Question about Strong Law of Large Numbers in Breiman

Question

I was reading through Breiman's Probability on Strong Large Numbers and got stuck on a part of its proof. It goes:

Let $\Omega$={0,1}$^\mathbb N$ and $S_n$=$\sum_{k=1}^n\omega_k$.

If $C$={$\omega\in\Omega$ : $\lim_{n\to \infty}$$S_n$($\omega$)/$n$=${1\over 2}$}, and $\hat C$={$\omega\in\Omega$ : $\lim_{m\to \infty}$$S_{m^2}$($\omega$)/$n$=${1\over 2}$}, then $C=\hat C$.

I tried to prove this on my own, but cannot seem to make any progress, or find where to start. Any help on how to start this proof or the proof itself would be greatly appreciated. Thank you.

Edit: Grammar, formatting

Answer 1

I have Breiman's proof in front of me, so I can attempt to explain it more clearly.

The assertion that $C=\hat C$ is an assertion about sequences of zeros and ones, so we can suppress the $\omega$ by focusing on just one sequence.

The inclusion $C\subset \hat C$ is trivial: if a sequence $x_n$ converges to zero as $n\to\infty$, then it also converges along the sequence $n_k:=k^2$. So the hard direction is proving that if $S_{m^2}/m^2\to\frac12$, then so does $S_n/n$.

Suppose $S_{m^2}/m^2\to\frac12$. Let $\epsilon>0$. To show $S_n/n\to\frac12$, the obvious first step is to write, using the triangle inequality, $$ \left|\frac{S_n}n-\frac12\right|\le\left|\frac{S_n}n-\frac{S_{m^2}}{m^2}\right|+ \left|\frac{S_{m^2}}{m^2}-\frac12\right|.\tag1 $$ But the second term on the RHS of (1) tends to $0$ as $m\to\infty$. So for all large $m$, say $m\ge M$, we have $$ \left|\frac{S_{m^2}}{m^2}-\frac12\right|<\epsilon.\tag2 $$ Now pick $n\ge M^2$. For this $n$, find an $m\ge M$ such that $m^2\le n<(m+1)^2$. Confirm that this implies that $|n-m^2|\le 2m$. By the triangle inequality, we can bound the first term on the RHS of (1): $$ \left|\frac{S_n}n-\frac{S_{m^2}}{m^2}\right|\le \left|\frac{S_n}n-\frac{S_n}{m^2}\right|+ \left|\frac{S_n}{m^2}-\frac{S_{m^2}}{m^2}\right|.\tag3 $$ Since $|S_n|\le n$, bound the first term on the RHS of (3) by $n\left|\frac 1n-\frac1{m^2}\right|\le\frac2m$. Bound the second term by $\frac1{m^2}|n-m^2|$, since the distance between $S_n$ and $S_{m^2}$ is at most the number of steps from time $m^2$ to time $n$. Adding these, it follows that the RHS of (3) is at most $\frac4m\le\frac4M$. Putting (2) and (3) together, we have $$ \left|\frac{S_n}n-\frac12\right|\le\frac4M+\epsilon $$ whenever $n\ge M^2$. This implies that (1) is less than $2\epsilon$ for all large $n$, and we're done.

Answer 2

If $m^{2} \leq n \leq (m+1)^{2}$ then $\frac {S_m^{2}} {m^{2}} \frac {m^{2}} n \leq \frac {S_n} n \leq \frac {S_{m+1}^{2}} {(m+1)^{2}} \frac {(m+1)^{2}} n $. Just verify that $\frac {m^{2}} n$ and $\frac {(m+1)^{2}} n $ tend to $1$ as $ n\to \infty$. [$\frac {m^{2}} {(m+1)^{2}} \leq\frac {m^{2}} n\leq 1$ and $\frac {m^{2}} {(m+1)^{2}} \to 1$ because $m \to \infty$ as $n \to \infty$. Similar argument for $\frac {(m+1)^{2}} n $.