= Total variation distance under a measurable map
{title2=$d_{\mathrm{TV}}(P\circ f^{-1},Q\circ f^{-1})\leq d_{\mathrm{TV}}(P,Q)$}
Applying the same <measurable function> $f$ to observations from two <probability distributions> cannot increase their <total variation distance>. Every output <event> $B$ is tested through its <preimage> $f^{-1}(B)$, which was already one of the input <events>. Therefore
$$
d_{\mathrm{TV}}(P\circ f^{-1},Q\circ f^{-1})\leq d_{\mathrm{TV}}(P,Q).
$$
For the four-outcome example, recording only whether the result is $D$ makes the output <probability distributions> identical: each is a <Bernoulli distribution> with parameter $0.3$. The original <total variation distance> was $0.3$, but this coarse measurement leaves zero discrepancy. Discarding information can make two sources less distinguishable.
The same contraction holds for a common <Markov kernel> $K$: $d_{\mathrm{TV}}(PK,QK)\leq d_{\mathrm{TV}}(P,Q)$. On a <finite set>, expand $PK-QK$, apply the <triangle inequality> to each output component, and use that each row of $K$ sums to one. A common randomized postprocessing step cannot increase distinguishability either.
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