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Define the total variation distance between two signed measures $\mu,\nu$ on a measure space $(X,\mathcal{A})$ as $\|\mu-\nu\|=|\mu-\nu|(X)$. If we have a sequence of measures $\mu_n$ converging in total variation to a measure $\mu$, i.e. $\|\mu_n-\mu\|\to 0$, is it true that $\mu_n(A)\to\mu(A)$ for every $A\in\mathcal{A}$? I think the answer should be yes, because $|\mu_n-\mu|$ is an unsigned measure and hence $$ 0\leq|\mu_n-\mu|(A)\leq |\mu_n-\mu|(X)\to 0$$

But this seems to imply that two measures such that $\mu(X)=\nu(X)$ are actually the same measure, which seems wrong. For example, define $\nu(A)=\int_A 2xd\mu $ where $A\subset X=[0,1]$ and $\mu$ is the Lebesgue measure. We have $\nu(X)=\mu(X)=1$ (and countable additivity by the monotone convergence theorem, so $\nu$ is a measure), but $\nu([0,1/2])=1/4\neq \mu([0,1/2])$. Is there something wrong with my reasoning above, or with this counterexample?

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  • $\begingroup$ YEs it does. As you pointed out $|\mu_n(A)-\mu(A)|\leq|\mu_n-\mu|(A)\leq|\mu_n-\mu|(X)=\|\mu_n-\mu\|_{TV}\xrightarrow{n\rightarrow\infty}0$. You second inference is wrong. $\endgroup$
    – Mittens
    Oct 4, 2021 at 1:25

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It is true that $\mu_n(A) \to \mu(A)$ for all $A \in \mathcal A$ but this does not imply by any means that $\mu(X)=\nu(X)$ implies $\mu=\nu$. Perhaps, you are thinking that $\mu -\nu$ is a positive measure so $0\leq (\mu-\nu)(A)\leq (\mu-\nu) (X)=0$ for all $A \in \mathcal A$. This is flawed since $\mu-\nu$ is not a positive measure.

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