Lebesgue measure isnot G-invariant

A transformation may have many invariant measures, but some are more improtant than others.

Example 1: To nd the digits in the standardcontinued fraction expansion of a real numberx:

x =a₀+ 1 a₁+ ¹

a₂+ ¹

a3+ 1

...

= [a_o :a₁,a₂,a₃, . . .]

we need theGauÿ map G : [0,1)→(0,1]:

G(x) = ¹_x − b¹_xc.

a₀ =bxc x₁ =x−a₀.

a_i =b1/x_ic x_i+1 =G(x_i) (stop ifx_i+1=0).

For example:

π= [3;7,15,1,292,1,1,1, . . .] and

e = [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10, . . .] Lebesgue measure isnot G-invariant.

Denition: A measure µis called absolutely continuousw.r.t. the measureν (notation: µν) if ν(A) =0 impliesµ(A) =0. If both µν andν µ, thenµandν are calledequivalent.

Theorem of Radon-Nikodym: If µis a probability measure and µν then there is a functionh ∈L¹(ν)(calledRadon-Nikodym derivativeordensity) such that µ(A) =R

Ah(x) dν(x) for every measurable setA.

Notation: h(x) = ^dµ(x)_dν(x).

Example 1: Lebesgue measure is notG-invariant, but there is a probability measureµ_G that is absolutely continuous w.r.t.

Lebesgue.

µG(A) = Z

A

h(x)dx for h(x) = 1 log2

1 1+x.

Proposition 1. Suppose that µν are bothT-invariant

probability measures, with a commonσ-algebraBof measurable sets. If ν is ergodic, then µ=ν.

Proof: First we show thatµis ergodic. Indeed, otherwise there is a T-invariant setAsuch that µ(A)>0 and µ(A^c)>0. By

ergodicity ofν at least one ofAor A^c must haveν-measure 0, but this would contradict thatµν.

Now letA∈ B and letY ⊂X be the set of ν-typical points. Then ν(Y^c) =0 and hence µ(Y^c) =0. Applying Birkho's Ergodic Theorem toµ andν separately for ψ=1A and some µ-typical y∈Y, we get

µ(A) = lim

n

1 n

n−1

X

i=0

ψ◦T(y) =ν(A).

ButA∈ B was arbitrary, soµ=ν.

Exercise 4.1: Show that the condition of ergodicity is essential for Proposition 1?

Ifµν andµis ergodic. Does it follow that µ=ν?

Proposition 2: LetT :U ⊂Rⁿ→U be (piecewise) dierentiable, andµ is absolutely continuous w.r.t. Lebesgue. Thenµis

T-invariant if and only if its densityh = ^dµ_dx satises h(x) = X

T(y)=x

h(y)

|detDT(y)| (1) for everyx.

Proof of Proposition 2: TheT-invariance means that dµ(x) =dµ(T⁻¹(x)), but we need to be aware thatT⁻¹ is multivalued. So it is more careful to split the spaceU into pieces U_n such that the restrictionsT_n:=T|U_n are dieomorphic (onto their images) and writeyn=T_n⁻¹(x) =T⁻¹(x)∩Un. Then we obtain (using the change of coordinates)

h(x) dx = dµ(x) =dµ(T⁻¹(x)) =X

n

dµ◦T_n⁻¹(x)

= X

n

h(yn)|det(DT_n⁻¹)(x)|dy_n=X

n

h(yn)

det|DT(y_n)|dyn. Conversely, if (1) holds, then the above computation gives

dµ(x) =dµ◦T⁻¹(x).

Example 1 continued: TheGauÿ map has invariant density h(x) = _log¹_21+x¹ . Here _log¹₂ is just the normalising factor (so that R₁

0 h(x)dx =1).

LetI_n= (_n+¹₁,¹_n]for n=1,2,3, . . . be the domains of the branches ofG, and forx ∈(0,1), and y_n:=G⁻¹(x)∩I_n= _x+n¹ . AlsoG⁰(yn) =−_y¹₂

n. Therefore X

n≥1

h(yn)

|G⁰(yn)| = 1 log2

X

n≥1

y_n² 1+yn

= 1 log2

X

n≥1

(x+n)1 ²

1+_x+n¹

= 1

log2 X

n≥1

1

x+n · 1 x+n+1

= 1

log2 X

n≥1

1

x+n − 1

x+n+1 telescoping series

= 1

log2 1

x+1 =h(x).

Exercise 4.2: Compute the average frequency of the digit 1 for points that are normal w.r.t. the standard continued fraction.

Exercise 4.3: Show that for each integern≥2, the interval map given by

T_n(x) =

(nx if 0≤x ≤ ¹_n,

x1 − b_x¹c if ¹_n <x ≤1, has invariant densityh(x) = _log¹_21+x¹ .

Example 2: The mapT :R\ {0} →R, T(x) =x−¹_x is called the Boole transformation. It is 2-to-1; the two preimages of x∈Rare y±= ¹₂(x±√

x²+4). Clearly T⁰(x) =1+_x¹₂. It can be shown

that 1

|T⁰(y−)|+ 1

|T⁰(y+)| =1. Indeed,

|T⁰(y±)|=1+ 2 x²+2±x√

x²+4

and 1

|T⁰(y±)| = x²+2±x√ x²+4 x²+4±x√

x²+4, and

1

|T⁰(y−)|+ 1

|T⁰(y+)|

= x²+2−x√ x²+4 x²+4−x√

x²+4+x²+2+x√ x²+4 x²+4+x√

x²+4

= (x²+2−x√

x²+4)(x²+4+x√ x²+4) (x²+4)²−x²(x²+4)

+ (x²+2+x√

x²+4)(x²+4−x√ x²+4) (x²+4)²−x²(x²+4)

= (x²+2)²−x²(x²+4) +2(x²+2)−2x√ x²+4

4(x²+4) +

(x²+2)²−x²(x²+4) +2(x²+2) +2x√ x²+4 4(x²+4)

= 4(x²+2) +8 4(x²+4) =1.

