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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 $h: Y \to [0,1]$?

geodude
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    Just a question : is the above definition of the TV distance equivalent to the usual one i.e. $d(p,q) = \sup_{A} |p(A) - q(A)|$? I'd like to think so but I'm not sure. For this particular distance, for each $A$ you can find a product-type set arbitrarily close to it in the product measure, and so $f,g$ can be constructed. Hence, I believe for this definition that the theorem is true. Using a monotone class argument for functions, I would like to imagine that your statement is true as well. – Sarvesh Ravichandran Iyer Sep 03 '21 at 10:48
  • Are you sure $d(p,q)$ is well-defined?. For example if $p$ and $q$ are point masses (dirac measures). And define a function $f_n(x)=n$ for point mass of $p$ (or $q$). then $d(p,q)$ becomes infinity. and i believe for any 2 measure that is not the same we could find such sequence of $f_n$'s. But maybe i'm wrong. – Oğuzhan Kılıç Sep 03 '21 at 12:20
  • @Oguzhan The maps are required to take values in $[0,1]$. – WoolierThanThou Sep 03 '21 at 12:26
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    My impression is that the answer is negative: the functions of the form $f(x,y)=g(x)h(y)$ form a total family in $L^1(p+q)$, not a dense family. I suggest to search for a counterexample with two probability measures on ${0,1}^2$. – Christophe Leuridan May 19 '22 at 21:24

2 Answers2

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We can construct a counterexample:

Take $X=Y=[0,1]$, $\tilde f(x,y) = \begin{Bmatrix}1, x\leq y \\ -1 , x>y\end{Bmatrix}$,

$p(A)=\lambda_2(A \cap \{x\leq y\})$, $q(A)=\lambda_2(A \cap \{x\ > y\})$.

Now $$\sup_{f}|\int f \; dp - \int f \; dq|$$ is $1$ and is achieved with $f = \tilde f$.

Now suppose we have $g_n$ and $h_n$ such that $$\sup_{f}|\int g_nh_n\; dp - \int g_nh_n \;dq| = \sup_{f}|\int g_nh_n\tilde f \; d\lambda_2| \rightarrow 1$$

The function under the last integral is bounded by $|g_nh_n\tilde f|$, which is in turn bounded by $1$. For the integral to approach $1$ the $|g_nh_n\tilde f|$ must also approach $1$ almost everywhere. Furthermore $\lambda_2(g_nh_n\tilde f > 0)$ or $\lambda_2(g_nh_n\tilde f < 0)$ must tend to $1$.

In other words, either $g_nh_n\tilde f\rightarrow 1$ or $g_nh_n\tilde f\rightarrow -1$ almost everywhere.

However, that is impossible. To convince yourself of this consider the sets $$G_+ = \{x|g_n(x) \geq 0\}$$$$G_- = \{x|g_n(x) \leq 0\}$$ $$H_+ = \{y|h_n(y) \geq 0\}$$$$H_- = \{y|h_n(y) \leq 0\}$$ and show that both $\{g_nh_n\tilde f \geq 0\}$ and $\{g_nh_n\tilde f \leq 0\}$ have positive measure.

[![this should give you the idea why that is][1]][1] [1]: https://i.sstatic.net/7EkSJ.png

Dood
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  • I'm pretty sure that if g and h are nowhere 0, then the last two sets will always have a measure of 1/2 each. – Dood May 24 '22 at 18:39
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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}

Yuval Peres
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