$ \newcommand{\abs}[1]{| #1 |} \newcommand{\bigabs}[1]{\left| #1 \right|} \newcommand{\R}{\mathbb{R}} \newcommand{\B}{\mathcal{B}} \newcommand{\norm}[1]{\lVert #1 \rVert} $ Here is a well-known identity for the total variation distance between two distributions.

Lemma: Let $\mu,\nu$ be two probability measures on a measure space $(X, \B)$. Let $p,q$ be the Radon-Nikodym derivatives of $\mu,\nu$ with respect to some base measure $\tau$ (note that this is without loss of generality since $\mu,\nu$ are absolutely continuous w.r.t. $\frac{1}{2}(\mu + \nu)$). Then, $$ \norm{\mu - \nu}_{\mathrm{tv}} := \sup_{A \in \B} \abs{\mu(A) - \nu(A)} = \frac{1}{2} \norm{ p - q }_{L_1(\tau)} \:. $$

Proof: We partition $X$ into the subsets $I_+, I_-$, where $$ I_+ := \{ x \in X : p(x) \geq q(x) \} \:, \\ I_- := \{ x \in X : p(x) < q(x) \} \:. $$ Next, we argue that, $$ \int_{I_+} \abs{p(x) - q(x)} \; d\tau = \int_{I_-} \abs{p(x) - q(x)} \; d\tau \:. $$ To see this, we use the identities, $$ \begin{align*} 1 &= \int p(x) \; d\tau = \int_{I_+} p(x) \; d\tau + \int_{I_-} p(x) \; d\tau \:, \\ 1 &= \int q(x) \; d\tau = \int_{I_+} q(x) \; d\tau + \int_{I_-} q(x) \; d\tau \:, \end{align*} $$ from which we conclude, $$ \begin{align*} \int_{I_+} \abs{p(x) - q(x)} \; d\tau &= \int_{I_+} (p(x) - q(x)) \; d\tau \\ &= \int_{I_+} p(x) \; d\tau - \int_{I_+} q(x) \; d\tau \\ &= 1 - \int_{I_-} p(x) \; d\tau - \left( 1 - \int_{I_-} q(x) \; d\tau \right) \\ &= \int_{I_-} -(p(x) - q(x)) \; d\tau \\ &= \int_{I_-} \abs{p(x) - q(x)} \; d\tau \:. \end{align*} $$ But this means that, $$ \begin{align*} \int \abs{p(x) - q(x)} \; d\tau &= \int_{I_+} \abs{p(x) - q(x)} \; d\tau + \int_{I_-} \abs{p(x) - q(x)} \; d\tau \\ &= 2 \int_{I_+} \abs{p(x) - q(x)} \; d\tau \:. \end{align*} $$ An identical argument for $I_-$ holds, from which we conclude $$ \frac{1}{2} \int \abs{p(x) - q(x)} \; d\tau = \int_{I_+} \abs{p(x) - q(x)} \; d\tau = \int_{I_-} \abs{p(x) - q(x)} \; d\tau \:. $$ Since $I_+ \in \B$, this shows that $$ \sup_{A \in \B} \abs{\mu(A) - \nu(A)} \geq \abs{\mu(I_+) - \nu(I_+)} = \int_{I_+} \abs{p(x) - q(x)} \; d\tau = \frac{1}{2} \norm{p-q}_{L_1(\tau)} \:. $$ On the other hand, for any measurable set $A \in \B$, $$ \begin{align*} \abs{\mu(A) - \nu(A)} &= \bigabs{ \int_{A \cap I_+} \abs{p(x) - q(x)} \; d\tau - \int_{A \cap I_-} \abs{p(x) - q(x)} \; d\tau } \\ &\stackrel{(a)}{\leq} \max\left( \int_{A \cap I_+} \abs{p(x) - q(x)} \; d\tau, \int_{A \cap I_-} \abs{p(x) - q(x)} \; d\tau \right) \\ &\stackrel{(b)}{\leq} \max\left( \int_{I_+} \abs{p(x) - q(x)} \; d\tau, \int_{I_-} \abs{p(x) - q(x)} \; d\tau \right) \\ &= \frac{1}{2} \norm{p-q}_{L_1(\tau)} \:. \end{align*} $$ Above, (a) uses the fact that for any two non-negative real numbers $x,y$, we have $\abs{x-y} \leq \max(x,y)$, and (b) follows since the integral of a non-negative function over a larger set can only increase the value. Since this inequality holds uniformly for every $A \in \B$, it holds also for the supremum. Hence, we have shown $$ \frac{1}{2} \norm{p-q}_{L_1(\tau)} \leq \sup_{A \in \B} \abs{\mu(A) - \nu(A)} \leq \frac{1}{2} \norm{p-q}_{L_1(\tau)} \:, $$ from which the claim follows. $\square$