## Fubini foiled

An important issue in hyperbolic dynamics is that of absolute continuity. Suppose some neighbourhood ${U}$ of a smooth manifold ${M}$ is foliated by a collection of smooth submanifolds ${\{W_\alpha \mid \alpha\in A\}}$, where ${A}$ is some indexing set. (Here “smooth” may mean ${C^1}$, or ${C^2}$, or even more regularity depending on the context.)

Fixing a Riemannian metric on ${M}$ gives a notion of volume ${m}$ on the manifold ${M}$, as well as a notion of “leaf volume” ${m_\alpha}$ on each ${W_\alpha}$, which is just the volume form coming from the induced metric. Then one wants to understand the relationship between ${m}$ and ${m_\alpha}$.

The simplest example of this comes when ${U=(0,1)^2}$ is the (open) unit square and ${\{W_\alpha\}}$ is the foliation into horizontal lines, so ${I=(0,1)}$ and ${W_\alpha = (0,1)\times \{\alpha\}}$. Then Fubini’s theorem says that if ${E}$ is any measurable set, then ${m(E)}$ can be found by integrating the leaf volumes ${m_\alpha(E)}$. In particular, if ${m(E)=0}$, then ${m_\alpha(E)=0}$ for almost every ${\alpha}$, and conversely, if ${E}$ has full measure, then ${m_\alpha(E)=1}$ for almost every ${\alpha}$.

For the sake of simplicity, let us continue to think of a foliation of the two-dimensional unit square by one-dimensional curves, and assume that these curves are graphs of smooth functions ${\phi_\alpha\colon (0,1)\rightarrow (0,1)}$. The story in other dimensions is similar.

Writing ${\Phi(x,y) = \phi_y(x)}$ gives a map ${\Phi\colon (0,1)^2 \rightarrow (0,1)^2}$, and we see that our foliation is the image under ${\Phi}$ of the foliation by horizontal lines. Now we must make a very important point about regularity of foliations. The leaves of the foliation are smooth, and so ${\Phi}$ depends smoothly on ${x}$. However, no assumption has been made so far on the transverse direction — that is, dependence on ${y}$. In particular, we cannot assume that ${\Phi}$ depends smoothly on ${y}$.

If ${\Phi}$ depends smoothly on both ${x}$ and ${y}$, then we have a smooth foliation, not just smooth leaves, and in this case basic results from calculus show that a version of Fubini’s theorem holds:

$\displaystyle m(E) = \int_A \int_{W_\alpha} \rho_\alpha(z) {\mathbf{1}}_E(z) \,dm_\alpha(z)\,d\alpha,$

where the density functions ${\rho_\alpha\in L^1(W_\alpha,m_\alpha)}$ can be determined in terms of the derivative of ${\Phi}$.

It turns out that when the foliation is by local stable and unstable manifolds of a hyperbolic map, smoothness of the foliation is too much to ask for, but it is still possible to find density functions ${\rho_\alpha}$ such that Fubini’s theorem holds, and the relationship between ${m}$-null sets and ${m_\alpha}$-null sets is as expected.

However, there are foliations where the leaves are smooth but the foliation as a whole does not even have the absolute continuity properties described in the previous paragraph. This includes a number of dynamically important examples, namely intermediate foliations (weak stable and unstable, centre directions in partial hyperbolicity) for certain systems. These take some time to set up and study. There is an elementary non-dynamical example first described by Katok; a similar example was described by Milnor (Math. Intelligencer 19(2), 1997), and it is this construction which I want to outline here, since it is elegant, does not require much machinery to understand, and neatly illustrates the possibility that absolute continuity may fail.

The example consists of two constructions: a set ${E}$ and a foliation of the square by a collection of smooth curves ${W_\alpha}$. These are built in such a way that ${E}$ has full Lebesgue measure (${m(E)=1}$) but intersects each curve ${W_\alpha}$ in at most a single point, so that in particular ${m_\alpha(E)=0}$ for all ${\alpha}$.

1. The set

Given ${p\in (0,1)}$, consider the piecewise linear map ${f_p\colon (0,1)\rightarrow (0,1]}$ given by

$\displaystyle f_p(y) = \begin{cases} \frac yp & y\leq p \\ \frac {y-p}{1-p} &y >p \end{cases}$

It is not hard to see that ${f_p}$ preserves Lebesgue measure ${dy}$ and that ${(f_p,dx)}$ is measure-theoretically conjugate to ${(\sigma,\mu_p)}$, where ${\sigma}$ is the shift map on ${\Sigma=\{0,1\}^{\mathbb N}}$ and ${\mu_p}$ is the ${(p,1-p)}$-Bernoulli measure. The conjugacy ${\pi_p\colon (0,1) \rightarrow \Sigma}$ is given by coding trajectories according to whether the ${n}$th iterate falls into ${I_0 = (0,p]}$ or ${I_1=(p,1)}$.

