Variational distance basic properties The variational distance between two probability distributions $X$ and $Y$ taking values on the same alphabet $\mathcal A$ is defined as 
\begin{equation}
\delta (X,Y)=1/2\sum_{a\in A} |p_X(a)-p_Y(a)|$
\end{equation} 
There are two very basic claims with regard to the variational distance that I would like to formally prove. 
1) It cannot increase by the application of a function:
\begin{equation}
\delta (X,Y)\geq \delta (f(X),f(Y))
\end{equation} 
2) 
\begin{equation}
\frac{1}{2}\sum_{a\in A} |p_X(a)-p_Y(a)| = 1 -\sum_{a\in A} \min (p_X(a),p_Y(a))
\end{equation}

$\def\abs#1{\left|#1\right|}$
Let $f\colon \mathcal A \to \mathcal A$. Then 
\begin{align*}
  \delta\bigl(f(X), f(Y)\bigr) &= \frac 12 \sum_{a \in \mathcal A} \abs{p_{f(X)}(a) - p_{f(Y)}(a)}\\
   &= \frac 12 \sum_{a \in \mathcal A} \abs{\sum_{b\in f^{-1}(a)} \bigl(p_X(b) - p_Y(b)\bigr)}\\
   &\le \frac 12 \sum_{a \in \mathcal A} \sum_{b \in f^{-1}(a)} \abs{p_X(b) - p_Y(b)}\\
   &= \frac 12 \sum_{b \in \mathcal A} \abs{p_X(b) - p_Y(b)}\\
   &= \delta(X,Y)
\end{align*}

Using did's hint from below we have for 2)
\begin{align*}
 \delta(X, Y) &= \frac 12 \sum_{a \in \mathcal A} \abs{p_X(a) - p_{Y}(a)}\\
    &= \sum_{a \in \mathcal A} \frac 12 p_X(a) + \frac 12 p_Y(a) - \min\{p_X(a), p_Y(a)\}\\
    &= \frac 12 \sum_{a \in \mathcal A} p_X(a) + \frac 12 \sum_{a \in \mathcal A} p_Y(a) - \sum_{a\in \mathcal A} \min\{p_X(a), p_Y(a)\}\\
   &=  1 - \sum_{a\in \mathcal A} \min\{p_X(a), p_Y(a)\}.
\end{align*}
