Let $\Omega$ be a finite probability space and define the Total Variation as $TV(\mu,\nu):=\sup\{|\mu(A)-\nu(A)|: A\subseteq\Omega\}$. 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.
The supremum is reached at $A=\{x\in\Omega\,;\,\mu(x)\ge\nu(x)\}$.