This is related to a certain question I have about whether of a sequence on [0,1] that is itself a partition of two equi-distributed sequences, L1, L2 on [0,1] is equi-distributed. It depends on what those sequences L1, L2 are. I don't believe it holds for all such equi-distributed sequences on [0,1].

Where above this denomator equals (|{Am:Am\in B1\and  1

Well its possible, if by that you mean the upper and lower bounds of the relative frequency limits, in the inequality here,

limrelfreq(a\inB1)

Rather see attached; the claim directly  above  lemma 3; in definition 2, the end of the proof where it says 'thus the proposition below tells us'.....an inequality " which suffices for this claim". I am asking about the name of this theorem, where presumably it follows on the basis that the upper and lower limits approximately equal each other.

I am not sure if research gate has the facility to code things up in a latex style so i refer to this lyx document, as attached

What I am talking about, is for example a sequence A, consisting only of numbers in [0,1]. This can be partitioned into two mutually exhaustive and exclusive subsequences, B1, and B2, so every element in A is in B1 or in B2. Each of, B1. B2, are are sequences of real numbers equidistributed on [0,1] and I am  considering whether sequence A is equi-distributed on [0,1].

So \a here would is an interval such [c,d], where  d>c; c>0dinfinity. N being the nth member of B1. So I am computing the proportion/relative frequency of the first n elements of B1, that are in [c,d], as n ->infinity  . Likewise for B2; the first n elements of B2 (which are different elements of A the aforementioned first n elements of B1)).

The basic question then is, if these two limits, are equal (presumably what you call the lim sup, and lim inf), ie the relative frequency of elements in B1\cap[c,d] equals or approximately equals the relative frequency of elements in B2\incap [c,d]=d-c, so that the the rhs and lhs are equal, is the middle term, (the limiting relative frequency of elements of A in [c,d] equal to this value and thus to d-c), and if so, or if when it can be applied, is there a name for this  squeezing/asymptotic intermediate value like theorem.  Clearly the value of, th

yes the numbers are confined to real numbers on [0,1]; i hope that was mentioned in the attached document. I will need to check

Ok, if by that you mean the proportion of the first n elements of sequence A that are in subinterval [i] that is correct rendition of the relative frequency. By equi-distribution we want to show that as n goes to infinity, this value max value(I)-min value in I. ie if the interval is [c,d] then the limit relative frequency of An as n goes to infinity should be d-c. Where what is being shown, is that A is equi-distributed [0,1], as result of it being a partition of two equi-distributed sequence on [0,1]. This is more clear in the document. What I am dubious about, are issues of conglomerability and relative frequencies not being closed under countable union, (disjunction) or intersection, that even if the two partiions limit to d-c the entire sequence may not. This is in the document in a more technical format (its only about one page and half)- smaller then most of my posts. I just want to make sure the proof presented works

Hello Peter,

Thanks for this. I myself was a little suspicious of the proof at least originally. I presumed that if if the limit existed rf(A|I) one would expect it to be bounded by, and thus equal to, approximately, RF(B1,I)=Rf(B2|I). So

long asn1 and n2  go to infinity as n goes to infinity (which i also realized); and that n1/n, and n2/n need to be well defined (which I want for various other reasons) even if rf(B1,I)  and rf(B2,I) converges to an approximately similar value (as one can never say that they never converge to precisely the same value at any finite point), in virtue of their equi-distribution.

When you say that for free one has rf(B1,I)+rf(B2,I)>=rf(A,I).; wouldnt it be the case, that given if one of n1 remains finite as n->inf, that n2->inf, n2/n would converge presumably to one, in which case, presumably, the limiting relative frequency of rf(A,I) would approximately match rf(B2,1), or whichever of the Bi is such that ni/n goes to one, ie ni goes to infinity; so the sum inequality would hold in a restricted sense .

Also I presume that the argument that n1/n should be well defined, is an additional fact over and above, n1 n2->inf, and the inequality holding, that is required for the limit to precisely exist. So the inequality itself is not sufficient, for the limit frequency to exist; even if, I presume both the rf(Bi,I) and rf(B2, I) converge to an equal value. There is an issue between going from the finite case where t relative frequency(A,i)= [(freq(b,I) + freq(B2,I[)/(n1+n2) and the infinite case, a presumably n1/n=n1/(n1+n2) must be defined so that can form the fraction with a common denominator a rf(B,I)=[ freq(B,I) \times (n1 +n2)/n1]\[n1\times (n1+n2)/n1)= (freq(B,I)\times (1 + n2/n1))/(n1+n2) and so #{x | x

Hello Peter, thanks for this; by

ith f(n,xn) = 0 if the nth element of the sequence is in the B1 subsequence, 1 otherwise, you get the limit n1/n, the proportion of the sequence that is B". I presume you mean that n denotes the length of the subsequence of A (the first elements, and xn or xi denotes the ith element of that subsequence and thus the ith element of A). SO that when we compute the sum; it is  [sum_{i=1)^(i=n)][f(n,xi)]]/n; where f itself is an indicator function not a frequency function, it only considers one element at a time.

