Let $\{X_n\}_{n=1}^\infty$ an i.i.d. sequence of random variables. Assume that $\limsup_{n \to \infty} |\sum_{i=1}^n X_i|/\sqrt{n \log \log n} < \infty$ almost surely.
My book claims that by Kolmogorov's $0-1$-law, there is a positive constant $K $ such that
$$\limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i|}{\sqrt{n \log\log n}} \leq K < \infty \quad a.s.$$
I'm not sure how to apply this $0-1$ law here.
Put $$Z := \limsup_{n\to \infty} \frac{|\sum_{i=1}^n X_i|}{\sqrt{n \log\log n}}$$
and define the tail-sigma algebra$$\mathcal{T}:= \bigcap_{m=1}^\infty \sigma\{X_m, X_{m+1}, \dots\}$$
I want that $Z$ is $\mathcal{T}$-measurable (or at least that the events $\{Z \leq K\}$ are in the tail-sigma-algebra, but why is this the case?
Best Answer
You should have defined $Z := \limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i|}{\sqrt{n \log \log n}}$. For any $K$, the event $\{Z \le K\}$ lies in $\mathcal{T}$ because $Z$ does not depend on any finite subset of the $X_i$.
Explicitly, let us show $\{Z \le K\} \in \sigma\{X_m, X_{m+1}, \ldots\}$.
Note that for $n > m$, $$ \frac{|\sum_{i=m}^n X_i|}{\sqrt{n \log \log n}} - \frac{|\sum_{i=1}^{m-1} X_i|}{\sqrt{n \log \log n}} \le \frac{|\sum_{i=1}^n X_i|}{\sqrt{n \log \log n}} \le \frac{|\sum_{i=m}^n X_i|}{\sqrt{n \log \log n}} + \frac{|\sum_{i=1}^{m-1} X_i|}{\sqrt{n \log \log n}}$$ Since $\frac{|\sum_{i=1}^{m-1} X_i|}{\sqrt{n \log \log n}} \to 0$ almost surely as $n \to \infty$, $$\limsup_{n \to \infty} \frac{|\sum_{i=1}^n X_i|}{\sqrt{n \log \log n}} = \limsup_{n\to \infty} \frac{|\sum_{i=m}^n X_i|}{\sqrt{n \log \log n}}$$ almost surely.