# Does total variation convergence imply pointwise convergence of measures?

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?

• 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. Oct 4, 2021 at 1:25

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.