Do you mean by n, the length of the sub-sequence of A, (which will be the first n elements of A)and xn, xi the ith member of sequence. So that n denotes the fact we are considering the first n elements of sequence A, and Xn being the nth element of sequence of A (the last element of that subsequence, and xi being the ith element of that subsequence, and thus the ith member of A).

So for example when there are two partitions and two functions f1, f2 (f1 for B1, and f3 for B2),you mean that the ith element of the sequence A (or rather the ith member of the first n elements of sequence A, which will be the ith member of A) , is in sequence B1, and otherwise if f^1(n,xi)=0 its in B2, and thus f^2(n,xi)=1

I presume by the second statement one is then considering the number of elements in A as a whole, that are in I, where n denotes the fact we are considering the first n elements of A, and xn, or xi, denotes the ith element of that sub-sequence, and thus the ith element of A. Or do you mean the number of elements in one of B1 or B2.

So what what we are really have are two functions, one denoting membership of A in a subsequence B1 or B2, and then another function denoting whether that member of B1 or B2 is in the interval,so the sum you speak of is a sum over all sub-sequence functions (for B1, and B2) computing relative frequency of elements in B1 & interval and the the relative frequency of B2 and interval. So that I need that the [freq(the n elements of A, that are in B1 and in I) +freq (the first n elements of A, that are in B2 and in I)]/n converges

I presume you are talking about two distinct functions, one, f (or rather f1 and f2, f1 for B1 and f2 for B2) that denotes whether a member of the first n elements of A is in B1, or instead in B2.

And another function lets call it G,or rather G1 and G2, which denotes whether that element same element (which is in B1 or B2, G1 for B1, and G2 for B2) is also in I. So that we really have a sum (over all sub-sequences functions) of the form \sum_(j=1)^(j=n)[\sum_{i\in{1,2}f^i(n,xj)G^i(n,xj) ]]/n=\sum_(j=1)^(j=n)[f1(n,xj)\timesG1(n,xj)+[f2(n,xj)\timesG2(n,xj)]/n. And we want that, \sum_(j=1)[(j=n)[f1(n,xj)]/n converges and likewise for f2; denoting the proportion of elements in B1 and B2 respectively.

We also want \sum_(j=1)^(j=n)[f1(n,xj)\timesG1(n,xj) converges as well; which means I do not need to have two different functions G1 and G2, to denote the conditional relative frequency, 'given an element of the first n A's is in B1,'the relative frequency of such elements that are in I is such and such (, ie" the proportion 'the first n A's that are in B1', which are in I") elements of B1 that are in the interval is such and such. As f1*G will only have value 1, otherwise 'only if'if f1 has value 1,; so it does not really matter if ones uses uses the same indicator function on A (as you use in the last sentence) which determines if an element, xi, the ith element in the first n A's (and thus A) is in I, in each of these products.

It will come out in the wash that the sum of this indicator function/n will have a defined limiting   value e ;ie converge. That is, limiting value of the relative frequency of elements in I in the first n elements of A is well defined if relfrequency of elements in first n elements in A,.which are in both I and B1, is well defined and likewise for the other partition; which will presumably be the case or at least have as a necessary condition, that these conditional relative frequencies are defined; plus more (ie the products are well defined as well).

"W

I also noticed this https://en.wikipedia.org/wiki/Squeeze_theorem; which is presumably the kind of theorem i was talking about by intermediate value like limit convergence rule'.

As to whether you the functions, are measurable, positive, bounded, it depends on what you means; I mean the indicator functions you mentioned are just bounded in value between zero and one, are non-negative, and I presume therefore measurable; by which I presume you mean that the sets are measurable- that it is possible to determine whether any element is an element of that set, f they just determine whether a singular element is a a certain set or not.

The indicator functions you use is what I have in mind; given what I mean by relative frequency (ie proportions of truths, or proportions of elements in a sequence A which are elements of some subseqeuence, B where B is a proper subset of the first sequence A, or where the every element in both partitions is in [0,1] so the elements in each partition and in the whole sequence that are in some sub-interval of [0,1] is also subset; by equidistribution, unless one mentions certain irrational numbers.

Likewise for the partition proportions, which are given by the same process, at a previous stage; the number of element in B1 is equal to the number of elements in the entire sequence at a previous stage, which are in some proper sub-interval. [m,n] of [0,1], where |m-n| is positive and finite (not null or infinitesimal; so as to yield a positive limiting  relative frequency as n goes to infinity, and thus an infinite number of elements in [mn] as n goes to infinity.

So it may not be a proper-subset- although by construction it should be so long, unless something goes wrong after many induction steps, and the equ-distributed sequences, being countable long, cannot give a non zero limiting relative frequency to all of the many (tiny) intervals.That is given any one partition B1, all of the  other partitions Bi, i\neq1 should be  non-empty, and contain countable infinite many elements; So  B1  should not contain all or almost all of the elements of A