Measurability of a total variation metric

Question

Let $X$ be a standard Borel space and let us denote by $\mathcal P(X)$ the space of Borel probability measures on $X$ endowed with the topology of weak convergence. Define $d:\mathcal P(X)\times \mathcal P(X)\to [0,1]$ by $$ d(p,q) := \sup_{A\in \mathcal B(X)}|p(A) - q(A)| $$ to be the total variation metric on $\mathcal P(X)$. I wonder whether $d$ is a measurable function.

Since the topology induced by $d$ is stronger than the weak convergence, it is not a continuous function on $\mathcal P(X)$ and hence I can't use this argument to show the measurability. Perhaps, there is a way of showing measurability of $d$ based on the fact that the $\sigma$-algebra of $\mathcal P(X)$ can be equivalently defined as the one generated by evaluation maps $\theta_A(p):=p(A)$.

Answer 1

We first assume that $X$ is a separable metric space.

Let $\mathcal O$ denote the collection of open subsets of $X$. Then for each $(p,q)\in\mathcal P(X)\times\mathcal P(X)$, we have $d(p,q)=\sup_{O\in\mathcal O}|p(O)-q(O)|$.

It's indeed standard: we define $\mathcal S:=\{B\subset X,\forall\varepsilon>0, \exists F\mbox{ closed}, O\mbox{ open}, (p+q)(O\setminus F)\lt\varepsilon, F\subset X\subset O\}$ and we can check it's a $\sigma$-algebra containing the open sets, hence the Borel $\sigma$-algebra.

Since $X$ is a separable metric space, we can find $(O_n)_{n\geqslant 1}$ a sequence of open sets such that if $O$ is an open set, there is $I\subset \mathbb N$ such that $O=\bigcup_{i\in\mathbb N}O_i$.

Fix an integer $n$, and take $U_n$ such that $$|p(U_n)-q(U_n)|\geqslant d(p,q)-n^{-1}.$$ We have $U_n=\bigcup_{i\in I}O_i$. For $N\geqslant 1$, define $V_N:=\bigcup_{i\in I\cap [1,N]}O_i$. Then $V_N\uparrow U_n$ hence there is an integer $N$ such that $p(U_n\setminus V_N)\lt n^{-1}$ and $q(U_n\setminus V_N)\lt n^{-1}$. We thus have that $$|p(V_N)-q(V_N)|\geqslant d(p,q)-3n^{-1}.$$ Define $$\mathcal F:=\left\{\bigcup_{i\in I}O_i,I\subset\mathbb N,I\mbox{ finite}\right\}.$$ We proved that $$d(P,Q)=\sup_{O\in \mathcal F}|p(O)-q(O)|$$ and since $\mathcal f$ is countable and $(p,q)\mapsto |p(O)-q(O)|$ is measurable for each $O$, we are done.

---

In the general case, we use an isomorphism with a separable metric space. We denote $X$ the Borel space and $Y$ the associated metric space, $\varphi\colon X\to Y$ the isomorphism. This provides a homeomorphism between $\mathcal P(X)$ and $\mathcal P(Y)$.

Answer 2

Following up on the idea of Davide to express $d$ as a supremum over a countable collection of sets, let me give a purely measure-theoretical proof, which also holds true for a slightly more general case, and is based on Davide's question here.

The metric $d$ is a measurable function if the underlying measurable space $(X,\mathfrak B_X)$ is separable.

The fact that $(X,\mathfrak B_X)$ is separable means that there exists a countable collection of sets that generates $\mathfrak B_X$. In particular, the algebra $\mathfrak A$ generated by this collection is countable as well. It follows that for any finite positive measure $\mu$ on $(X,\mathfrak B_X)$, set $B\in \mathfrak B_X$ and positive real $\epsilon>0$ there exists a set $A\in \mathfrak A$ such that $\mu(A\Delta B)\leq\epsilon$. Note in particular, that the latter condition implies that $|\mu(A) - \mu(B)|\leq\epsilon$. Indeed, $$ \mu(A) - \mu(B) = \mu(A\setminus B) - \mu(B\setminus A) \leq \mu(A\setminus B) + \mu(B\setminus A) = \mu(A\Delta B)\leq \epsilon $$ and the same applies to $\mu(B) - \mu(A)$.

Let us define $d'(p,q):= \sup_{A\in \mathfrak A}|p(A) - q(A)|$. Since $\mathfrak A$ is a countable collection, and $|p(A) - q(A)|$ is a measurable function of $(p,q)$ for any $A\in \mathfrak A$, we obtain that $d'$ is a measurable function as well. Let us show now that $d = d'$: clearly, $d\geq d'$. For the reverse inequality, let probability measures $p,q$ and a set $B\in \mathfrak B$ be arbitrary. Denote $\mu:= p+q$; for any $\epsilon>0$ there exists $A_\epsilon\in \mathfrak A$ such that $\mu(A_\epsilon\Delta B)\leq \epsilon$. In particular, $p(A_\epsilon\Delta B)\leq \epsilon$ and $q(A_\epsilon\Delta B)\leq \epsilon$ so that as above $|p(A_\epsilon) - p(B)|\leq\epsilon$ and $|q(A_\epsilon) - q(B)|\leq\epsilon$. Thus $$ |p(B) - q(B)|\leq |p(B) - p(A_\epsilon)| + |p(A_\epsilon) - q(A_\epsilon)| + |q(A_\epsilon) - q(B)|\leq d'(p,q) + 2\epsilon. $$ Since $\epsilon>0$ is arbitrary here, we obtain $|p(B) - q(B)|\leq d'(p,q)$ and thus $d(p,q)\leq d'(p,q)$ as desired.