Let $x_1,\dots,x_n,y_1,\dots,y_n\in \mathbb{R}^k$ be distinct and $0\leq a_i,b_i\leq 1$ be such that $\sum_{i=1}^n a_i = 1 = \sum_{i=1}^n b_i$. Define the finite measures $\mu=\sum_{i=1}^n a_i \delta_{x_i}$ and $\nu=\sum_{i=1}^n b_i y_i$. What is the total variation measure
$|\mu-\nu|$ explicitly?
The underlying measurable space is $(\mathbb{R}^k,\mathcal{B}(\mathbb{R}^k))$ where $\mathcal{B}(\mathbb{R}^k)$ is the Borel $\sigma$-algebra on $\mathbb{R}^k$.
I expect that it should be something like
$$
|\mu|(A) := \sup_{i,j:(x_i,y_i)\in A^2} \sum_{i,j\leq N}|a_i-b_j| \delta_{(x_i,y_j)}
$$
Best Answer
Let $X:=\bigcup_{i=1}^n\{x_i\}$ and $Y:=\bigcup_{i=1}^n\{y_i\}$. By definition, for $E\in\mathcal{B}(\mathbb{R}^k)$, $$ |\mu-\nu|(E)=\sup\{|(\mu-\nu)(A)|+|(\mu-\nu)(B)|:A,B\in\mathcal{B}(\mathbb{R}^k),A\cap B=\emptyset,A\cup B\subseteq E\}. $$ Since $X\cap Y=\emptyset$, we have, $$ |\mu-\nu|(E)\ge \mu(E\cap X)+\nu(E\cap Y). $$ On the other hand, for any $A,B\in \mathcal{B}(\mathbb{R}^k)$ s.t. $A\cap B=\emptyset$ and $A\cup B\subseteq E$, \begin{align} |(\mu-\nu)(A)|+|(\mu-\nu)(B)|&\le \mu(A)+\nu(A)+\mu(B)+\nu(B) \\ &=\mu(A\cup B)+\nu(A\cup B) \\ &\le \mu(E\cap X)+\nu(E\cap Y). \end{align} Therefore, $$ |\mu-\nu|(E)=\mu(E\cap X)+\nu(E\cap Y). $$