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!
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3$\begingroup$ Which topology do employ on the space of measures? The one induced by total variation norm? $\endgroup$– shuhaloCommented Feb 19, 2011 at 8:46
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$\begingroup$ And measures on which space, by the way ? $\endgroup$– BS.Commented Feb 19, 2011 at 12:19
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$\begingroup$ Sorry for the late response. I'm concerned with the equality in the post which looks like a continuity property, so I used the term "continuous" in the title. Of course it would be better if we can establish a setting in which topology is equipped and then continuity of total variation of signed measures has the usual meaning. But currently I have no idea about this. $\endgroup$– zzzhhhCommented Feb 21, 2011 at 10:59
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2 Answers
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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 $\nu_i=\mu_i-\mu_{i-1}$).
answered Feb 20, 2011 at 6:00
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$\begingroup$ Sorry but what is the uniform measure? $\endgroup$– zzzhhhCommented Feb 21, 2011 at 11:00
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1$\begingroup$ The uniform measure is the probability measure of total measure 1, and for each $i$, $\mathbb{P}(x_i=0)=\frac12$ independently. $\endgroup$ Commented Feb 21, 2011 at 18:57
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$\begingroup$ I refrained from asking what is total measure. Could you please refer me to a textbook containing conceptions like uniform measure, total measure so that I can understand your reply after reading it? Thanks! $\endgroup$– zzzhhhCommented Feb 22, 2011 at 10:40
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1$\begingroup$ I'm really nor very good with books recommendations, but any rigorous book on probability theory should be helpful. Perhaps the following reformulation will help: the space is just $I=[0,1]$. $\mu$ is the Lebesgue measure restricted to $I$. $I_n$ is the set of all reals in $I$ whose $n$-th binary digit is 0. $\mu_n$ is twice the Lebesgue restricted to $I_n$, i.e. $\mu_n(A)=2\mu(A \cap I_n)$. Then for every measurable $A$ we have $\mu_n(A) \to \mu(A)$, but the total variation distance between any $\mu_n$ and $\mu$ is 1. $\endgroup$ Commented Feb 22, 2011 at 18:22
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Am I misunderstanding your question? Because it seems to me that if you consider the $\nu_n = \frac{(-1)^n}{n} \delta$ where $\delta$ is the Dirac mass, then we've got the usual situation with the alternating harmonic series.
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$\begingroup$ Yes, this is an example that the equality holds. But do we have counterexamples? $\endgroup$– zzzhhhCommented Feb 21, 2011 at 11:00