Let $X$ be a measurable space. Given probability measures $p$ and $q$ on $X$, define their total variation distance as
$$
d(p,q) = \sup_f \Big| \int f \, dp – \int f \, dq \Big| ,
$$
where $f$ varies over measurable functions $X \to [0,1]$.
If $Y$ is also a measurable space, let now $p$ and $q$ be measures on the product $X \times Y$. Again,
$$
d(p,q) = \sup_f \Big| \int f \, dp – \int f \, dq \Big| ,
$$
where now $f$ varies over measurable functions $X \times Y\to [0,1]$.
However, can we equivalently test the distance with functions in the form $f(x,y)=g(x)h(y)$?
That is, can we write
$$
d(p,q) = \sup_{g,h} \Big| \int gh \, dp – \int gh \, dq \Big| ,
$$
where $g: X \to [0,1]$ and $h: Y \to [0,1]$?
Best Answer
Let $X=Y=\{0,1\}$. Suppose $p$ is uniform on the two points $\{(0,0),(1,1)\}$ and $q$ is uniform on $\{(0,1),(1,0)\}$.
The function $F:X \times Y \to [0,1]$ that is the indicator of $\{(0,0),(1,1)\}$ shows that $$d(p,q)=1\,.$$ However, given $g:X \to [0,1]$ and $h: Y \to [0,1]$, we have \begin{eqnarray} \Bigl|\int gh \,dp-\int gh \, dq\Bigr| &=&\frac12\Bigl|g(0)h(0)+g(1)h(1)-g(0)h(1)-g(1)h(0)\Bigr|\\ &=&\frac12|g(0)-g(1)| \cdot |h(0)-h(1)| \le 1/2. \end{eqnarray}