Prove the existence of uniformly distributed sequence of reals in $[0,1]$ 
A sequence of reals $\{x_n\}$ in $[0,1]$ is uniformly distributed iff for every $a<b$ in [0,1], $$\lim_{n\to\infty}[\sum_{j=1}^n 1_{(a,b]}(x_j)]/n=b-a$$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 $a<b, a,b\in [0,1]$. 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 $x_j\in (a,b]$ 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: 


*

*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?

*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!
 A: 
Lemma. $(x_j)$ is equidistributed if and only if, for each rationals $0 \leq a < b \leq 1$,
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(x_j) = b-a. \tag{*} $$

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 $k$,
$$ \limsup_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(x_j)
\leq \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a_k, b_k]}(x_j) = b_k - a_k. $$
Letting $k \to \infty$ shows that
$$ \limsup_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(x_j)
\leq b - a. $$

*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,
$$ \liminf_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(x_j)
\geq \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a_k, b_k]}(x_j) = b_k - a_k $$
and letting $k \to \infty$ gives
$$ \liminf_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(x_j)
\geq b - a. $$
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
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(U_j) = b-a \quad \mathbb{P}\text{-a.s.}$$
So, if we consider the event 
$$ \Omega_0 := \bigg\{ \omega : \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \mathbf{1}_{(a, b]}(U_j(\omega)) = b-a \text{ for all } a, b \in \mathbb{Q} \text{ with } 0 \leq a < b \leq 1 \bigg\}, $$
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 $[0, 1]$.
