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====== The total variation norm between two probability measures ======


Let $\mu, \nu  \in \meas{1}(\Xset)$ where $(\Xset,\Xsigma)$. The total variation norm between $\mu$ and $\nu$ is defined by 
$$
\tv{\mu-\nu}=\sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|=\gamma(|f-g|)
$$ 
where $\mu\preceq \gamma$, $\nu \preceq \gamma$ and $\mu= f \cdot \gamma$, $\nu= g \cdot \gamma$. 

====== Proof ======

Indeed for all $h \in \mcf(\Xset)$, 
$$
|\mu(h)-\nu(h)|=|\gamma( (f-g)h)| \leq \supnorm{h} \gamma|f-g|
$$
showing that $\sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|\leq \gamma(|f-g|)$

Moreover, 
$$
\gamma(|f-g|)=\gamma( (f-g)\ \underbrace{\sg(f-g)}_{h})=\mu(h) -\nu(h)
$$
where $h=\sg(f-g)$ and since $\supnorm{h}=1$, we thus have 
$\gamma(|f-g|) \leq \sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|$. The proof is completed.