Finite KL divergence agrees with measurable-space KL

By [F9], the finite vectors define probability spaces on $(S,\mathcal{P}(S))$, with every singleton measurable; [F2] gives their event weights. If $p_i>0=q_i$, then $P(\{i\})>0=Q(\{i\})$, so [F3] gives $P\ll ̸Q$. The general definition [F6] and finite convention [F1] both give $+\mathrm{\infty }$.

Suppose $p_i=0$ whenever $q_i=0$. If $Q(A)=0$, the finite sum ${\sum }_{i\in A}{q}_i=0$ and nonnegativity force $q_i=0$ for every $i\in A$; hence $p_i=0$ for those $i$ and $P(A)=0$. Thus $P\ll Q$ by [F3]. Define $r_i=p_i\mathrm{/}{q}_i$ when $q_i>0$ and $r_i=0$ when $q_i=0$. Since every subset of $S$ is measurable, this finite function is measurable. For every $A\subseteq S$, [F5] gives ${\int }_{A}rdQ={\sum }_{i\in A}{r}_i{q}_i={\sum }_{i\in A}{p}_i=P(A)$, so [F4] identifies $r$ as a Radon–Nikodym derivative. It is explicitly constructed; no choice is used in this step.

Under the AC interface assumption [A1], substitute the density from step 1.2 into [F8], using the logarithm convention [F7]. Applying [F5] to $\mathrm{\ell }(i)=\ln r_i$ for $r_i>0$ and $\mathrm{\ell }(i)=0$ for $r_i=0$ gives $D(P\mathrm{\parallel }Q)={\sum }_{i:p_i>0}{p}_i\ln r_i={\sum }_{i:p_i>0}{p}_i\ln (p_i\mathrm{/}{q}_i)$; coordinates with $p_i=0$ contribute $0$ by [F1].

Step 1.1 handles support failure, while step 2.1 proves equality with the finite sum when $p$ is supported by $q$. These cases exhaust the finite vectors in the Statement.