measure-theorysigned-measuresweak-convergence
Let $X$ be a Polish space and $\mu, \mu_n$ finite signed Borel measures on $X$. Assume that $\mu_n \overset{\ast}{\rightharpoonup} \mu$, i.e.,
$$
\int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu
$$
for all bounded continuous functions $f:X \to \mathbb R$. Let $\mu = \mu^+ – \mu^-$ and $\mu_n = \mu_n^+ – \mu_n^-$ be their Jordan decompositions.
Is it true that $\mu^+_n \overset{\ast}{\rightharpoonup} \mu^+$ and $\mu^-_n \overset{\ast}{\rightharpoonup} \mu^-$?
This is in response to a comment by the OP. It is the version of the Theorem in his posting under the additional assumption that $X$ is a metric space:
Theorem A: Suppose $(S,d)$ is a locally compact separable metric space. Let $(\mu_n,\mu)$ a sequence of finite nonnegative measures. The following statement are equivalent:
- $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu$ (convergence in the topology $\sigma(\mathcal{M}(S),\mathcal{C}_b(S))$)
- $\mu_n\stackrel{v}{\longrightarrow}\mu$ as $n\rightarrow\infty$ (convergence in the topology $\sigma(\mathcal{M}(S),\mathcal{C}_{00}(S))$, and $\mu_n(S)\xrightarrow{n\rightarrow\infty}\mu(S)$.
That (1) implies (2) is obvious.
The proof that (2) implies (1) is based in a well known result:
Theorem L: Let $(S,d)$ be a metric space. For any net $\{\mu_\alpha:\alpha\in D\}\subset\mathcal{M}^+(S)$ and $\mu\in \mathcal{M}^+(S)$,
- $\mu_\alpha\Rightarrow\mu$ if and only if \begin{align} \liminf_\alpha\int f\,d\mu_\alpha\geq \int f\,d\mu\tag{1}\label{one} \end{align} for all $f\in L_b(S)$.
If in addition $(S,d)$ is a locally compact separable metric space,- If $\mu_\alpha\stackrel{v}{\longrightarrow} \mu$, then \eqref{one} holds for all $0\leq f\in L_b(S)$.
Here, $L_b(S)$ the set of all lower semicontinuous functions that are bounded below.
Let $f\in L_b(S)$ with $c\leq f$ for some constant $c$. Then $0\leq f-c\in L_b(S)$ and by the second part of Theorem L, $\liminf_n\int (f-c)\,d\mu_n\geq \int (f-c)\,d\mu$. The assumption $\mu_n(S)\rightarrow\mu(S)$ implies that \begin{align} \liminf_n\int f\,d\mu_n\geq \int f\,d\mu . \end{align} The implication $(2)\Rightarrow(1)$ in Theorem A follows from the first part of Theorem L.
Here is a proof of Theorem L.
Suppose that $\mu_\alpha\Rightarrow\mu$ and let $g\in L_b(S)$ with $g\geq c$. There is a sequence $g_k$ of bounded Lipschitz functions such that $c\leq g_k\leq g_{k+1}\nearrow g$. Hence, for each $k$ \begin{align} \liminf_\alpha\int g\,d\mu_\alpha\geq \liminf_\alpha \int g_k\,d\mu_\alpha =\int g_k\,d\mu. \end{align} As $\mu(S)<\infty$, $\liminf_\alpha\int g\,d\mu_\alpha\geq \int g\,d\mu$ by monotone convergence.
Conversely, suppose $f\in\mathcal{C}_b(S)$. Since $\mathcal{C}_b(S)\subset L_b(S)$, both $f$ and $-f$ are in $L_b(S)$, so \begin{align} \liminf_\alpha\int f\,d\mu_\alpha&\geq \int f\,d\mu\\ \liminf_\alpha\int -f\,d\mu_\alpha&\geq \int -f\,d\mu \end{align} Therefore, $\lim_\alpha \int f\,d\mu_\alpha =\int f\,d\mu$.