Given ${y}$, let ${\beta_n(y)}$ be the number of times that ${y,f_p(y),\dots,f_p^{n-1}(y)}$ lands in ${I_1}$. Then by the strong law of large numbers, Lebesgue almost every ${y\in (0,1)}$ has ${\frac 1n \beta_n(y) \rightarrow 1-p}$. Let ${E}$ be the set of all pairs ${(p,y)}$ such that this is true; then ${m(E) = 1}$ by Fubini’s theorem (applied to the foliation of the unit square into vertical lines). It only remains to construct a foliation that intersects ${E}$ in a “strange” way.

2. The foliation

The curves ${W_\alpha}$ will be chosen so that two points ${(p,y)}$ and ${(p',y')}$ lie on the same ${W_\alpha}$ if and only if they are coded by the same sequence in ${\Sigma}$ — that is, if and only if ${\pi_p(y) = \pi_{p'}(y')}$. Then because the limit of ${\frac 1n\beta_n}$ must be constant on ${W_\alpha}$ if it exists, we deduce that each ${W_\alpha}$ intersects the set ${E}$ at most once.

It remains to see that this condition defines smooth curves. Given ${\alpha\in (0,1)}$, let ${a=\pi_{1/2}(\alpha)=a_1a_2\cdots}$ be the binary expansion of ${\alpha}$, so that ${\alpha = \sum a_n 2^{-n}}$. Now let ${W_\alpha = \{(p,y) \mid \pi_p(y)=a\}}$.

Fix ${p\in (0,1)}$, let ${y_1=y = \phi_\alpha(p)}$ be such that ${(p,y)\in W_\alpha}$, and let ${y_{n+1}=f_p(y_n)}$, so that ${y_n\in I_{a_n}}$ for all ${n}$. Thus

$\displaystyle y_{n+1} = \begin{cases} \frac {y_n}{p} & a_n=0 \\ \frac{y_n-p}{1-p} & a_n=1 \end{cases}$

For convenience of notation, write ${p(0)=p}$ and ${p(1)=1-p}$. Then the relationship between ${y_{n+1}}$ and ${y_n}$ can be written as

$\displaystyle y_{n+1} = \frac{y_n - a_n p(0)}{p(a_n)} \qquad\Rightarrow\qquad y_n = p(0)a_n + p(a_n)y_{n+1}.$

This can be used to write ${y=\phi_\alpha(p)}$ in terms of the sequence ${a_n}$. Indeed,

\displaystyle \begin{aligned} \phi_\alpha(p) &=y = y_1 = p(0)a_1 + p(a_1)y_2 \\ &= p(0)a_1 + p(a_1)\big(p(0) a_2 + p(a_2)y_3\big) \\ &= p(0)\big(a_1 + p(a_1)a_2\big) + p(a_1)p(a_2)y_3 \\ &= p(0)\big(a_1 + p(a_1)a_2 + p(a_1)p(a_2)a_3\big) + p(a_1)p(a_2)p(a_3)y_4, \end{aligned}

and so on, so that writing ${\psi_n(p) = p(a_1)p(a_2)\cdots p(a_{n-1})}$, one has

$\displaystyle \phi_\alpha(p) = p(0) \sum_{n=1}^\infty \psi_n(p) a_n.$

The summands are analytic functions of ${p}$, and the sum converges uniformly on each interval ${[\epsilon,1-\epsilon]}$, since for ${p\in [\epsilon, 1-\epsilon]}$ we have ${\psi_n(p) \leq (1-\epsilon)^n}$. In fact, this uniform convergence extends to complex values of ${p}$ in the disc of radius ${\frac 12 - \epsilon}$ centred at ${\frac 12}$, and so by the Weierstrass uniform convergence theorem, the function ${\phi_\alpha}$ is analytic in ${p}$.

As discussed above, ${W_\alpha}$ is the graph of the analytic function ${\phi_\alpha}$, and each ${W_\alpha}$ intersects ${E}$ at most once, despite the fact that ${E}$ has Lebesgue measure 1 in the unit square. This demonstrates the failure of absolute continuity.

2. Unfortunately I’m not aware of any place where the details are clearly spelled out in a manner that could be easily referenced: most papers/books that I’ve seen refer to this just include a line roughly equivalent to what I wrote. That said, it only takes a few lines to complete the argument: writing $J$ for the Jacobian determinant of $\Phi$ and writing $(x',y')=\Phi(x,y)$ gives $\displaystyle m(E) = \int_{(0,1)} \int_{(0,1)} J(x,y)^{-1} \mathbf{1}_E(x,y) \,dx' \,dy'$ by Fubini’s theorem. Let $J_\alpha$ be the Jacobian derivative of $\Phi|_{W_\alpha}$, so $dx' = J_\alpha dm_\alpha$. Finally, observe that $\Phi(W_\alpha) = (0,1)\times \{y(\alpha)\}$ where $\alpha\mapsto y(\alpha)$ is differentiable, and so $dy = y'(\alpha) d\alpha$. This gives the formula in the post by putting $\displaystyle \rho_\alpha = \frac{J_\alpha \cdot y'}{J}$.