DO WE HAVE ALMOST EVERYWHERE CONVERGENCE OF CONVOLUTION POWERS FOR L1 FUNCTIONS?

Let (X,Σ,m) denote a complete non-atomic probability space, and let τ be a measurable measure preserving invertible point transformation from X to itself. Let µ be a probability measure on Z and define the operator µ by µ(f)(x) = k=−∞ µ(k)f(τkx). We can apply the operator µ to the function µ(f), and more generally for n> 1, define µn(f)(x) = µ(µn−1f)(x). For t ∈ [0, 1), let µ̂(t) =∑∞k=− ∞ µ(k)e2piikt. Definition 0.1. A probability measure µ on Z is said to satisfy the bounded angu-lar ratio condition if µ is strictly aperiodic (that is, the support of µ is not contained in a proper coset of Z) and there is a finite constant B such that sup t6=0 |1 − µ̂(t)| 1 − |µ̂(t) | < B.(0.1) In [2] it was shown that if µ satisfies the bounded angular ratio condition, then limn→ ∞ µn(f)(x) exists a.e. for all f ∈ Lp(X), 1 < p ≤ ∞. Subsequently it was shown that if condition (0.1) fails, then the averages can diverge, even for bounded functions. This left open the case f ∈ L1(X). This case was partially solved by A. Bellow and A.P. Calderón [1], who showed if µ satisfies the bounded angular ratio condition and k=− ∞ k 2µ(k) < ∞ then convergence holds for f ∈ L1. We now have the following problem: Problem 0.2. Let µ satisfy the bounded angular ratio condition, but have k=−∞ k2µ(k) =∞. Does µnf converge a.e. for all f ∈ L1(X)? If the answer to the above problem is no, we can then ask what additional conditions can be placed on the measure µ so that the answer is yes? We already know that for such µ we have convergence on a dense class in L1(X), since we have convergence for all f ∈ Lp, p> 1. Consequently it will be enough t