Regularity of limit measure and prove that $|\mu_h|\stackrel{*}{\rightharpoonup}|\mu|$

Question

I have some questions.

First of all, let $\mu_h$ a sequence of Radon measures and suppose that $\mu_h$ weakly-converge to another measure $\mu$.

Now, this limit measure $\mu$ is still Borel? Is it a Radon measure?

Then, I have an exercise on weak-convergence of measures, but I don't know if what I've done is correct. We know that this proposition holds true (Prop 4.30 Sets of finite perimeter and geometric variational problems, F. Maggi):

If $\mu_h$ is a sequence of $\mathbb R^m$-vector valued Radon measures with $\mu_h\stackrel{*}{\rightharpoonup}\mu$, $|\mu|(\mathbb R^n)< \infty$ $|\mu_h|(\mathbb R^n)\to |\mu|(\mathbb R^n)$, then $|\mu_h|\stackrel{*}{\rightharpoonup}|\mu|$.

Having said this, suppose we have a sequence of $\mathbb R^m$-vector valued Radon measures with $\mu_h\stackrel{*}{\rightharpoonup}\mu$, a sequence of radii $r_h\to \infty$ and $|\mu_h|(B_{r_h})\to |\mu|(B_{r_h})$ as $h\to \infty$ for each $k\in \mathbb N$. We want to prove that $|\mu_h|\stackrel{*}{\rightharpoonup}|\mu|$.

Here is what I've done:

We want to prove that $|\mu|(\mathbb R^n)< \infty$ $|\mu_h|(\mathbb R^n)\to |\mu|(\mathbb R^n)$, so that by the previous proposition we have the claim.

Since $|\mu_h|$ is a Borel measure, we have that $|\mu_h|(\mathbb R^n)=\lim_{k\to \infty} |\mu_h|(B_{r_k})$. Taking the limit as $h\to \infty$ we have $$\lim_{h\to \infty}|\mu_h|(\mathbb R^n)\to \lim_{k\to \infty} |\mu|(B_{r_h})=|\mu|(\mathbb R^n).$$

This proof doesen't convince me much. Firstly, I don't know anything on $|\mu|$, so I can't conclude that $|\mu|(B_{r_h})\to |\mu|(\mathbb R^n)$ as $k\to \infty$. Then I don't know if it is correct to exchange limits (in $k$ and in $h$) as I've done. Then it remains also to prove that $|\mu|(\mathbb R^n)< \infty$

Can someone help me? I would really appreciate it.

Answer 1

Some comments on your questions:

If $\mu_h$ is a sequence of Radon measures which weakly converges to a measure $\mu$, is $\mu$ necessarily Radon?

I don't think this is necessarily true: If $\mathcal{L}$ is Lebesgue measure on $\mathbb{R}$, define $\mu_h = h\cdot \mathcal{L}$, and \begin{equation} \mu(E) := \begin{cases} +\infty & \text{if } \mathcal{L}(E) > 0\\ 0 & \text{otherwise.} \end{cases} \end{equation} If I am not mistaken then, with the convention that $0\cdot \infty = 0$, we have \begin{equation} \lim_{h \to \infty}\int_{\mathbb{R}}f d\mu_h = \int_{\mathbb{R}}f d\mu \end{equation} for all $f \in C_c(\mathbb{R})$. But $\mu$ is not a Radon measure.

For your question,

Suppose we have a sequence of $\mathbb{R}^m$-vector valued Radon measures with $\mu_h \rightharpoonup^* \mu$, a sequence of radii $r_k\to \infty$ and $|\mu_h|(B_{r_k})\to|\mu|(B_{r_k})$ as $h \to \infty$ for each $k \in \mathbb{N}$. We want to prove that $|\mu_h|\rightharpoonup^*|\mu|$.

We want to prove that $|\mu|(\mathbb{R}^n)< \infty$ and $|\mu_h|(\mathbb{R}^n)→|\mu|(\mathbb{R}^n)$, so that by the previous proposition we have the claim.

Take $\mu_h = \mu = \mathcal{L}^n$, Lebesgue measure on $\mathbb{R}^n$. The hypotheses (and conclusion) are satisfied, but obviously $|\mu|(\mathbb{R}^n) = \infty$. The point is that your proof idea to show $|\mu|(\mathbb{R}^n) < \infty$ may not be possible.

Also for this exercise, I believe we should also assume that $\mu$ is a Radon measure.

Hint: Recall, that to show the weak convergence, it is sufficient to prove \begin{equation} \limsup_{h \to \infty}|\mu_h|(K) \leq |\mu|(K) \end{equation} for all compact $K$. Now imitate the proof of Prop. 4.30 (the proposition you wrote above), but use a suitable $B_{r_k}$ instead of $\mathbb{R}^n$.