Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 55 / 3

A qualification is necessary: an arbitrary symbolic Gibbs measure does not satisfy the claim. Consider the doubling map with its two binary branches and a Bernoulli symbolic Gibbs measure with probabilities $2/3$ and $1/3$. This is the Gibbs law for a locally constant symbolic potential, while every level-$n$ interval has length $2^{-n}$. Along the nonboundary alternating itinerary, for even $n$,No finite $g$ can approximate these $g_n$ with a decaying uniform error. This proves the reusable fact that an arbitrary Gibbs measure need not be absolutely continuous.

For the intended geometric version, take the Gibbs measure for the geometric potential of an expanding map $-\log |T^{\prime }|$, with the standard sufficiently smooth finite full-branch hypotheses. The allowed transfer-operator result supplies a strictly positive Lipschitz continuous invariant probability density function $h$, normalized by $\int h=1$, and its symbolic weights of cylinder sets are $\nu (C_w)={\int }_{{\Delta }_w}h(x)dx$. State the needed regularity explicitly: $h\ge h_{\ast }>0$ and $|h(x)-h(y)|\le L_h|x-y|$. Expansion gives $\mathrm{diam}{\Delta }_w\le {\lambda }^{-n}$.

Let $\pi (ϵ)$ denote the point coded by the infinite sequence, and define $g(ϵ)=\log h(\pi (ϵ))$. The expression defining $g_n$ is the logarithm of the average of $h$ over the cylinder. For $x=\pi (ϵ)\in {\Delta }_w$,Since the logarithm is Lipschitz continuous with constant $1/h_{\ast }$ on this positive range, the cylinder averages of an invariant density giveThis proves the intended estimate. Expansion alone and an unspecified Gibbs symbolic potential are insufficient; the geometric choice and the regularity producing a positive Lipschitz continuous probability density function are substantive hypotheses.

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 givesFix a level-$m$ cylinder $C_v$ and subdivide its interval into level-$n$ intervals, with $n\ge m$. Integrating these inequalities and summing the subcylinder masses yieldsLetting $n\to \infty$ proves the exact equality ${\int }_{{\Delta }_v}pdx=\nu (C_v)$. 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 ${\pi }_{\ast }\nu$.

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 $B$,Consequentlyis 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.