    # On absolutely normal numbers and their discrepancy estimate

We construct the base 2 expansion of an absolutely normal real number x so that, for every integer b greater than or equal to 2, the discrepancy modulo 1 of the sequence (b^0 x, b^1 x, b^2 x , ...) is essentially the same as that realized by almost all real numbers.

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## 1 Primary definitions and results

We use some tools from  and . For non-negative integers and , for a sequence of real numbers and for real numbers such that , we define

 F(M,N,α1,α2,(xj)j≥1)=∣∣ #{j:M≤j

We write to denote Lebesgue measure.

###### Lemma 2 ([3, Lemma 8], adapted from Hardy and Wright [7, Theorem 148]).

Let be an integer greater than or equal to . Let and be positive integers and let be a real such that . Then, for any non-negative integer and for any integer such that ,

 μ{x∈(0,1):|F(M,N,ab−m,(a+1)b−m,{bjx}j≥0)|>εN}

is less than

The next lemma is similar to Lemma 2 but it considers dyadic intervals instead of -adic intervals.

###### Lemma 3.

Let be an integer greater than or equal to , let and be positive integers and let be a real. Then, for any pair of integers and such that and ,

 μ{x∈(0,1):F(M,N,a2−k,(a+1)2−k,{bjx}j≥0)≥εN}

is less than .

###### Remark 4.

In , Philipp proves a proposition more general than Lemma 3. His result yields the same order of magnitude but does not make explicit the underlying constant while Lemma 3 does.

Clearly, for arbitrary reals such that , for any sequence and for any non-negative integers , and ,

 |F(0,N,α1,α2,(xj)j≥1)|≤N/2k−1+k∑m=1max0≤a<2m|F(0,N,a2−m,(a+1)2−m,(xj)j≥1)|.
###### Lemma 5 ([9, Lemma 4], adapted from [6, Lemma 3.10]).

Let be an integer greater than or equal to , let be a positive integer and let be such that . Then, there are integers with for , such that for any positive integer and any , with ,

 F(0,N,a2−h,(a+1)2−h,{bjx}j≥0) ≤N1/3+F(0,2n,a2−h,(a+1)2−h,{bjx}j≥0) + n∑ℓ=n/2F(2n+mℓ2ℓ,2ℓ−1,a2−h,(a+1)2−h,{bjx}j≥0).

Let and be positive reals. For each integer greater than or equal to and for each positive integer let

 ~Cb =1/2+2/(√b−1), ϕ(N) =2(1+2δ)~Cb(NloglogN)1/2, T(N) =⌊logN/log4⌋+1.

For integers and such that

 b≥2, n≥1, 0≤a<2T(2n), 1≤h≤T(2n), n/2≤ℓ≤n, and 1≤m≤2n/2,

define the following sets

 G(b,n,a,h)= {x∈(0,1):F(0,2n,α1,α2,{bjx}j≥0)≥2−h/8ϕ(2n)}, where α1=a2−(h+1), α2=(a+1)2−(h+1), if 1≤h
###### Lemma 6.

Let and be positive real numbers. For each and for every ,

 μ(Gb,n)=n−1−4δ,μ(Hb,n)=2n−1−3δ,

and there is such that

 μ(⋃n≥n0(Gb,n∪Hb,n))<η

and such that for every real outside ,

 limsupN→∞DN({bnx}n≥0)√N√loglogN<(1+4δ)Cb,

where is Philipp’s constant, .

###### Proof.

To bound we apply twice Lemma 3, first with , and , and then with , and . We write to denote and we write instead of . Assuming ,

 μGb,n≤ μ⎛⎝2T−1⋃a=0G(b,n,a,T)⎞⎠+T−1∑h=1μ⎛⎝2h−1⋃a=0G(b,n,a,h)⎞⎠ ≤ T−1∑h=12h9⋅22(h+1)(h+3)exp(−2−h/4ϕ2(2n)2−nbh+1b26(h+3)) +2T9⋅22(T+2)(T+2)exp(−2−T/4ϕ2(2n)2−nbTb26(T+2)) ≤ 9⋅23T+5(T+2)exp(−2−T/4loglog(2n)4(1+ 2δ)2bT+216(T+2)~C2b) ≤ n−(1+4δ).

To bound we apply twice Lemma 3 first letting , and , and then letting , and . Again we write instead of . Assuming ,

 μ Hb,n= μ⎛⎝T⋃h=12h−1⋃a=0n⋃ℓ=n/22n−ℓ⋃m=1H(b,n,a,h,ℓ,m)⎞⎠ ≤ n∑ℓ=n/22n−ℓT−1∑h=19 23h+6(h+3)exp(−2−h/4bh+322(n−ℓ)/3loglog(2n)(1+δ)246(h+3)~C2b) +n∑ℓ=n/22n−ℓ9 23T+4(T+2)exp(−2−T/4bT+222(n−ℓ)/3loglog(2n)(1+δ)246(T+2)~C2b) ≤ n∑ℓ=n/22n−ℓexp(−2−1/422(n−ℓ)/3loglog(2n)(1+4δ)b424)T−1∑h=12−h +n∑ℓ=n/22n−ℓ9 23T+4(T+2)exp(−2−T/4bT+222(n−ℓ)/3loglog(2n)(1+δ)216(T+2)) ≤ exp(−2−1/4loglog(2n)(1+3δ)b424)n∑ℓ=n/22n/2−ℓ−1 +exp(−2−T/4bT+2loglog(2n)(1+3δ)16(T+2))n∑ℓ=n/22n/2−ℓ−1 ≤ 2 n−(1+3δ).

