Definition of the total variation distance: $ V(P,Q) = \frac{1}{2} \int |p-q|d\nu$?

Let $B = \{p \ge q\}$. Note that \begin{align*} \int_\Omega \def\abs#1{\left|#1\right|}\abs{p-q}\, d\nu &= \int_B (p - q) \, d\nu + \int_{\Omega \setminus B} (q- p)\, d\nu\\ &\le 2 \sup_A \abs{\int_A (p-q) \, d\nu} \end{align*} On the other side, note first that $$ \int_\Omega (p-q) \,d\nu = P(\Omega) - Q(\Omega) = 0 $$ and hence $$ \int_B (p-q) \, d\nu = \int_{\Omega \setminus B} (q-p) \, d\nu $$ Now for any $A \in \mathscr F$, we have \begin{align*} \abs{\int_A (p-q)\, d\nu} &= \max\left\{\int_A (p-q)\, d\nu, \int_A (q-p)\, d\nu\right\}\\ &\le\max\left\{ \int_{A\cap B} (p-q)\, d\nu, \int_{A \cap (\Omega \setminus B)} (q-p)\, d\nu\right\}\\ &\le \max\left\{ \int_{B} (p-q)\, d\nu, \int_{\Omega \setminus B} (q-p)\, d\nu\right\}\\ &= \int_B (p-q)\, d\nu\\ &= \frac 12 \int_\Omega \abs{p-q}\,d\nu \end{align*} Taking the supremum over $A \in \mathscr F$, gives $$ \sup_A \abs{\int_A (p-q)\, d\nu} \le \frac 12 \int_\Omega \abs{p-q}\, d\nu $$ which is the other needed inequality.