Here is a classical question about a.s. convergence and GC Theorem(Uniform L.L.N):
Suppose that $X_1$, $X_2$,....,$X_n$ are i.i.d. with $E|X| < \infty$, and let $D_n := \frac{1}{n} \Sigma_{i=1}^{n} |X_{i} - \bar X_{n}|^r$ with $0<r<1$ fixed. Show that $D_n \rightarrow_{a.s.} d := E{|X_1 - \mu|^{r}}$ with $\mu = E(X_1)$.
Proof (in brief):
Let $\delta >0$. Then $\bar X_n \in [\mu -\delta, \mu +\delta]$ for n sufficiently large with probability 1, so we can consider the collection of functions $f(x,t) = f_t(x)=|x-t|^r$ with $t \in [\mu-\delta, \mu+\delta]$, namely:
$\tilde F = \{ f_t(x)=|x-t|^r: \mu-\delta \leq t \leq \mu+\delta \}$
Then apply uniform Law of large number and $C_r$- inequality :
$~~~~|D_n - d| \\ = |\frac{1}{n}\Sigma_{i=1}^{n} |X_{i} - \bar X_{n}|^r - E|X_1 - \mu|^{r} | \\ = |\mathbb{P}_n f_{\bar X_{n}} - P_n f_{\bar X_{n}} + P_n f_{\bar X_{n}} - P f_{\mu} | \\ \leq sup_{\mu-\delta \leq t \leq \mu+\delta}|\mathbb{P}_n f_t - P f_t| + |P f_{\bar X_{n}} - Pf_{\mu}| \\ \rightarrow_{a.s} 0+0 = 0 $
I can understand the later part, but cannot understand the first part(the bold part): why after we know that $\bar X_n$ located in $[\mu-\delta, \mu+\delta]$ with probability 1, then we could just focus only on the function class $\tilde F$ and still ensure the conclusion work under almost surely convergence? (namely, how to ensure that the case $\bar X_n \notin [\mu-\delta, \mu+\delta]$ would not influence the convergence a.s ??)
I know this question may be simple to many of yours, but as a beginner in measure theory and statistics, it play an important role in understanding various modes of conv. Thank you for your help!!
Perhaps this will clear your doubt: if $x_n \to \mu$ and $\bar {x}_n =\frac {x_1+x_2+...+x_n} n$ then $\frac 1 n \sum_1 ^{n} |x_j -\bar {x}_n| \to 0$ and this is what is being proved here. This is a deterministic result and random variables are not involved in this argument. In fact you can also give a simpler proof using the fact that $(a+b)^{r} \leq 2^{r} (a^{r} +b^{r})$.