Is a uniformly tight weakly convergent net of signed measures uniformly bounded?

Question

Let $X$ be a non-compact locally compact Hausdorff space and $(\mu_\alpha)_{\alpha \in A}$ a net of signed Radon measures with bounded variation that converges weakly (= relative to $C_b(X)$) to some signed Radon measure $\mu$. Assume that $(\mu_\alpha)_{\alpha \in A}$ is uniformly tight, i.e. for all $\varepsilon > 0$ there is a compact $K \subseteq X$ such that $\sup_\alpha |\mu_\alpha|(X \setminus K) < \varepsilon$. Does it follow that $(\mu_\alpha)_{\alpha \in A}$ is (eventually) uniformly bounded, i.e. $\sup_{\alpha \geq \alpha_0} \lVert \mu_\alpha \rVert < \infty$ for some $\alpha_0$?

Note that

• a set of measures that is uniformly tight need not be uniformly bounded, e.g. $\{ n \delta_x \mid n \in \mathbb{N} \}$. But the sequence forming this set is not weakly convergent.
• if the $\mu_\alpha$ are positive (no need for tightness) then $(\mu_\alpha)_{\alpha \in A}$ is eventually uniformly bounded (because $\lVert \mu_\alpha \rVert = \mu_\alpha(X) = \mu_\alpha 1_X$ converges ($1_X \in C_b(X)$)
• if $(\mu_\alpha)_{\alpha \in \mathbb{N}}$ is a sequence (no need for tightness) then $(\mu_\alpha)_{\alpha \in \mathbb{N}}$ is uniformly bounded (convergent sequences with their limit form a weakly compact set and these are uniformly bounded)
• without tighness, a weakly convergent net need not be eventually uniformly bounded, see here.

To prove the conjecture, one could take for $\varepsilon = 1$ a compact set $K$ such that $\sup_\alpha |\mu_\alpha|(X \setminus K) \leq 1$ and split $\lVert \mu_\alpha \rVert = |\mu_\alpha|(X) = |\mu_\alpha|(X \setminus K) + |\mu_\alpha|(K)$. If $|\mu_\alpha|(K)$ converges then we are done. One can also construct a $\psi \in C_b(X)$ with $\psi = 1$ on $K$ and $0 \leq \psi \leq 1$ such that $|\mu_\alpha|(K) \leq |\mu_\alpha| \psi = \int \psi d|\mu_\alpha|$ and check whether this upper bound converges. However, we only know that $\mu_\alpha \psi$ converges. It holds $|\mu_\alpha| \psi = \sup \{ \mu_\alpha \varphi \mid |\varphi| \leq \psi, \varphi \in C_b(X) \}$.

For a counterexample, $\mu_\alpha$ must be necessarily a net (not a sequence) and consist of signed measures (not positive). For $X = \mathbb{R}$ I tried a net such as $m (\delta_0 - \delta_{\frac{1}{n}})$ with index set $\mathbb{N} \times \mathbb{N}$ directed by $(n, m) \leq (n', m')$ defined as $n \leq n', m \in \mathbb{N}$ or as $n \leq n', m \leq m'$, but this net does not weakly converge.

Answer 1

It seems that $(\mu_\alpha)_{\alpha \in A}$ need not be eventually uniformly bounded. The counterexample is a refinement of the construction here.

Consider $X = \mathbb{R}$. For the compact set $K = [0,1] \subseteq X$ let us construct a net $(\mu_\alpha)_{\alpha \in A}$ such that each $\mu_\alpha$ is concentrated on $K$ (so that $(\mu_\alpha)_{\alpha \in A}$ is uniformly tight), $\mu_\alpha \to 0$ weakly, but no tail $(\mu_\alpha)_{\alpha \geq \alpha_0}$ is uniformly bounded.

Let $M(X)$ denote the space of signed Radon measures on $X$. For $\mu \in M(X)$ and $f \in C_b(X)$ set $\mu f := \int f \, d\mu$. Denote by $\mathcal{F}$ the collection of nonempty finite subsets of $C_b(X)$ and for $F = \{ f_1, \dots, f_n \} \in \mathcal{F}$ set $U_F = U_{f_1, \dots, f_n} := \{ \mu \in M(X) \mid |\mu f_1| < 1, \dots, |\mu f_n| < 1 \}$. Then $(U_F)_{F \in \mathcal{F}}$ forms a neighborhood base of $0$ for the weak topology in $M(X)$. The set $\mathcal{F}$ is directed by inclusion $F \leq G :\Leftrightarrow F \subseteq G$.

For each $F = \{ f_1, \dots, f_n \} \in \mathcal{F}$ with pairwise disjoint $f_i$ (so that $n = |F|$) let us explicitely construct a measure $\mu_F$ such that $\mathbb{R} \mu_F \subseteq U_F$. Choose $n+1$ points $x_1, \dots, x_{n+1} \in K$ arbitrarily ($x_i \neq x_j$ for $i \neq j$; note that $K$ is infinite) and form the vectors $v_i := (f_1(x_i), \dots, f_n(x_i)) \in \mathbb{R}^n$ for $i = 1, \dots, n+1$. Then these vectors $v_1, \dots, v_{n+1}$ are linearly dependent, so that there are $a_i \in \mathbb{R}$, $i = 1, \dots, n+1$ with some $a_i \neq 0$ such that $a_1 v_1 + \dots + a_{n+1} v_{n+1} = 0$. (Note that it is possible that $v_i = v_j$ for some $i \neq j$). Set $\mu_F := a_1 \delta_{x_1} + \dots + a_{n+1} \delta_{x_{n+1}}$. Then $\mu_F \neq 0$ and $\mu_F f_j = a_1 f_j(x_1) + \dots + a_{n+1} f_j(x_{n+1}) = 0$ for all $j = 1, \dots, n$. Therefore, $\mu_F \in U_F$ and moreover $\mathbb{R} \mu_F \subseteq U_F$. Since the $x_i$ are pairwise disjoint, the total variation of $\mu_F$ is $\lVert \mu_F \rVert = |a_1| + \dots + |a_{n+1}| > 0$.

Set $\nu_F := \frac{n}{\lVert \mu_F \rVert} \mu_F$ so that $\lVert \nu_F \rVert = n = |F|$. Then

• $\nu_F \to 0$ weakly: Let $U$ be a neighborhood of $0$. Then there is $F_0 \in \mathcal{F}$ such that $U_{F_0} \subseteq U$ (the sets $U_F$ form a neighborhood base at $0$). Then for all $F \geq F_0$ (i.e. $F \supseteq F_0$) it holds $a \mu_F \in U_F \subseteq U_{F_0}$ for all $a \in \mathbb{R}$ and therefore $\nu_F \in U_F \subseteq U_{F_0} \subseteq U$.
• $(\nu_F)_{F \in \mathcal{F}}$ is uniformly tight: all the measures $\nu_F$ are concentrated on $K = [0,1]$, so that $|\nu_F|(X \setminus K) = 0$ for all $F \in \mathcal{F}$ and therefore $\sup_{F \in \mathcal{F}} |\nu_F|(X \setminus K) = 0$.
• for any $F_0 \in \mathcal{F}$ the tail net $(\nu_F)_{F \geq F_0}$ is not uniformly bounded since $\lVert \nu_F \rVert = |F|$ is unbounded: for any $r > 0$ there is $F \geq F_0$ such that $|F| \geq r$; therefore $\sup_{F \geq F_0} \lVert \nu_F \rVert = \infty$.