Restating the law of the iterated logarithm If $x$ is a real number, denote by $x_n$ the $n$'th bit of $x$ after the radix.
Define $S := \{x: x \in [0,1) \wedge \limsup_{n \rightarrow \infty}{\frac{\sum_{k \lt n}(-1)^{x_k}}{\sqrt{n \cdot \log{\log{n}}}}} = \sqrt{2}\}$ 
Is the statement that $S$ has Lebesgue measure $1$ equivalent to the law of the iterated logarithm?
I think so but I'm not sure.  I'm trying to check if I understand how to apply the concept of "almost surely" as used in the Wikipedia definition.
 A: The crucial fact here is that the distribution of $x$ is the Lebesgue measure on $[0,1]$ if and only if the distribution of $(x_n)$ is the product of uniform Bernoulli measures on $\{0,1\}$. Thus, setting $y_n=(-1)^{x_n}$, one sees that $(y_n)$ is an i.i.d. sequence of Bernoulli random variables on $\{-1,1\}$. In particular $E[y_n]=0$ and $\mathrm{var}(y_n)=1$, hence everything is in place to apply the LIL to $(y_n)$.
The LIL says that, almost surely for the product of uniform Bernoulli measures on $\{-1,1\}$, $(y_n)$ satisfies the property-you-know. Equivalently, almost surely for the Lebesgue measure on $[0,1]$, $x$ satisfies the property-you-know. The set $S$ being the set of $x$ such that the property-you-know holds, what we are saying is that $\mathrm{Leb}(S)=1$.
A: If you meant $\sum_{k=0}^n (-1)^{x_k}$ instead of $(-1)^{x_n}$ on the numerator, then yes.
An event "happens almost surely" means the same as "has probability $1$".
If not, $S$ is empty because the sequence converge to $0$ for any $x$, and so its limsup is always $0$, which is not $\sqrt 2$.
