Total Variation Between Finite Measures

Question

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 $$ |\mu|(A) := \sup_{i,j:(x_i,y_i)\in A^2} \sum_{i,j\leq N}|a_i-b_j| \delta_{(x_i,y_j)} $$

Answer 1

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, $$ |\mu-\nu|(E)=\mu(E\cap X)+\nu(E\cap Y). $$

Answer 2

Edit: I misread the question and thought we were talking about the TV distance, not the TV measure. Leaving this answer for "historical purposes".

If all the "base points" are distinct, the total variation between these measures is 1.

Take, for instance, $A = \{x_1,\ldots,x_n\}$ (the support of $\mu$). Then $\mu(A) = 1$ but $\nu(A) = 0$, so $\vert \mu(A)-\nu(A)\vert = 1$; this is the largest possible value of $\vert \mu(A)-\nu(A)\vert$, since any other set might include a support point of $\nu$, hence $\nu(A)>0$ and so $\mu(A)-\nu(A)\leq 1$.

If the base points are allowed to be shared between the measures, things are a little more interesting, but I'll let you work out the details.