Cfrgtkky

A sequence of reals ${x_n}$ in $[0,1]$ is uniformly distributed iff for every $a<b$ in [0,1], $$lim_{ntoinfty}[sum_{j=1}^n 1_{(a,b]}(x_j)]/n=b-a$$Prove that such a sequence exists. (Hint: Show that it suffices to show the above result holds for rational values $a$ and $b$.

Here is my approach:

Take $a<b, a,bin [0,1]$. Let ${x_j}$ be a sequence of real numbers in $[0,1]$.

Define random variables $X_j=1_{(a,b]}(x_j)$, thus getting a sequence of random variables ${X_j}$. Then $P(X_j=1)=b-a, P(X_j=0)=1-(b-a)$ for every $j$, and they should be independent since $x_jin (a,b]$ for any $j$ should be independent events. Then we get $E(X_j)=b-a$ for every $j$. Define, $S_n=sum_{j=1}^n X_j$ and we get $frac{S_n}{n}to b-a$ almost surely by Strong Law of Large Numbers, which is the condition we want to show.

Now I have two questions:

1. Is my approach to this question correct? I'm a bit unsure that such random variables should exist in [0,1] though I feel that they should. Is this something I need to prove separately?




2. After I got through the question, I realized that I didn't use the hint at all, and I think the hint comes into play because the convergence in Strong Law of Large Numbers is almost surely. Therefore I need to take care of the null sets where the above result doesn't hold, but I'm a bit stumped on how to do that, and not sure how to show that it "suffices" to prove the above for rational $a,b$.

I would like some feedback on my proof and help on the remaining questions that I have! Thanks!

Lemma. $(x_j)$ is equidistributed if and only if, for each rationals $0 leq a < b leq 1$,

$$ lim_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(x_j) = b-a. tag{*} $$

Proof. Assume that $text{(*)}$ holds for all rationals $0 leq a < b leq 1$. We want to show that $text{(*)}$ holds for any reals $0 leq a < b leq 1$. To this end, we adopt the usual squeezing argument.

• 
  

  Choose $a_k, b_k in mathbb{Q} cap [0, 1]$ such that $a_k uparrow a$ and $b_k downarrow b$ as $k to infty$. Then $mathbf{1}_{(a, b]}(x) leq mathbf{1}_{(a_k, b_k]}(x)$ for all $x in mathbb{R}$, and so, for each fixed $k$,

  
  
  

  $$ limsup_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(x_j)
  leq lim_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a_k, b_k]}(x_j) = b_k - a_k. $$

  
  
  

  Letting $k to infty$ shows that

  
  
  

  $$ limsup_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(x_j)
  leq b - a. $$

  
  


• 
  

  In this time, we choose $a_k, b_k in mathbb{Q}cap [0, 1]$ such that $a_k downarrow a$ and $b_k uparrow b$ as $k to infty$. Then, as before,

  
  
  

  $$ liminf_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(x_j)
  geq lim_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a_k, b_k]}(x_j) = b_k - a_k $$

  
  
  

  and letting $k to infty$ gives

  
  
  

  $$ liminf_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(x_j)
  geq b - a. $$

  
  

Combining two estimates shows that $text{(*)}$ does hold as required. ////

Let $(U_j)$ be a sequence of i.i.d. random variables uniformly distributed on $[0, 1]$. Then for each rationals $0 leq a < b leq 1$, SLLN tells that

$$ lim_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(U_j) = b-a quad mathbb{P}text{-a.s.}$$

So, if we consider the event

$$ Omega_0 := bigg{ omega : lim_{ntoinfty} frac{1}{n} sum_{j=1}^{n} mathbf{1}_{(a, b]}(U_j(omega)) = b-a text{ for all } a, b in mathbb{Q} text{ with } 0 leq a < b leq 1 bigg}, $$

then $Omega_0$ is the countable intersection of events having full probability, hence $mathbb{P}[Omega_0] = 1$. Moreover, by Lemma, for each $omega in Omega_0$ the sequence $(U_j(omega))_{j=1}^{infty}$ is equidistributed. So by setting $x_j = U_j(omega)$ for any choice of $omega in Omega_0$, we obtain an equidistributed sequence.

As a totally different approach, one may give a concrete example. Indeed, choose $alpha in mathbb{R} setminus mathbb{Q}$ and set $x_n = alpha n text{ mod } 1$. Weyl's criterion immediately tells that $(x_j)$ is equidistributed over $[0, 1]$.