0
$\begingroup$

Given $\mu_n$ and $\mu$ finite signed Radon measures on the domain $\Omega$. We assume $\mu_n\to \mu$ in weak* sense, i.e. $\int_{\Omega}\phi \,d\mu_n \to \int_{\Omega} \phi\, d\mu$ for all test functions $\phi \in C_c(\Omega)$.

Generally we should NOT expect that $\mu_n^+\to \mu^+$ and $\mu_n^-\to \mu^-$, where $\mu_n = \mu_n^+-\mu_n^-$, the Jordan decomposition of $\mu_n$.

However, if we in additionally assume that the total variation of $\mu_n$, $|\mu_n|=\mu_n^+ + \mu_n^-$, satisfies $|\mu_n|(\Omega)\to |\mu|(\Omega)$, we have that $|\mu_n|\to|\mu|$ in weak* sense as well.

Hence we could conclude that $\mu_n^+\to \mu^+$ and $\mu_n^-\to \mu^-$ as $\mu^+=\frac{(|\mu|+\mu)}{2}$ and $\mu^-=\frac{(|\mu|-\mu)}{2}$.

I think I saw this result somewhere but I can not find it now. Could someone help me to verify my conclusion, and if possible, point out where in the literature this result is stated (maybe as an exercise)?

Thank you very much!

$\endgroup$
  • $\begingroup$ You can find this result in Bogachev, "Measure Theory", vol. 2, Corollary 8.4.8 (which also holds in a slightly greater generality). $\endgroup$ – yadaddy Oct 31 '15 at 15:06
0
$\begingroup$

Don't know a reference.

Compactness. If $\mu_n^+$ does not converge to $\mu^+$ there is a subnet converging to something else: $$\mu_{n_j}^+\to\lambda\ne \mu^+.$$

So $$\mu_{n_j}^-\to\nu\ne \mu^-.$$It follows that $$\lambda-\nu=\mu,$$hence $$\lambda\ge\mu^+,\quad\nu\ge\mu^-.$$And also $\mu_{n_j}^++\mu_{n_j}^-\to\lambda+\nu,$ which implies $$\limsup||\mu_{n_j}||\ge||\lambda+\nu||>||\mu||,$$contradicting the fact that $||\mu_n||\to||\mu||$.

$\endgroup$
  • $\begingroup$ I second your proof. It is simple and nice. I just though, for a chance, that there is a name for this lemma, but it is ok if not. $\endgroup$ – spatially Aug 20 '15 at 13:08
  • $\begingroup$ There may be a name for it. I'd never heard of the result until you posted the question, but that proves nothing - there are many many things I've never heard of. $\endgroup$ – David C. Ullrich Aug 20 '15 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.