Convergence in total variation

Question

There are the very basic convergence types in probability theory: almost sure, in $L^p$-norm, in measure and in distribution. Besides that there is the concept of convergence in total variation norm. It is obvious that convergence in TV norm implies convergence in distribution. But is there anything we can say about relations between convergence in total variation norm and convergence a.s.,in $L^p$ or in measure?- I did not get any direct hints by googling this, so maybe there is nothing to say about.

Answer 1

Total variation norm is a norm on measures whereas all the other types of convergence/norms/topologies that you mentioned are on spaces of functions.

Nonetheless functions can be measures too. If you have a sequence of measures $\mu_n$ and say all of them are absolutely continuous with respect to a third measure $\lambda$, then the measures converge in TV norm iff their radon-nikodym derivadives w.r.t. $\lambda$ converge in $L^1$.

Answer 2

To complement the answer of User guest, here is the proof that TV-norm convergence is equivalent to $L^1$ convergence. This follows from Scheffé's identity: The TV-norm is twice the L1 norm.

Proof of Scheffé's identity: Let $p,q$ two functions which integrate to $1$, i.e. $\int_{\mathbb{R}^D} p(x)dx = \int_{\mathbb{R}^D} q(x)dx = 1$, then

\begin{align} ||p-q||_{\rm{TV}} =& \sup_{B\in \mathcal{B}(\mathbb{R}^{D})} | \int_{B} p(x) dx - \int_{B} q(x) dx | \notag \\ =& \int_{ \{x':p(x')> q(x')\} } p(x) - q(x) dx \notag \\ =& \int_{ \{x':p(x') < q(x')\} } q(x) - p(x) dx \notag \\ =& \frac12 \int_{\mathbb{R}^{D}}|p(x)-q(x)| dx \notag \\ =& \frac12 ||p-q||_{L^1} \notag \end{align} where the second equality follows from taking the supremum over all Borel sets $B$, and the third equality follows from the fact that $p$ and $q$ are probability densities such that, at the one hand, \begin{equation} \int_{\mathbb{R}^D} p(x) dx = \int_{ \{x':p(x')> q(x')\} } p(x) dx + \int_{ \{x':p(x') < q(x')\} } p(x) dx = 1, \end{equation} and, at the other hand, \begin{equation} \int_{\mathbb{R}^D} q(x) dx = \int_{ \{x':p(x')> q(x')\} } q(x) dx + \int_{ \{x':p(x') < q(x')\} } q(x) dx = 1. \end{equation}