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Let $\Omega$ be a finite probability space and define the Total Variation as $TV(\mu,\nu):=\sup\{|\mu(A)-\nu(A)|: A\in\Omega\}$. One can also derive that

$TV(\mu,\nu)=\frac{1}{2}\sum_{x\in\Omega}|\mu(x)-\nu(x)|$

I know how to derive this fact by showing $\leq$ and $\geq$, however, I am interested if there is a direct way of deriving this. My hope is that such a direct proof would hint at the intuition behind maximizing over the largest event, compared to looking at the total sum of differences.

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Did you get something out of the answer below? If this is so, you might accept it. –  Did Apr 7 '11 at 8:26

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up vote 2 down vote accepted

The supremum is reached at $A=\{x\in\Omega\,;\,\mu(x)\ge\nu(x)\}$.

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Or it's complement. I see, thanks! –  SKS Mar 21 '11 at 20:45
    
@Sam At both, and at a few others like $A'=\{x\in\Omega\,;\,\mu(x)>\nu(x)\}$, really. –  Did Mar 21 '11 at 21:05

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