I'm reading about the Law of Iterated Logarithm for standard Brownian motion.
As a Corollary of this we have the next:
Suppose that $\{B(t)\}_{t\geq 0}$ is a standard Brownian Motion. Then almost surely $$\displaystyle\limsup_{t\rightarrow 0}\frac{|B(t)|}{\sqrt{2t\log\log\frac{1}{t}}}=1.$$
The proof is as follows:
The process $\{X(t)\}_{t\geq 0}$ defined by $X(t)=tB(1/t)$ is a Brownian Motion for $t>0.$ Then, for LIL we have $$\displaystyle\limsup_{t\rightarrow 0}\frac{|B(t)|}{\sqrt{2t\log\log\frac{1}{t}}}=\displaystyle\limsup_{h\rightarrow\infty}\frac{|X(h)|}{\sqrt{2h\log\log h}}=1.$$
I don't understand it. To apply L.I.L. we need that $|X(h)|$ be Brownian Motion. So I suppose that $|X(h)|$ it is,but I can't see it.
Any kind of help is thanked in advanced.