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Let $E$ be a complete locally compact separable metric space and $\mu,\mu_n$ be probability measures on $\mathcal B(E)$. We say $\mu_n\to\mu$ weakly if $$\int f\:{\rm d}\mu_n\to\int f\:{\rm d}\mu\tag1$$ for all bounded continuous $f:E\to\mathbb R$ (write $f\in C_b(E)$) and we say $\mu_n\to\mu$ vaguely if $(1)$ for all compactly supported continuous $f:E\to\mathbb R$ (write $f\in C_c(E)$).

By the Portmanteau theorem, $\mu_n\to\mu$ weakly if and only if $\mu_n\to\mu$ vaguely and $\mu(E)=\lim_{n\to\infty}\mu_n(E)$.

Now, we say that $M\subseteq C_b(E)$ is convergence determining if the assertion that $(1)$ holds for all $f\in M$ implies $\mu_n\to\mu$ weakly. By Proposition 4.4 of Chapter 3 in the book of Ethier/Kurtz, $C_c(E)$ is convergence determining. But that means (by our definition) $\mu_n\to\mu$ vaguely.

Question 1: Am I missing something or does that mean that in the Portmanteau theorem we are able to conclude $\mu_n\to\mu$ vaguely and $\mu(E)=\lim_{n\to\infty}\mu_n(E)$ from $\mu_n\to\mu$ weakly, while we only need to know $\mu_n\to\mu$ vaguely in order to conclude $\mu_n\to\mu$ weakly?

Now assume we have complete locally compact separable metric spaces $E_1,\ldots,E_k$. I would like to show that $$\tilde M:=\left\{\prod_{i=1}^k(f_i\circ\pi_i):f_i\in C_0(E_i)\right\},$$ where $\pi_i:\times_{j=1}^dE_j\to E_i$ is the projection onto the $i$th coordinate) is convergence determining (on $\times_{i=1}^kE_i$). Again in Ethier/Kurtz, Proposition 4.6 of Chapter 3, we find the claim that if $M_i\subseteq C_b(E_i)$ is convergence determining (on $E_i$), then $$M:=\left\{\prod_{i=1}^k(f_i\circ\pi_i):f_i\in M_i\cup\color{red}{\left\{1\right\}}\right\}$$ is convergence determining. I would clearly take $M_i=C_0(E_i)$ (which is convergence determining by the former result and since $C_c(E_i)$ is dense in $C_0(E_i)$ by definition). However, Ethier/Kurtz allow $f_i$ to be the constant function $1$ (which is clearly not in $C_0(E_i)$, unless $E_i$ is compact). They state Proposition 4.6 in the more general case of a countable product, but I don't think that this is the reason why we need to allow $f_i\equiv 1$. On the other hand, taking $k=1$ yields a result weaker than what we assume beforehand (we start with $M_1$ being convergence determining and conclude $M_1\cup\left\{1\right\}$ is convergence determining).

Question 2: What am I missing here? Is my $\tilde M$ convergence determining?

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  • $\begingroup$ The vague topology uses continuous functions that vanish at infinity, not only continuous functions with compact support. Notice that the two classes of "test functions" give different topologies. Where did you take your definition from? Also, where did you take that version of the Portmanteau theorem from? The one I know does not say what yours says. $\endgroup$
    – Alex M.
    Apr 20 '19 at 13:45
  • $\begingroup$ For question 1, note that $\mu_n(E) = \mu(E) = 1$ since these things are assumed to be probability measures. $\endgroup$ Apr 20 '19 at 20:16
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    $\begingroup$ @AlexM. It's true that the two classes of "test functions" give different topologies on the space complex Radon measures but as far as convergence goes, when restricted to probability measures the two classes of "test functions" at least give the same notion of convergence which is possibly the reason OP (or their source) isn't careful here (you can prove this using a similar technique as in my answer to the second question of OP). $\endgroup$ Apr 20 '19 at 22:53
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Question 1: The point is that in the definition of a set being convergence determining you make the additional assumption that $\mu_n$ and $\mu$ are probability measures and so the thing you call the Portmanteau Theorem (which is true but is not the thing I would call the Portmanteau Theorem) says that $\mu_n \to \mu$ vaguely if and only if $\mu_n \to \mu$ weakly (under the extra assumption). Unfortunately, the space of probability measures isn't closed in the vague topology on the space of, say, finite measures but is closed in the weak topology. To make this concrete, take $E = \mathbb{R}$ and $\mu_n = \delta_n$ to be the point mass at $n$. Then $\mu_n \to 0$ vaguely and not weakly, the problem being that $C_0(E)$ is failing to capture "the escape of mass to infinity" which is exactly what the extra condition in what you call the Portmanteau Theorem fixes.

