$\frac 1 2$ in the definition of total variation distance between two probability measures

Solution 1:

It is not a matter of adding a factor of $\frac{1}{2}$ in the finite case. The second expression is a sum over all elements of the underlying set, while the first expression is not a sum, but a sup over all events in the space. The reason for the $\frac{1}{2}$ in the second expression is that it can be proved that in the finite case, the two quantities are equal. See for example Proposition 4.2 on page 48 of Markov chains and mixing times by Levin, Peres, and Wilmer. I do not know the full extent of analogies to the second expression for cases when the underlying set is infinite, but the sum would have to become an integral. See cardinal's comments for more information.