An admissible product has a finite coordinate normalizer

Put $\alpha =q_j{\mu }_j$ and $\nu =v{\mu }_{-j}$. They are probabilities and the law with density $q$ is their product by [F1]–[F3]. A density is positive and finite almost surely under its own probability, since its zero set has mass zero and its infinite set has reference measure zero. Thus $A(x)=\log q_j(x)$ and $B(y)=\log v(y)$ are finite almost surely, and [F8] gives $\log q=A+B$ almost surely. Choose a finite integer $K$ for which $p:=\nu (\mathrm{\mid }B\mathrm{\mid }\le K)>0$; such a $K$ exists because those events increase to a full-measure set. On that event $\mathrm{\mid }A\mathrm{\mid }\le \mathrm{\mid }A+B\mathrm{\mid }+K$, so product integration [F2]–[F3] yields $p\int \mathrm{\mid }A\mathrm{\mid }d\alpha \le \int \mathrm{\mid }A+B\mathrm{\mid }d(\alpha \otimes \nu )+Kp<\mathrm{\infty }$. Interchanging $A$ and $B$ and choosing an analogous positive-probability bounded event proves $\int \mathrm{\mid }B\mathrm{\mid }d\nu <\mathrm{\infty }$. Hence $H=\int Bd\nu$ is finite, including $H=0$ for the unit complementary factor.

By [F2] and $Z<\mathrm{\infty }$, the marginal $G(x):=\int \gamma (x,y)d{\mu }_{-j}(y)$ is finite for ${\mu }_j$-almost every $x$. Fix such an $x$. Under $\nu$, define $g(y)=\gamma (x,y)\mathrm{/}v(y)$ on $\{0<v<\mathrm{\infty }\}$; arbitrary values on its $\nu$-null complement do not matter. Then $\int gd\nu ={\int }_{\{0<v<\mathrm{\infty }\}}\gamma (x,y)d{\mu }_{-j}(y)\le G(x)$ by [F3]. In particular $g$ is finite almost surely. Write $L=\log g$, allowing $L=-\mathrm{\infty }$ at $g=0$. Its positive part is integrable since ${\log }^{+}g\le g$, and the defined expected-log convention and integrability of $B$ give $\int Ld\nu =h_j(x)-H$ in the extended sense; in particular $h_j(x)$ cannot equal $+\mathrm{\infty }$.

Let $L_n=\max (L,-n)$. Its negative part is bounded and its positive part is integrable by step 2.1, so $L_n$ is integrable. Also $e^{L_n}=\max (g,e^{-n})\le g+1$ is integrable. Apply [F4] with the convex exponential [F5] to obtain $\exp (\int L_{n}d\nu )\le \int e^{L_n}d\nu$. The negative parts increase to $L^{-}$, so [F6] gives $\int L_{n}d\nu \downarrow \int Ld\nu$, possibly $-\mathrm{\infty }$. By [F7], the right side tends to $\int gd\nu$, with domination $g+1$; continuity of the exponential and $e^{-\mathrm{\infty }}=0$ therefore give $\exp (h_j(x)-H)\le \int gd\nu \le G(x)$.

The inequality in step 3.1 holds for ${\mu }_j$-almost every $x$. Multiply it by $e^H$ and integrate using [F2] to obtain $c_j\le e^H\int Gd{\mu }_j=e^{H}Z<\mathrm{\infty }$. Since $h_j$ is finite $\alpha$-almost surely, $e^{h_j}>0$ on an $\alpha$-full measurable set. If $c_j=0$, [F9] would make this set ${\mu }_j$-null, hence $\alpha$-null by [F3], contradicting $\alpha ({\mathrm{\Omega }}_j)=1$. Thus $c_j>0$. This does not prove admissibility of the normalized replacement, which remains a separate requirement.