= Solution
Assume the uniform estimate obtained in the geometric setting of part (i). The limit $g$ is bounded and continuous on the finite-alphabet symbolic space, so $p(x)=e^{g(E(x))}$ is measurable, positive, and bounded. Possible duplicate itineraries at partition endpoints do not affect its <Lebesgue measure> density.
For a level-$n$ interval $\Delta_w$, the estimate gives
$$
e^{-C\lambda^{-n}}\frac{\nu(C_w)}{|\Delta_w|}
\leq p(x)\leq
e^{C\lambda^{-n}}\frac{\nu(C_w)}{|\Delta_w|},\qquad x\in\Delta_w.
$$
Fix a level-$m$ cylinder $C_v$ and subdivide its interval into level-$n$ intervals, with $n\geq m$. Integrating these inequalities and summing the subcylinder masses yields
$$
e^{-C\lambda^{-n}}\nu(C_v)\leq\int_{\Delta_v}p(x)\,dx\leq e^{C\lambda^{-n}}\nu(C_v).
$$
Letting $n\to\infty$ proves the exact equality $\int_{\Delta_v}p\,dx=\nu(C_v)$. The same argument on the entire interval gives $\int p=1$. Shrinking cylinders generate the interval Borel sigma-algebra, so the <Monotone class theorem> identifies $p(x)\,dx$ with the pushforward $\pi_*\nu$.
The symbolic <Gibbs measure> in this setting is shift invariant, and coding satisfies $T\circ\pi=\pi\circ\sigma$ away from the endpoint exceptions. These exceptions form a <null set>: bounded $p$ and shrinking cylinders give zero mass to individual endpoints. Therefore, for every <Borel set> $B$,
$$
(\pi_*\nu)(T^{-1}B)=\nu(\sigma^{-1}\pi^{-1}B)=\nu(\pi^{-1}B)=(\pi_*\nu)(B).
$$
Consequently
$$
\boxed{d\mu(x)=e^{g(E(x))}\,dx}
$$
is an absolutely continuous <invariant measure>. This implication uses the uniform cylinder estimate directly and does not rely on naming an invariance theorem in place of the requested argument.
Back to article page
