Additivity of KL divergence for finite product laws

If $P_i\ll ̸Q_i$ for some $i<n$, choose $A_i\in {\mathcal{A}}_i$ with $Q_i(A_i)=0<P_i(A_i)$. Let $E={\prod }_{j<n}{E}_j$, where $E_i=A_i$ and $E_j=X_j$ for $j\mathrm{\ne }i$. The product rectangle has $Q(E)=0$ and $P(E)=P_i(A_i)>0$ by [F2, F3]. Hence $P\ll ̸Q$, so the left side is $+\mathrm{\infty }$; the corresponding term on the right is also $+\mathrm{\infty }$ and the equality holds.

Now suppose $P_i\ll Q_i$ for every $i<n$, and choose nonnegative representatives $r_i=d{P}_i\mathrm{/}d{Q}_i$. Define $r(x_0,\dots ,x_{n-1})={\prod }_{i<n}{r}_i(x_i)$. Each coordinate function $x\mapsto r_i(x_i)$ is measurable by the finite product sigma-algebra definition: its level-set preimages are measurable rectangles. Hence the vector of coordinate values is measurable, since rectangles generate the Borel sigma-algebra of ${\mathbb{R}}^n$; composing with the continuous multiplication map gives measurable $r$. Repeated Tonelli and the rectangle formulas give ${\int }_{E}rdQ={\prod }_{i<n}{\int }_{A_i}{r}_{i}d{Q}_i={\prod }_{i<n}{P}_i(A_i)=P(E)$ for every measurable rectangle $E={\prod }_{i<n}{A}_i$; applying the same calculation to the full space gives $\int rdQ=1$.

The measure $rdQ$ is a probability measure by [F4] and step 1.2. The product rectangles form a generating pi-system and the finite product spaces are sigma-finite, so [F3] and the rectangle identity in step 1.2 imply $rdQ=P$ on the entire product sigma-algebra. Therefore $r=dP\mathrm{/}dQ$ as a Radon–Nikodym class.

Each $r_i$ is finite $Q_i$-almost everywhere because $\int r_{i}d{Q}_i=1$, and $P_i(\{r_i=0\})={\int }_{\{r_i=0\}}{r}_{i}d{Q}_i=0$; thus $0<r_i<\mathrm{\infty }$ $P_i$-almost everywhere. Therefore [F6] gives $\log r={\sum }_{i<n}\log r_i(x_i)$ $P$-almost everywhere. Each $\log r_i$ has integrable negative part by [F1], and the negative part of their finite sum is bounded by the sum of those negative parts. Split each summand into positive and negative parts: Tonelli factors each nonnegative part through the product law, and Fubini applies to the negative parts, which are integrable. Hence, allowing $+\mathrm{\infty }$ for positive parts, $\int \log rdP={\sum }_{i<n}\int \log r_{i}d{P}_i={\sum }_{i<n}D(P_i\mathrm{\parallel }{Q}_i)\mathrm{.}$ By [F1] and step 2.1 the left side is $D(P\mathrm{\parallel }Q)$, proving the claim.