On the convergence in total variation

Question

$\newcommand\R{\mathbb R}$For a probability measure $\mu$ over $\R^2$ and a unit vector $u\in\R^2$, let $\mu^u$ denote the pushforward of $\mu$ under the projection map $\R^2\ni x\mapsto u\cdot x\in\R$, where $\cdot$ denotes the dot product.

By the Lévy-Cramér continuity theorem (see e.g. Proposition 7.3.17), a sequence $(\mu_n)$ of probability measures over $\R^2$ converges weakly to a probability measure $\mu$ over $\R^2$ if and only if the sequence $(\mu_n^u)$ converges weakly to $\mu^u$ for each unit vector $u\in\R^2$.

Is it true that a sequence $(\mu_n)$ of probability measures over $\R^2$ converges in total variation to a probability measure $\mu$ over $\R^2$ if and only if the sequence $(\mu_n^u)$ converges in total variation to $\mu^u$ for each unit vector $u\in\R^2$?

The answer to this question is presumably no, but I do not know a counterexample. (Of course, the "only if" part here is trivial. So, the question is really only about the "if" part.)

I had thought the polar chessboard construction could provide such a counterexample, but now this does not seem to be the case. Maybe some modification of that construction will do?

Answer 1

Take a uniform distribution on the circle of radius $1 - 2^{-n}$. These have disjoint support & so total variation distance is 1, but for a given measure the projections are all the same, and they are absolutely continuous 1-d measures & quite similar.