Given a sequence of signed measures $<\nu\_j>$, if it happens that $\nu=\sum\limits\_{j = 1}^\infty \nu\_j$ is still a valid signed measure (then it can be proved that each partial sum $\nu\_n=\sum\limits\_{j = 1}^n \nu\_j$ is valid signed measure), do we have $\lim\limits\_{n\to \infty}|\sum\limits\_{j = 1}^n \nu\_j|=|\sum\limits\_{j = 1}^\infty \nu\_j|$ ($=|\lim\limits\_{n\to \infty} \sum\limits\_{j = 1}^n \nu\_j|$)? Thanks!



Is total variation continuous?
Assuming I understood the question correctly, the answer is no. Consider measures on $\{0,1\}^\omega$ with the product topology and Borel $\sigma$-algebra. Let $\mu\_i$ be the uniform measure on the set with $i$-th coordinate equal to 0. This sequence converges by your definition to the uniform measure, but all $\mu\_i$ are far (in total variation) from the uniform measure. (To fit your description we can take $\nu\_i=\mu\_i-\mu\_{i-1}$).