Thus, there is such that for every integer greater than or equal to ,

 μ(⋃n≥n0(Gb,n∪Hb,n))<∑n≥n0(n−1−4δ+2n−1−3δ)<η.

It follows from Philipp’s proof of [9, Theorem 1] that for every real outside ,

 limsupN→∞DN({bjx}j≥0)√N√loglogN<(1+4δ)Cb,

where .∎

## 2 Proof of Theorem 1

We give an algorithm to compute a real outside the set . The technique is similar to that used in the computable reformulation of Sierpinski’s construction given in .

The next definition introduces finite approximations to this set. Recall that by Lemma 6, for every integer , provided and ,

 μ(⋃n≥n0(Gb,n∪Hb,n))≤∑n≥n0n−(1+4δ)+2n−(1+3δ)≤∑n≥n0n−2<η.
###### Definition 7.

Fix and fix . For each integer , let be the least integer greater than such that

 ∞∑k=zb1k2<η2b.

We define

 Δ= ∞⋃b=2∞⋃m=zb(Gb,m∪Hb,m), s= ∞∑b=2∞∑k=zb1k2.

Observe that .

For each , let

 bn= max(2,⌊log2n⌋), Δn= bn⋃b=2n⋃m=zb(Gb,m∪Hb,m), sn= bn∑b=2n∑k=zb1k2, rn= s−sn=bn∑b=2∞∑k=max(n+1,zb)1k2+∞∑b=bn+1∞∑k=zb1k2, pn= 22n+2.

The next propositions follow immediately from these definitions.

For every ,

###### Proposition 9.

For every and such that ,

###### Proposition 10.

For any interval and any ,

The proof of Theorem 1 follows from the next lemma.

###### Lemma 11.

There is a computable sequence of nested dyadic intervals such that for each , and .

###### Proof.

Proposition 10 establishes, for any interval and any ,

 μ(Δ∩I)<μ(Δm∩I)+rm.

Then, to prove the lemma it suffices to give a computable sequence of nested dyadic intervals such that for each , and . We establish

 pn=22n+2.

This value of is large enough so that the error is sufficiently small to guarantee that even if all the intervals in fall in the half of that will be chosen as , will not be completely covered by . We define the inductively.

Base case, . Let . We need to check that . Since , and ,

 Δp0=bp0⋃b=2p0⋃n=zb(Gb,n∪Hb,n)=∅.

Since and , . Then,

 rp0=s=∞∑b=2∞∑k=zb1k2.

We conclude .

Inductive case, . Assume that for each ,

 μ(Δpm∩Im)+rpm<12m(η+m∑j=12j−1⋅rpj),

where . Note that for , is the empty sum. We split the interval in two halves of measure , given with binary representations of their endpoints as

 I0n=[0.d1…dn−1 , 0.d1…dn−11] and I1n=[0.d1…dn−11 , 0.d1…dn−1111111…].

Since is equal to interval , we have

 μ(Δpn∩I0n)+μ(Δpn∩I1n)=μ(Δpn∩In−1).

Since , we obtain

 μ(Δpn∩I0n)+μ(Δpn∩I1n)≤μ(Δpn−1∩In−1)+rpn−1−rpn.

Adding to both sides of this inequality we obtain

 (μ(Δpn∩I0n)+rpn)+(μ(Δpn∩I1n)+rpn)≤μ(Δpn−1∩In−1)+rpn−1+rpn.

Then, by the inductive condition for ,

 (μ(Δpn∩I0n)+rpn)+(μ(Δpn∩I1n)+rpn)<12n−1(η+n∑j=12j−1⋅rpj).

Hence, it is impossible that the terms

 μ(Δpn∩I0n)+rpn and μ(Δpn∩I1n)+rpn

be both greater than or equal to

 12n(η+n∑j=12j−1⋅rpj).

Let be smallest such that

 μ(Δpn∩Idn)+rpn<12n(η+n∑j=12j−1⋅rpj)

and define

 In=Idn.

To verify that satisfies the inductive condition it suffices to verify that

 η+n∑j=12j−1⋅rpj<1.

Developing the definition of we obtain

 n∑j=12j−1⋅rpj= n∑j=12j−1⎛⎜⎝bpj∑b=2∞∑k=max(zb,pj+1)1k2+∞∑b=bpj+1∞∑k=zb1k2⎞⎟⎠ = ⎛⎜⎝n∑j=12j−1bpj∑b=2∞∑k=max(zb,pj+1)1k2⎞⎟⎠+⎛⎜⎝n∑j=12j−1∞∑b=bpj+1∞∑k=zb1k2⎞⎟⎠ < ⎛⎝n∑j=12j−1bpj∞∑k=pj+11k2⎞⎠+⎛⎜⎝n∑j=12j−1∞∑b=bpj+1η2b⎞⎟⎠ < (n∑j=12j−1bpjpj+1)+(n∑j=12j−1η2bpj) < < 34+η4 < 78.

Then, using that we obtain the desired result,

 μ(Δpn∩In)+rpn<12n(η+n∑j=12j−1⋅rpj)<12n(η+78)<12n.

Let’s see that the number obtained by the next Algorithm 12 is external to

 Δ=∞⋃b=2∞⋃n=nzb(Gb,n∪Hb,n)

Suppose not. Then, there must be an open interval in such that . Consider the intervals