Likelihood ratio martingale

Repair first the integral foundation inherited by RN. Augment any finite disjoint display of a nonnegative simple function by the complement with coefficient $0$. Pairwise intersections of two augmented displays partition the whole space and have equal coefficients on nonempty cells; finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants and the sets $\{f_j\ge cs\}$, $0<c<1$, give monotone convergence; increasing simple approximations give nonnegative additivity, and positive/negative plus real/imaginary decompositions give finite $L^1$ linearity. With these facts substituted for the affected foundation, the cited RN proof applies. Apply it to $\mu =P,\nu =Q$ with the constant exhaustion $X_j=\mathrm{\Omega }$. Both masses and the total variation of the positive $Q$ are one. Thus $L$ is real measurable, integrable and ${\int }_{A}LdP=Q(A)$ for every $A\in {\mathcal{F}}_N$. It is nonnegative a.s.: for each positive integer $j$, put $B_j=\{L\le -1\mathrm{/}j\}$. Then $0\le Q(B_j)={\int }_{B_j}LdP\le -P(B_j)\mathrm{/}j$, so each $B_j$ is null, and ${\bigcup }_{j\ge 1}{B}_j=\{L<0\}$. Testing $\mathrm{\Omega }$ gives $E_{P}L=1$.

Extend the filtration constantly after $N$ to apply [F3]; up to $N$, it makes $Z$ a martingale. Conditional positivity gives $Z_n\ge 0$ a.s. For every $A\in {\mathcal{F}}_n$, the defining event identity gives ${\int }_A{Z}_{n}dP={\int }_{A}LdP=Q(A)$, exactly the restricted density assertion. At $n=N$ known-variable conditioning gives $Z_N=L$. If ${\mathcal{F}}_0$ is trivial, the constant one has the same integrals as $L$ on its two events, so it is the conditional class $Z_0$. AC covers RN and CE existence and the finite choice of versions.

For instance let $N=1$, $\mathrm{\Omega }=\{a,b\}$, $P(a)=P(b)=1\mathrm{/}2$, $Q(a)=3\mathrm{/}4$, $Q(b)=1\mathrm{/}4$, and ${\mathcal{F}}_0$ trivial. Then $L(a)=3\mathrm{/}2$, $L(b)=1\mathrm{/}2$, and $Z_0=1$, $Z_1=L$. The average $(3\mathrm{/}2+1\mathrm{/}2)\mathrm{/}2=1$ verifies the martingale equality, and $(1\mathrm{/}2)(3\mathrm{/}2)=3\mathrm{/}4$ verifies the restricted density on $\{a\}$.