0
$\begingroup$

How to describe the set $\{ 1 \leq n \leq N: \alpha_n \in(a,b)\}$ when $(a,b) \subset [0,1)$ and you have following information:

a sequence of numbers $(\alpha_1,\alpha_2, \alpha_3,...)$, where $a_j \in [0,1)$, $j \in \mathbb{N}$, is equidistributed, if for every interval $(a,b) \subset [0,1)$ $\lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}=b-a$ holds, where $\# A$ is the cardinality of set $A$ (number of elements in the set).

I think $A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}\Rightarrow\#A = N\;,$ but then how do you get $\lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}=b-a\;?$

  • 0
    I have restored the OP’s original post, merely improving the formatting and making a very few minor changes in the wording. The previous edits were far too extreme and distorted the post beyond recognition. I agree with @did that it’s not at all clear what is being asked.2012-10-29

2 Answers 2

0

You defined $A_{\alpha,N}(a,b)=\{1 \leq n \leq N : \alpha_n \in ]a,b[\}$

This set contains exactly the indices $n \in [|1,N|]$ for which $a<\alpha_n. Since $0 and $\alpha_n\in[0,1[$, there is no reason why your statement would be true in general.

  • 0
    #A is at most N. Moreover, you can always find counter-examples since not every sequence is equidistributed. Let us consider for instance $a=\frac{1}{3}$ and $b=\frac{2}{3}$, and the sequence $\alpha$ such that $\forall n, \alpha_n=0$. Then the limit is 0, not $b-a$, which means the sequence is not equidistributed.2012-10-30
2

There's one mistake : your statement

$A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\} = \{1,2,3,4, \dots, N \}$

is false.

You can think at this problem as a probability distribution problem : Let's consider some probability distribution $\mathcal{P}$. You draw $n$ samples in $[0,1)$ called $\alpha_1,\dots,\alpha_n$. For some arbitrary interval $(a,b)$, your set $A=\{ 1 \leq n \leq N: \alpha_n \in(a,b)\}$ are the set of points that are drawn inside $(a,b)$. Of course, you won't get all the $N$ points inside $(a,b)$. The probability that a point is inside $(a,b)$ can be written as :

$\mathbb{P}(\alpha \in (a,b) | \alpha \sim \mathcal{P}) = \lim_{N \rightarrow \infty} \frac{ \# \{1\leq n \leq N: \alpha_n \in (a,b) \}}{N}$

(when you are drawing an infinite quantity of points, this is the proportion of points that are $(a,b)$).

You can say that $\mathcal{P}$ is an uniform law if this probability is $\frac{measure (a,b)}{measure(0,1)}= b-a$. You will say that the sample $\alpha_1,\dots,\alpha_n$ is equidistributed if it comes from an uniform law.

(I know that these explanations aren't really rigorous, sorry for that but I hope this "sketch" explanation can help)