by AMathsStudent1
Last Updated January 16, 2018 13:20 PM

I recently came accross the definition of a sequence of real numbers being uniformly distributed modulo $1$. The definition is:

For every choice of $a,b$ with $0 \le a < b < 1$, we have:

$\frac{1}{n}\cdot \#\left\{j \in \{0,\ldots,n-1\} | \{x_j\} \in [a,b]\right\}$ converges to $b - a$ for $n\to \infty$.

{$x_j$} denotes the fractional part of $x_j$ and $\#$ is the number of elements of the set.

Now, it is stated that we can replace $[a,b]$ with $[a, b)$ and get the same answer. I'm a bit stuck on how to actually prove this, am I missing something obvious?

Thanks in advance.

Let $x_j \sim \mathcal{U}\left([0,1]\right)$ be uniformly distributed on $[0,1]$ and independent, then $$P( x_j \in [a,b]) = P(x_j \in [a,b)) = b-a$$ where $P(\cdot)$ denotes the probability

Now let $$Y = \#\{ j \in \{1,\ldots, n-1\} | x_j \in A\}$$ be the number of $x_j$ till $n-1$ which are in the set $A$.

Then $Y$ can be written as: $$Y = \sum_{k=1}^{n-1} 1_{\{x_j \in A\}}$$

Now you already have for $A = [a,b]$ that $$\frac{1}{n} Y = \frac{1}{n}\sum_{k=1}^{n-1} 1_{\{x_j \in A\}} \to (b-a)$$

and want to know if it also holds for $A = [a,b)$.

But this is pretty obvious by considering that $$1_{\{x_j \in [a,b]\}} = 1_{\{x_j \in [a,b)\}} + 1_{\{x_j = b\}}$$

But $$1_{\{x_j = b\}} = 0$$ $P$-almost surely because the $x_j$ are uniformly distributed.

And so we get: $$\#\{ j \in \{1,\ldots, n-1\} | x_j \in [a,b]\} = \#\{ j \in \{1,\ldots, n-1\} | x_j \in [a,b)\} \quad P-\text{almost surely}$$

Without loss we can assume that $x_i\in[0,1)$ by replacing them with their mod $1$ residues at the beginning. Let $\mu$ be Lebesgue measure on $[0,1]$. For each $n$ let the "empirical measure" $\mu_n$ be the average of the point masses at the first $n$ of the $x_i$, defined by: $$\mu_n(A) = \frac 1 n \operatorname{card}\{i: x_i\in A, 1\le i\le n\},$$ which is to say $$\mu_n = \frac 1 n \sum_{i=1}^n\delta_{x_i},$$ where $\delta_p$ denotes the unit point mass at $p$.

The question is, suppose $\mu_n([a,b))\to\mu([a,b))$ for all $0\le a\le b<1)$. Does this imply $\mu_n([a,b])\to\mu([a,b])$ for all $a\le b\le1$?

Answer: yes. For if it failed, there would be some $a$ and $b$ for which $\lim_{n\to\infty}\mu_n([a,b])$ did not exist or did not equal $b-a$. That would imply that $\mu_n(\{b\})$ failed to converge to $0$. But that cannot be, since for every $\epsilon>0$, $b\in[b-\epsilon,b+\epsilon)$ and it is known that $$\limsup_{n\to\infty} \mu_n(\{b\}) \le \lim_{n\to\infty} \mu_n([b-\epsilon,b+\epsilon))=2\epsilon.$$

This state of affairs is usually called equidistribution and amounts to the weak convergence of the measures $\mu_n$ to $\mu$. The "Portmanteau theorem" gives many equivalent sets of necessary and sufficient conditions for this to happen.

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