Weak convergence of measure and their total variation measures.

Asked May 31 '24 at 09:20

Active Jul 22 '24 at 13:34

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Does weak convergence of Radon measures implies weak convergence of their total variation measures? Precisely speaking, I want to know whether the following proposition is correct.

Let $X$ be a locally compact Hausdorff space, and $\mu_k$ is a sequence of signed(or vector valued) Radon measures such that $\mu_k$ converges weakly to $\mu_0$. Namely, for any $\phi \in C_c(X)$, \begin{equation} \int_{X}\phi d\mu_k \to \int_{X} \phi d\mu_0. \end{equation} Then $|\mu_k|$ converges weakly to $|\mu_0|$.

edited Jul 22 '24 at 13:34

Bruno B

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asked May 31 '24 at 09:20

gaoqiang

$\mu_k=\delta_k, \mu_0=0$. – Kavi Rama Murthy May 31 '24 at 09:22
@geetha290krm I mean total variation measure, not total variation. Updated. – gaoqiang May 31 '24 at 09:26
I'm guessing this should be doable using the polar decomposition $d\mu = h d|\mu|$ with $\forall x,, |h(x)| = 1$, in some fashion? – Bruno B May 31 '24 at 10:06
(Decided to turn my answer into a comment) There appear to exist counterexamples to the fact that the hypothesis would imply the convergence of $(\mu_k^+)_k$ to $\mu_0^+$, see the answer by gerw to this post here If $\mu_n \overset{\ast}{\rightharpoonup}\mu$, then $\mu^+_n \overset{\ast}{\rightharpoonup} \mu^+$ and $\mu^-_n \overset{\ast}{\rightharpoonup}\mu^-$, and we have $\mu_k^+ = \frac{1}{2}(|\mu_k| + \mu_k)$ so it would also be a counterexample to your claim.
It also seems however that upon placing an additional condition you do have conver – Bruno B Jul 22 '24 at 13:31
@geetha290krm and OP, it's been a while, but I've decided to turn my previous answer into a comment due to its current lack of verifiability. – Bruno B Jul 22 '24 at 13:32

Weak convergence of measure and their total variation measures.

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