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Fix a measurable space $(X,\Sigma)$. Let $\mu:\Sigma\to\mathbb{R}$ be some signed measure. Then the total variation norm is defined by $$ \|\mu\|_{\textrm{TV}}=\mu_{+}(X)+\mu_{-}(X),\quad (*) $$ where $\mu=\mu_{+}-\mu_{-}$ is the Hahn decomposition of $\mu$. Now, let $\mu,\nu:\Sigma\to[0,1]$ be two probability measures. Then $$ \|\mu-\nu\|_{\textrm{TV}}=2\max_{A\in\Sigma}(\mu(A)-\nu(A)),\quad (**) $$ To see this, recall that the Hahn decomposition $$ \mu-\nu=(\mu-\nu)_{+}-(\mu-\nu)_{-} $$ has the form $$ (\mu-\nu)_{\pm}(A)=(\mu-\nu)(A\cap A_{\pm}),\quad A\in\Sigma, $$ where $A_{+}\cup A_{-}=X$, and the choice of the sets $A_{+}$, $A_{-}$ is unique up to a set of zero measures $$ |\mu-\nu|=(\mu-\nu)_{+}+(\mu-\nu)_{-}. $$ This apparently immediataly implies (**) with the maximum attained at the set $A_{+}$.

My question: I'm having a hard time connecting the definition of total variation norm in (*) with the arguments that lead to (**). Any help is greatly appreciated.

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  • $\begingroup$ what is the reference you are using here? $\endgroup$
    – Ommo
    Commented Sep 14, 2023 at 8:41

1 Answer 1

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From the Hahn decomposition theorem, you can check that the definition

$$ \|\mu\|_{\textrm{TV}}=\mu_{+}(X)+\mu_{-}(X),\quad (*) $$

can be rewritten as

$$ \|\mu\|_{\textrm{TV}}=\max_{A \in \Sigma}\mu(A)-\min_{A \in \Sigma}\mu(A) $$

with the max and min achieved at $A_+$ and $A_-$ respectively.

So for the difference of two probability measures,

\begin{align*} \|\mu-\nu\|_{\textrm{TV}}&=\max_{A \in \Sigma}(\mu(A)-\nu(A))-\min_{A \in \Sigma}(\mu(A)-\nu(A))\\ &=(\mu(A_+) - \nu(A_+)) - (\mu(A_-)-\nu(A_-))\\ &=(\mu(A_+) - \nu(A_+)) - ((1-\mu(A_+))-(1-\nu(A_+)))\\ &=2(\mu(A_+) - \nu(A_+))\\ &=2\max_{A \in \Sigma}(\mu(A)-\nu(A)) \quad (**) \end{align*}

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  • $\begingroup$ any reference for this proof? $\endgroup$
    – Ommo
    Commented Sep 18, 2023 at 10:12
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    $\begingroup$ No, sorry; I'd assume it's in a textbook somewhere, but I don't have a citation. @limone $\endgroup$
    – Adam
    Commented Sep 18, 2023 at 14:54

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