A set $S$ is said to be dense in $\Omega$ if for given $\omega \in \Omega$ and $\epsilon >0$, there exists $s \in S$ such that $|s-\omega|<\epsilon$.
Let $N=\{\omega: \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n d_i(\omega)=\frac{1}{2}\}$ where $d_i(\omega)$'s are terms in the dyadic expansion of $\omega \in (0,1]$ and is either $0$ or $1$.
Show that $N$ is dense in $\Omega=(0,1]$
My idea:- Let $\omega \in (0,1]$. If $\omega \in N$ it is trivial. If $\omega \in N^c$ then we have to come up with a $s \in N$ such that $|s-\omega|<\epsilon$. In contradiction, suppose there exist no such $s$ in $N$ such that $|s-\omega|<\epsilon$. In other words $$\{s:|s-\omega|<\epsilon\} \not \subset N$$ This would simply mean that $\{s:|s-\omega|<\epsilon\} \subset N^c$. Thus $A=(\omega-\epsilon, \omega+\epsilon)$ is a negligible set, because $N^c$ is itself negligible. But choosing $\delta <\frac{\epsilon}{4}$ we get that $|A|>\delta$. Thus $A$ can't be negligible as it can't be covered by any collection of intervals whose sum of lengths are less than $\delta$
Is it okay? I will be happy to be pointed out if there is any lack of completeness.