For the last statement, let $0\leq f\in L_b(S)$ and let $f_k\in C_b(S)$ be such that $0\leq f_k\nearrow f$ pointwise. Since $S$ is locally compact and separable, there is a sequence of open sets $V_j$ with compact closure such that $\overline{V}_j\subset V_{j+1}\nearrow S$. Choose $v_j\in C_{00}(S)$ so that $\mathbb{1}_{\overline{V}_j}\leq v_j\leq \mathbb{1}_{V_{j+1}}$ and $\operatorname{supp}(v_j)\subset V_{j+1}$. Let $f_{kj}=f_kv_j$; clearly $f_{kj}\in C_{00}(S)$ and $f_{kj}\nearrow f_k$ as $j\nearrow\infty$. Then for all $k$ and $j$ \begin{align} \liminf_\alpha\int f\,d\mu_\alpha\geq \liminf_\alpha\int f_k\,d\mu_\alpha\geq \liminf_\alpha\int f_{kj}\,d\mu_\alpha=\int f_{kj}\,d\mu \end{align} The result follows now by monotone convergence by letting $j\nearrow\infty$ and then $k\nearrow\infty$.
All the arguments presented here depend heavily on the metrizability of $S$. I don't think they can be generalize easily into the more general setting where $(S,\tau)$ is a l.c.H space.
This was initially intended to be a reply to geetha290krm's supposed counterexample to (2) (showing it is $\textbf{not}$ in fact a valid counterexample, and that in the context of $X$ being locally compact Hausdorff, (2) will always in fact be true), but was too long for a comment, so I wrote it up as an answer.
Lets work in $\mathbb{R}_{> 1}$, for simplicity (with the Lebesgue measure).
Let $0 \leq \chi_{n} \leq \frac{1}{n}$ be a bump function equal to $1/n$ on $[n + \epsilon_{n}, n+1 - \epsilon_{n}] $, with support contained in $[n + \epsilon_n^2 ,n + 1 - \epsilon_n^2] \subset (n,n+1)$ for each $n$ , where $\frac{1}{4}> \epsilon_{n} > 0$ are chosen sufficiently small so that $\epsilon_{n} \downarrow 0$ as $n \uparrow \infty$, finally notice that because $\frac{1}{4} > \epsilon_{n} > 0$, $(1 - 2 \epsilon_{n}) \geq \frac{1}{2}$ for all $n = 1,2,3,...$
Clearly then, $\chi_n \in C_c(\mathbb{R}_{>1}) \subset C_0(\mathbb{R}_{>1})$ for each $n$, and we have $|| \sum_{k=m}^n \chi_{k} ||_{u} \leq \frac{1}{m}$, for all $1 \leq m < n$, so that $\sum_{k \geq 1} \chi_{k} \rightarrow_{||.||_{u}} f$, where $f \in C_0(\mathbb{R}_{>1})$ and is non-negative everywhere.
Writing $\mathbb{R}_{>1} = (1,2) \cup (2,3) \cup (3,4) \cup ...$, we see that $\sum_{k \geq 1} \chi_k \restriction (n,n+1) = \chi_n$, as $\text{supp}(\chi_{k}) \subset (k,k+1)$ for all $k \geq 1$, so that only $\chi_n$ may be non-zero when $f$ is restricted to $(n,n+1) \subset \mathbb{R}_{>1}$.
geetha290krm claims that the measures $\mu_n = n d \lambda \restriction (n,n+1) $, which are indeed in $\mathcal{M}(\mathbb{R}_{>1})$ converge to $0$ in the weak$^{\ast}$-sense.
This is false, as we will now show. Indeed, suppose it were true. Then by the definition of convergence in the weak$^{\ast}$-sense as defined by the OP, we should have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} 0 = \int f d\mu$ , here $\mu = 0$ is the zero measure on $\mathbb{R}_{>1}$.
