Let $\nu_n$ be a sequence of finite signed radon measure such that $\nu_n\to \nu$ strongly for a finite signed radon measure $\nu$. Let $|\nu_n|$ denote the total variation measure of $\nu_n$. We know that $|\nu_n|$ is a positive Radon measure.

My question: do I have $|\nu_n|\to |\nu|$ strongly? and do I have $||\nu|-|\mu||\leq |\nu-\mu|$ for two arbitrary finite signed measure $\nu$ and $\mu$?

I know the above statement is absolutely false if $\nu_n\to \nu$ only in weak star sense. But I somehow remembered for strong convergence it is true but I can not find the source. So, if it is true, please confirm it for me and directly me to a reference, if not... maybe a counter example?

Thank you!


The sequence of measures $\nu_n(E) = \int_E \sin nx\,dx$ (considered on an interval $[0,1]$) converges to zero strongly. Indeed, for every measurable $E\subset [0,1]$ the sequence $\nu_n(E)$ is (up to a factor) the since Fourier series of $\chi_E$. So it's square summable, and therefore converges to zero.

On the other hand, the variation $|\nu_n|(E) = \int_E |\sin nx|\,dx$ does not converge to $0$ strongly; it converges to $\frac12 \,dx$ instead.

This is the same example you probably had in mind for weak convergence; my point is that it works for strong convergence too.


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