# Compactness of signed measure?

This idea never comes to me but I just realize that I am making a serious mistake that the space of finite signed measure is weakly compact...

We all know that the space of finite Radon measure is weakly compact, i.e., test against compact supported continuous function. But do we have same properties for finite signed measure? I thought yes since the total variation of a sequence of signed measure $(\mu_n)$ is bounded, then the positive and negative part of it are bounded and hence each of them can have a weak star convergence subsequence, i.e., $\mu_n^+\to \lambda_1$ and $\mu_n^-\to \lambda_2$ and we define $\mu=\lambda_1-\lambda_2$. Hence we have $\mu_n\to\mu$ in weak star sense, and $\mu$ is a signed measure.

Am I making a mistake here?

The set of finite Radon measures is certainly not weakly compact. If we're talking about measures on $K$, and if $A>0$, the set of Radon measures $\mu$ with $\mu(K)\le A$ is weakly compact. Similarly the set of signed (Borel) measures with $||\mu||\le A$ is weakly compact.
• If you think the set of finite Radon measures is weakly compact you're certainly missing something. Say $m$ is Lebesgue measure on $[0,1]$. Let $\mu_n=nm$, the measure $m$ multiplied by $n$. Each $\mu_n$ is a finite Radon measure. There is no weakly convergent subsequence. – David C. Ullrich Aug 21 '15 at 17:58