Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\theta_A(p):=p(A)$.

share|cite|improve this question
up vote 1 down vote accepted

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 $\mathcal P(Y)$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.