Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 108 1 Solution Created 2026-10-03 Updated 2026-10-05
Use the probability-system convention . The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and ,where is the invariant sigma-algebra. The limit is integrable and has the same integral as . On a probability space the convergence also holds in , as in the allowed mean ergodic theorem. If is an ergodic transformation, is trivial modulo null sets, giving
The integer multiplication map on the circle preserves Lebesgue measure: for any integrable on ,To prove the ergodicity of integer multiplication on the circle, suppose satisfies . Let be its Fourier coefficients in the Fourier basis . Since , the Fourier coefficients of at index are zero if does not divide , and are otherwise. This identity holds for all functions by approximation with trigonometric polynomials and the isometry . Invariance givesEvery nonzero integer can be divided by only finitely often. Thus for every , and completeness of the Fourier basis makes constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so
A normal number in base has every word of base- digits occurring with limiting overlapping frequency . Use the expansion that is not eventually equal to when there are two expansions. The word corresponds to the half-open intervalA word starting at position occurs exactly when . The Birkhoff ergodic theorem, applied to , gives frequency almost everywhere. There are countably many pairs , so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base- intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.
For the growth assertion, put . For every , the Tonelli theorem gives the useful summability boundSince is a measure-preserving transformation, . The first Borel-Cantelli lemma shows that occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for proves the linear growth bound for integrable observables, . Multiplication by then gives
The threshold is sharp. For , choose with , and take the Bernoulli shift on with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define ; it is integrable becauseThe variables are independent. For any fixed ,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer even yields . For , the constant observable already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 108 3 Solution Created 2026-10-03 Updated 2026-10-05
Let denote the join of measurable partitions recording the first observations, and set , with . The entropy of a countable measurable partition is , using natural logarithms and . Null atoms can be discarded. Throughout, is a probability measure.
The stronger property needed for monotonicity of normalized block entropy is that its increments decrease. DefineSince conditioning reduces entropy, . The conditional entropy chain rule and measure preservation giveIt follows that , and consequentlyThis uses stationarity as well as the entropy chain rule; subadditivity alone would not establish monotonicity of every successive ratio.
The entropy rate of a measurable partition and the Kolmogorov-Sinai entropy are, respectively,One may equivalently take the supremum over countable finite-entropy measurable partitions.
The Shannon-McMillan-Breiman theorem states that for a countable measurable partition with , the normalized informationconverges almost everywhere and in to a invariant function with integral . Here is the atom containing . If is an ergodic transformation, then almost everywhere. We prove the general form, including an explicit formula for its limit.
Let be the sigma-algebra generated by , let be trivial, and put . For every atom , the probabilitiesform a bounded conditional-expectation martingale. The Martingale convergence theorem gives almost everywhere and in . Each of these probabilities is positive almost everywhere on : for example, integrating over the measurable set gives zero. Countability of lets us choose a common full-measure set for every atom.
The conditional information functionstherefore converge almost everywhere to . We need an integrable bound on this sequence; convergence of the probabilities alone would not supply one after taking logarithms. The allowed maximal inequality for conditional information functions gives, for ,Combining this with the bound by and the tail integral formula for moments yieldsThus , and the dominated convergence theorem gives in . In particular,because is the average of the decreasing sequence .
The information chain rule, applied from the final observation backwards, gives the exact identityIndeed, the symbol at time is conditioned on times ; pulling that conditional probability back by gives . Measure preservation is sufficient for this pullback identity.
We now prove the required triangular ergodic averaging lemma in this instance. Put . Then almost everywhere, , and . Split the difference between the displayed triangular sum and . Terms whose index is at least are bounded by . The remaining terms are at mostAfter division by , each of these finitely many end terms tends to zero almost everywhere by the linear growth bound for integrable observables proved in Question 1. Applying the Birkhoff ergodic theorem to , for every fixed , therefore givesThese conditional expectations decrease to zero almost everywhere: they decrease, and their integrals tend to zero. Letting , then applying the Birkhoff ergodic theorem to , proves
Triangular ergodic averaging lemma 2026-10-05
Suppose almost everywhere and in a probability measure-preserving system. Thenalmost everywhere and in , where is the invariant sigma-algebra. For the almost-everywhere assertion, bound the terms with index at least by , apply the Birkhoff ergodic theorem, and then let . The finitely many remaining end terms vanish by the linear growth bound for integrable observables.