This is from Billingsley's Probability and Measure, Chapter I. The following theorems are proved:
The Weak Law of Large Numbers: For every $\epsilon > 0$, $$\lim_{n \rightarrow \infty} P\left[\omega \in (0, 1]: \left|\frac{1}{n}s_n(\omega)\right|\geq \epsilon \right] = 0$$ where $s_n(\omega) = \sum_{i=1}^n r_i(w)$. The Rademacher functions $r_i$ are defined as $r_n(\omega) = \begin{cases}+1 &d_n(\omega) = 1 \\ -1 &d_n(\omega) = 0\end{cases}$ where $d_n(\omega)$ is the $n$th digit in the binary expansion of $\omega$ (non terminating representation).
And the following special case of the Strong Law of Large Numbers is proved.
Borel's Normal Number Theorem: $$\lim_{n \rightarrow \infty} \frac{1}{n}s_n(\omega) = 0 \quad \text{almost everywhere on } (0, 1] $$ where $s_n(\omega)$ is defined as above.
The book then mentions (without proof) that if $f_n$ is a step function on $(0, 1]$ for each $n$ then if $\lim_{n \rightarrow \infty} f_n(\omega) = 0$ almost everywhere on $(0, 1]$ we have $\lim_{n \rightarrow \infty} P\left[\omega \in (0, 1]: \left|f_n(\omega)\right|\geq \epsilon \right] = 0$ for every $\epsilon > 0$. In particular, taking $s_n$ to be our step functions $f_n$, we can conclude that Borel's normal number theorem implies the weak law of large numbers.
I have been trying to prove this general result without much luck.
If for every $n \in \mathbb{N}$, $f_n$ is a step function on $(0, 1]$ and if $\lim_{n \rightarrow \infty} f_n(\omega) = 0$ almost everywhere on $(0, 1]$, then for every $\epsilon > 0$, $\lim_{n \rightarrow \infty} P\left[\omega \in (0, 1]: \left|f_n(\omega)\right|\geq \epsilon \right] = 0$.
Fix $\epsilon > 0$. If $A_n = \left\{\omega: \left|f_n(\omega)\right|\geq \epsilon \right\}$, we need to show that given any $\epsilon_0 > 0$, there exists some $N \in \mathbb{N}$ such that for $n \geq N$, $ |A_n| < \epsilon_0$. As $f_n$ is a step function, each $A_n$ is the union of a finite number of intervals, so its length is well defined.
If $A = \left\{\omega: \lim_{n \rightarrow \infty} f_n(\omega) = 0\right\}$, then by hypothesis, the complement of $A$ has measure zero. If $\omega \in A$, then there exists some $N_\omega$ such that for $n \geq N_\omega$, $f_n(\omega) < \epsilon$. That is, $\omega \notin A_n$ for $n \geq N_\omega$. However, I'm struggling to see where to go from here -- how do we choose an $N$ large enough so that $A_n$ does not contain "most" of the interval $(0, 1]$ for $n \geq N$? I feel like the fact that the $f_n$'s are step functions is key here.
Any hints appreciated, I feel quite stuck.