ s.t. $[\mu_n] \to [\mu]$". If not, are there some mild conditions on $X$ so that such an extension is possoble?

Thank you so much for your elaboration!

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    $\begingroup$ Take $\mu_n = -\delta_{1/n}\to -\delta_0:=\mu$ and $O = (0,1)$. Then $\liminf_n \mu_n(O) = -1 < 0 = \mu(O) $. $\endgroup$ Commented Nov 4, 2022 at 21:05
  • $\begingroup$ @OtisChodosh Thank you for your counter-example to (5). What if we replace $\liminf_n \mu_n(O) \ge \mu(O)$ by $\liminf_n |\mu_n| (O) \ge |\mu| (O)$ where $|\mu_n|$ is the variation of $\mu$? $\endgroup$
    – Analyst
    Commented Nov 4, 2022 at 21:18
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    $\begingroup$ As Otis Chodosh has shown condition (3), (4), (5) and (6) don't make sense for non positive measures. Only (7) may be correct, if you replace $\mu(\partial A)$ by $|\mu|(\partial A)$. Trying to replace $\mu$ by $|\mu|$ gives results about $|\mu|$, not $\mu$. $\endgroup$ Commented Nov 4, 2022 at 22:23
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    $\begingroup$ @DieterKadelka Assume $X$ is locally compact. Surprisingly, Proposition 2.8. in this paper shows that if $\mu_n \rightharpoonup \mu$ and $\limsup_n[\mu_n] \le [\mu]$ then $\mu^+_n \rightharpoonup \mu^+$ and $\mu^-_n \rightharpoonup \mu^-$. $\endgroup$
    – Analyst
    Commented Nov 4, 2022 at 22:56
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    $\begingroup$ The extension to all of $\mathcal{M}(S)$ (complex measures or signed measures of total finite variation) does not hold in general. There is however Prohorov's theorem which can be found in for example, Bogachev's Measure Theory, vol 2. $\endgroup$ Commented Nov 4, 2022 at 22:57

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