Thereforeh(x)≡1 is an invariant density, so Lebesgue measure is preserved.

Example 3: If T : [0,1]→[0,1]is (countably) piecewise linear, and each branchT :I_n→[0,1](on whichT is ane) is onto, thenT preserves Lebesgue measure. Indeed, the intervalsIn have pairwise disjoint interiors, and their lengths add up to 1. Ifs_n is the slope of T :I_n→[0,1], thens_n=1/|I_n|. Therefore

X

n

1

DT(y_n) =X

n

1 s_n =X

n

|I_n|=1.

Folklore TheoremIfT :S¹ →S¹ is aC² expanding circle map, then it preserves a measureµequivalent to Lebesgue, andµis ergodic.

Expandingmeans that there is λ >1 such that|T⁰(x)| ≥λfor all x∈S¹. The above theorem can be proved in more generality, but in the stated version it conveys the ideas more clearly.

Proof: Using the Mean Value Theorem twice, we obtain log|T⁰(x)|

|T⁰(y)| = log(1+|T⁰(x)| − |T⁰(y)|

|T⁰(y)| )≤ |T⁰(x)| − |T⁰(y)|

|T⁰(y)|

= |T⁰⁰(ξ)| · |x−y|

|T⁰(y)| = |T⁰⁰(ξ)|

|T⁰(y)|

|Tx−Ty| T⁰(ζ) .

SinceT is expanding, the denominators are ≥λand sinceT isC² on a compact space, also|T⁰⁰(ξ)|is bounded. Therefore there is someK ≤sup|T⁰⁰(ξ)|/λ² such that

log|T⁰(x)|

|T⁰(y)| ≤K|T(x)−T(y)|.

The chain rule then gives:

log|DTⁿ(x)|

|DTⁿ(y)| =

n−1

X

i=0

log|T⁰(Tⁱx)|

|T⁰(Tⁱy)| ≤K

n

X

i=1

|Tⁱ(x)−Tⁱ(y)|.

SinceT is a continuous expanding map of the circle, it wraps the circled times around itself, and for each n, there aredⁿ pairwise disjoint intervalsZ_n such that Tⁿ:Z_n→S¹ is onto, with slope at leastλⁿ. If we take x,y above in one suchZn, then

|x−y| ≤λ⁻ⁿ|Tⁿ(x)−Tⁿ(y)|

and in fact

|Tⁱ(x)−Tⁱ(y)| ≤λ⁻⁽ⁿ⁻ⁱ⁾|Tⁿ(x)−Tⁿ(y)|.

Therefore we obtain log|DTⁿ(x)|

|DTⁿ(y)| = K

n

X

i=1

λ⁻⁽ⁿ⁻ⁱ⁾|Tⁿ(x)−Tⁿ(y)|

≤ K

λ−1|Tⁿ(x)−Tⁿ(y)| ≤logK⁰ for someK⁰∈(1,∞). This means that if A⊂Zn (so Tⁿ:A→Tⁿ(A) is a bijection), then

1 K⁰

m(A)

m(Zn) ≤ m(TⁿA)

m(TⁿZn) = m(TⁿA)

m(S¹) ≤K⁰ m(A)

m(Zn), (2) wherem is Lebesgue measure.

Construct theT-invariant measure µ. TakeB ⊂ B arbitrary, and setµ_n(B) = ¹_nPn−1

i=0 m(T⁻ⁱB). Then by (2), 1

K⁰m(B)≤µ_n(B)≤K⁰m(B).

We can take a weak^∗ limit of theµn's; call itµ. Then 1

K⁰m(B)≤µ(B)≤K⁰m(B),

and thereforeµ andmare equivalent. The T-invariance of µ proven in the same way as in the Theorem of Krylov-Bogul'ubov.

Now for the ergodicity ofµ, we need the Lebesgue Density Theorem, which says that ifm(A)>0, then for m-a.e. x∈A, the limit

ε→lim0

m(A∩B_ε(x)) m(Bε(x)) =1,

whereB_ε(x) is theε-balls aroundx. Pointsx with this property are called(Lebesgue) density pointsof A. (In fact, the above also holds, ifBε(x)is just a one-sided ε-neighbourhood ofx.)

Assume by contradiction thatµis not ergodic. TakeA∈ B a T-invariant set such that µ(A)>0 andµ(A^c)>0. By equivalence ofµandm, alsoδ:=m(A^c)>0. Letx be a density point of A, andZn be a neighbourhood of x such that Tⁿ:Zn→S¹ is a bijection. Asn→ ∞,Z_n→ {x}, and therefore we can choosen so large (henceZn so small) that

m(A∩Zn)

m(Z_n) >1−δ/K⁰. Therefore ^m(A_m(Z^c∩Z_n)ⁿ⁾ < δ/K⁰, and using (2),

m(Tⁿ(A^c∩Z_n))

m(Tⁿ(Zn)) ≤K⁰m(A^c∩Z_n)

m(Zn) <K⁰δ/K⁰=δ.

SinceTⁿ:A^c∩Zn→A^c is a bijection, and

m(TⁿZn) =m(S¹) =1, we getδ=m(A^c)< δ, a contraction.

Thereforeµis ergodic.