However, we have that $\int f d\mu_n \geq \int_{[n + \epsilon_n, n+1 - \epsilon_n]} f d\mu_n = (1 - 2 \epsilon_n)\frac{n}{n} \geq (1 - 2 \epsilon_n) \geq \frac{1}{2} $ for all $n = 1, 2 , 3 , ...$ with the inequalities all holding because $f$ is non-negative, its restriction to each interval $[n+\epsilon_n, n+1 - \epsilon_n] \subset (n,n+1)$ is just $\chi_n$ restricted to this interval, where it is identically equal to $\frac{1}{n}$, and because we chose $\frac{1}{4} > \epsilon_n > 0$ for all $n = 1,2,3,...$, we indeed have that $(1 - 2 \epsilon_n) \geq \frac{1}{2}$ for all $n = 1,2,3,...$ (as already established in the first paragraph).
This shows that we do not have $\int f d \mu_n \rightarrow_{n \rightarrow \infty} \int f d \mu$, so that geetha290krm's counterexample is invalid, because $\mu_n$ does not converge to $0$ in the weak$^{\ast}$ - sense, as I have demonstrated.
Indeed, there is no counterexample to (2), by the Riesz-Markov theorem , whenever $X$ is a locally compact Hausdorff space, since then your $\mathcal{M}(X)$ is precisely isometric to $C_0(X)^{\ast}$, i.e. the dual space to $C_0(X)$, consisting of all bounded $\mathbb{R}$-linear functionals $C_0(X) \rightarrow \mathbb{R}$, which is a Banach space with the operator norm. By equipping $C_0(X)^{\ast}$ with the weak$^{\ast}$-topology, and then transporting this topology onto $\mathcal{M}(X)$ via the isometric isomorphism $\mathcal{M}(X) \rightarrow C_0(X)$ given by $\mu \rightarrow I_{\mu}$, where $I_{\mu}(f) = (f \rightarrow \int_{X} f d \mu)$, $I_{\mu} : C_0(X) \rightarrow \mathbb{R}$, your question for (2) is answered in the affirmative, because it is a general fact that for a Banach space $X$, a weak$^{\ast}$-convergent sequence $f_n \in X^{\ast}$ is uniformly bounded in operator norm.
Therefore to seek a counterexample for (2), we need to find a (metrizable) Hausdorff topological space $X$, which is NOT locally compact.
One candidate is $X = (1,2) \times (2,3) \times (3,4) \times (4,5) \times ...$, with the usual product metric (inducing the product topology). This is NOT locally compact, so perhaps we could find a counterexample here.
Now perhaps it is possible to adapt geetha290krm's counterexample to this situation. But I have not figured out all the details yet and am not entirely convinced $X$ is sufficient to provide a counterexample to (2) for the following reason:
While this space $X$ is not locally compact, it is Borel isomorphic to $[0,1]$, which is indeed (locally) compact Hausdorff, so perhaps this $X$ may not work to produce a counterexample.
So for (2) a counterexample may need to come from a metrizable topological space $Y$, that is NOT locally compact, and is either
1: NOT second countable, or
2: NOT $\textbf{completely}$ metrizable if it $\textbf{is}$ second countable (i.e. if condition 1. does hold)
Either of these conditions will prohibit $Y$ from being Borel isomorphic to a Borel subset of $[0,1]$, where (in the latter space $[0,1]$) the usual Riesz-Markov theorem applies, and this may cause complications.
Best Answer
No, this is not true. Unless $X$ is trivial, you can find a (nonatomic) measure $\lambda$ on $X$ and a sequence $(f_n) \subset L^2(\lambda)$ with $f_n \rightharpoonup 0$ and (say) $f^+_n \rightharpoonup 1$, both in $L^2(\lambda)$. Now, consider $\mu_n = f_n \lambda$.
Finally, if $g_n \rightharpoonup g$ in $L^2(\lambda)$, then $$ \int_X (g_n - g) h \, \mathrm{d}\lambda \to 0 $$ for all $h \in L^2(\lambda)$. Thus, if $\lambda$ is finite, this holds for all bounded, continuous functions $h$. Hence, the measures $g_n \lambda$ converge weak-$*$ towards $g \lambda$.