Question 2: The only thing that we can't do in the same way when we don't allow ourselves the function $1$ in the proof of Proposition 4.6 of Chapter 3 of Ethier and Kurtz is immediately conclude that for $f_i \in M_i$, $$\int f_i d\mu_n \to \int f_i d\mu$$ whenever $\mu_n, \mu$ satisfy $(1)$ for all $f \in \tilde{M}$. The extra structure on your equivalent of $M_i$ allows you to get around this. For (notational) convenience, I will demonstrate the idea in the case $k = 2$ and $i = 1$. I imagine this proof can be simplified significantly.

Fix an $\varepsilon > 0$. Also fix strictly increasing sequences $(K_n^i)_{n \geq 1}$ of compact subsets of $E_i$ that cover $E_i$ and satisfy $d(\partial K_n^i, \partial K_{n+1}^i) > 0$ (such compact exhaustions exists since $E_i$ is a locally compact separable metric space, see here). Then since $\mu$ is a probability measure, there is an $N_\varepsilon$ such that $$\mu(K_{N_\varepsilon}^1 \times K_{N_\varepsilon}^2) \geq 1 - \frac{\varepsilon}{2}.$$

Now take continuous functions $f_i^{(\varepsilon)}$ valued in $[0,1]$ such that $f_i^{(\varepsilon)} = 1$ on $K_{N_\varepsilon}^i$ and $f_i^{(\varepsilon)} = 0 $ outside of $K_{N_\varepsilon+1}^i $. Then $f_i^{(\varepsilon)} \in C_c(E_i) \subseteq C_0(E_i)$ so $$\int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu_n \to \int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu \geq 1 - \frac{\varepsilon}{2}$$ so for large enough $n$, $$\mu_n(K_{N_\varepsilon+1}^1 \times K_{N_\varepsilon+1}^2) \geq \int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu_n \geq 1 - \varepsilon.$$ Now pick continuous, compactly supported functions $g_i^{(\varepsilon)}$ valued in $[0,1]$ such that $g_i^{(\varepsilon)} = 1$ on $K_{N_\varepsilon+1}^i$.

Then, we can bound \begin{align} \bigg | \int_{E_1 \times E_2} f_1 d \mu_n - \int_{E_1 \times E_2} f_1 d \mu \bigg| & \leq \bigg | \int f_1 - f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d \mu \bigg| + \bigg | \int f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d( \mu - \mu_n ) \bigg| \\ &+ \bigg | \int f_1 - f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d \mu_n \bigg| \end{align} Now the first and last terms are each bounded by $\|f\|_\infty \varepsilon$ for $n$ sufficiently large, by our choice of $g_i^{(\varepsilon)}$. The middle term goes to $0$ since $f_1 g_1^{(\varepsilon)} \in C_0(E_1)$ and $g_2^{(\varepsilon)} \in C_0(E_2)$ and so for large enough $n$ we have $$\bigg | \int_{E_1 \times E_2} f_1 d \mu_n - \int_{E_1 \times E_2} f_1 d \mu \bigg| \leq (2 \|f_1\|_\infty + 1) \varepsilon$$ which gives the desired convergence.

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    $\begingroup$ @0xbadf00d As I say in the answer, the point is that this is the only thing that is not evident in your case if we try to follow the proof of proposition 4.6 to see where it goes wrong (which should be the first thing to try when you have slightly different hypotheses like this). So once you've understood my answer just read that proof in Ethier and Kurtz but with your hypotheses and you'll see it just works. $\endgroup$ Apr 22 '19 at 7:26
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    $\begingroup$ Actually it is not hard to see that $\bigcup_{n \in \mathbb{N}} \times_{i=1}^k K_n^i = \times_{i=1}^k \bigcup_{n \in \mathbb{N}} K_n^i$ here because the $K_n^i$ are chosen to be increasing (that was one reason for that choice). For simplicity in the case $k = 2$, $\bigcup_{n \in \mathbb{N}} K_n^1 \times \bigcup_{n \in \mathbb{N}} K_n^2 = \bigcup_{(n,m) \in \mathbb{N}^2} K_n^1 \times K_m^2$ and e.g. if $n < m$ then $K_n^1 \subset K_m^1$ so $K_n^1 \times K_m^2 \subset K_m^1 \times K_m^2$. The case $m<n$ is almost identical so you get the nontrivial inclusion. $\endgroup$ May 14 '19 at 10:33
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    $\begingroup$ Fine, take $f_i^{(\varepsilon)} = \min \{1, d(K_{N_\varepsilon^i}, (K_{N_\varepsilon}^i)^c)^{-1} \cdot d(x, (K_{N_\varepsilon}^i)^c)\}$. I made a small error because I didn't want to spend any time on that comment. This is standard metric spaces stuff that you should think about for yourself. $\endgroup$ May 15 '19 at 17:23
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    $\begingroup$ Check the discussion immediately after the definitions of convergence determining and separating again. The first implies the second. $\endgroup$ May 17 '19 at 14:05
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    $\begingroup$ @0xbadf00d Yes, this proof is really for $C_c(E_i)$ instead of $C_0(E_i)$. I had that in mind when writing it since it is no more difficult so I wanted the proof to work in that case too. $\endgroup$ Jun 23 '19 at 12:29

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