[proofplan] The proof is a bookkeeping argument based on the assumed one-step Gibbs $u$-transformation rule. For two points on the same unstable plaque, the plaque-normalising constant cancels when the transformation rule is divided at the two points. Iterating this ratio identity backward gives a finite product multiplied by the ratio of the densities on the $n$th backward plaque. The assumed backward asymptotic normalisation of the densities sends the remaining density ratio to $1$, so the finite products converge to the stated density ratio, which is finite and strictly positive. [/proofplan]

[step:Choose a full-measure domain where all ratios are defined] Let $\Lambda_1 \subset \Lambda_0$ be a full $\mu$-measure subset such that, for every $x\in \Lambda_1$ with $f^{-1}x\in \Lambda_1$, the one-step transformation rule holds on an $m_x^u$-full subset of $V_x$. Since $\rho_x>0$ for $m_x^u$-a.e. point of $\xi(x)$, we further restrict these full-measure plaque subsets so that every density appearing in a denominator is strictly positive. Fix $x\in \Lambda_1$ in the full-measure set for which the backward density-ratio hypothesis holds. Let $(y,z)\in \xi(x)\times \xi(x)$ be a backward-compatible pair in the corresponding full $m_x^u\otimes m_x^u$-measure set. For each integer $k\geq 0$, define \begin{align*} x_k:=f^{-k}x,\qquad y_k:=f^{-k}y,\qquad z_k:=f^{-k}z. \end{align*} By backward compatibility, $y_k,z_k\in \xi(x_k)$ for every $k\geq 0$, and the one-step identity holds at both points $y_k$ and $z_k$ whenever it is used. [/step]

[step:Divide the one-step transformation rule so the plaque constant cancels]Fix an integer $k\geq 0$. Applying the one-step transformation rule to the plaque $\xi(x_k)$ at the points $y_k$ and $z_k$ gives \begin{align*} \rho_{x_k}(y_k)=c(x_{k+1})\frac{\rho_{x_{k+1}}(y_{k+1})}{J^u f(y_{k+1})}. \end{align*} and \begin{align*} \rho_{x_k}(z_k)=c(x_{k+1})\frac{\rho_{x_{k+1}}(z_{k+1})}{J^u f(z_{k+1})}. \end{align*} Here $c(x_{k+1})>0$ depends only on the source plaque $\xi(x_{k+1})$, not on the point in the image subplaque. Since all denominator densities and unstable Jacobians under consideration are strictly positive, dividing the first identity by the second yields \begin{align*} \frac{\rho_{x_k}(y_k)}{\rho_{x_k}(z_k)} = \frac{\rho_{x_{k+1}}(y_{k+1})}{\rho_{x_{k+1}}(z_{k+1})} \frac{J^u f(z_{k+1})}{J^u f(y_{k+1})}. \end{align*}[/step]

[guided]The important point is that the one-step identity contains a normalising constant, but this constant is attached to the whole source plaque rather than to an individual point. For the fixed source plaque $\xi(x_{k+1})$, the same number $c(x_{k+1})>0$ appears when the identity is evaluated at $y_k$ and at $z_k$. Thus the transformation rule gives \begin{align*} \rho_{x_k}(y_k)=c(x_{k+1})\frac{\rho_{x_{k+1}}(y_{k+1})}{J^u f(y_{k+1})}. \end{align*} It also gives \begin{align*} \rho_{x_k}(z_k)=c(x_{k+1})\frac{\rho_{x_{k+1}}(z_{k+1})}{J^u f(z_{k+1})}. \end{align*} The backward-compatible full-measure domain was chosen so that both identities hold and so that the density values in denominators are positive. The unstable Jacobian is also strictly positive along the unstable bundle because $df_p|_{E^u_p}:E^u_p\to E^u_{f(p)}$ is a linear isomorphism for each $p\in \Lambda$. Therefore division is legitimate. Dividing the two displayed formulas cancels the common plaque constant and gives \begin{align*} \frac{\rho_{x_k}(y_k)}{\rho_{x_k}(z_k)} = \frac{\rho_{x_{k+1}}(y_{k+1})}{\rho_{x_{k+1}}(z_{k+1})} \frac{J^u f(z_{k+1})}{J^u f(y_{k+1})}. \end{align*} This cancellation is the mechanism behind the density-ratio formula: the individual conditional densities depend on plaque normalisation, but ratios inside the same plaque do not.[/guided]

[step:Iterate the ratio identity along the backward orbit] For each integer $n\geq 1$, repeated application of the previous identity with $k=0,1,\dots,n-1$ gives \begin{align*} \frac{\rho_x(y)}{\rho_x(z)} = \frac{\rho_{x_n}(y_n)}{\rho_{x_n}(z_n)} \prod_{j=1}^{n}\frac{J^u f(z_j)}{J^u f(y_j)}. \end{align*} Since $x_n=f^{-n}x$, $y_n=f^{-n}y$, and $z_n=f^{-n}z$, this is equivalently \begin{align*} \frac{\rho_x(y)}{\rho_x(z)} = \frac{\rho_{f^{-n}x}(f^{-n}y)}{\rho_{f^{-n}x}(f^{-n}z)} \prod_{j=1}^{n}\frac{J^u f(f^{-j}z)}{J^u f(f^{-j}y)}. \end{align*} [/step]

[step:Pass to the limit and identify the infinite product] Define the $n$th finite product $P_n(y,z)$ by \begin{align*} P_n(y,z):=\prod_{j=1}^{n}\frac{J^u f(f^{-j}z)}{J^u f(f^{-j}y)}. \end{align*} The iterated identity can be written as \begin{align*} P_n(y,z) = \frac{\rho_x(y)}{\rho_x(z)} \frac{\rho_{f^{-n}x}(f^{-n}z)}{\rho_{f^{-n}x}(f^{-n}y)}. \end{align*} By the assumed backward asymptotic normalisation, \begin{align*} \lim_{n\to\infty}\frac{\rho_{f^{-n}x}(f^{-n}y)}{\rho_{f^{-n}x}(f^{-n}z)}=1. \end{align*} Taking reciprocals is valid because the ratios are strictly positive on the chosen full-measure domain, hence \begin{align*} \lim_{n\to\infty}\frac{\rho_{f^{-n}x}(f^{-n}z)}{\rho_{f^{-n}x}(f^{-n}y)}=1. \end{align*} Therefore the finite products converge and satisfy \begin{align*} \lim_{n\to\infty}P_n(y,z)=\frac{\rho_x(y)}{\rho_x(z)}. \end{align*} Since $\rho_x(y)>0$ and $\rho_x(z)>0$ on the selected full-measure pair domain, the limiting value is finite and strictly positive. This proves \begin{align*} \frac{\rho_x(y)}{\rho_x(z)} = \prod_{k=1}^{\infty}\frac{J^u f(f^{-k}z)}{J^u f(f^{-k}y)}. \end{align*} The product is precisely the limit of its finite partial products, so the asserted convergence, finiteness, and strict positivity all follow. [/step]