convergence almost surely that uses lim sup of probabilities What's wrong with my intuition here?
I read that given the following sequence of random numbers $X_i \in (0,1]$, with $I$ as indicator function:
$$
X_1 = I(0,1/2] \quad X_2 = I(1/2,1] \quad X_3 = I(0,1/4] \quad X_4 = I(1/4,1/2] \quad X_5 = I(1/2,3/4] \quad X_6 = I(3/4,1] \quad X_7 = I(0,1/8] \quad X_8 = I(1/8,1/4] \quad ...
$$
The sequence $X_i$ converges in probability as $i \rightarrow \infty$ since given any $\epsilon > 0$, we have:
$$
P[X_i \geq \epsilon] \rightarrow 0
$$
However, $X_i$ does not converge almost surely (or almost everywhere), since given any $i$, there is a $n \geq i$, s.t. $P[X_n] > \epsilon$ for some $\epsilon > 0$ (which means that $X_n$ is above $0$ infinitely)... However, I could not relate this with an alternative definition for almost sure convergence which is as follows given any $\epsilon > 0$:
$$
\text{lim sup}_{i \rightarrow \infty} \{ P[X_i \geq \epsilon] \} \rightarrow 0
$$
I guess my confusion in this definition is when it starts to use probabilities $P$. Since the 'size' of the intervals where $P[X_i] > \epsilon$ also approaches zero, $0$, i.e. it gets narrower and narrower, shouldn't it also follow that the Lebesgue measure of $P[X_i] > \epsilon \rightarrow 0$ such that its supremum $\rightarrow 0$?
 A: I think the "alternate definition" is wrong. An equivalent definition for "$X_i$ converges to $0$ almost surely" is "for every $\epsilon > 0$, $P[\limsup_{i \to \infty} X_i \ge \epsilon] = 0$", rather than "for every $\epsilon > 0$, $\limsup_{i \to \infty} P[X_i \ge \epsilon] = 0$". Here, $\limsup_{i \to \infty} X_i$ is a random variable, defined, at a point $\omega$, by $[\limsup_{i \to \infty} X_i](w) = \limsup_{i \to \infty} X_i(w)$.
Of course, since the intervals get narrower and narrower, $\limsup_{i \to \infty} P[X_i \ge \epsilon] = 0$ for any $\epsilon > 0$. However, almost sure (i.e. pointwise) convergence concerns what happens to an arbitrary particular point (of the underlying probability/sample space), and in this case, it is in infinitely many of the intervals (and therefore $X_i$ at that point is $1$ infinitely often). 
It seems like you might have meant the correct alternate definition (and not the one you wrote), but just are having issues with notation, due to your last sentence "shouldn't it also follow that the Lebesgue measure of $P[X_i] > \epsilon \to 0$ such that its supremum $\to 0$?" The notation here is off ("Lebesgue measure of $P[X_i] > \epsilon$" doesn't make sense, since $P[X_i]$ is a number, not a set), but I'm assuming you meant "Lebesgue measure of $\{\omega : X_i(\omega) > \epsilon\}$", or, more simply, "$P[X_i > \epsilon]$" and are asking why isn't it the case that $P[\limsup_{i \to \infty} X_i \ge \epsilon] = 0$. 
So the true question here is why the following implication does not hold: $P[X_i > \epsilon] \to 0$ for each $\epsilon > 0$ implies $P[\limsup_{i \to \infty} X_i > \epsilon] \to 0$. In other words, if each particular $X_i$ has small probability of being greater than $\epsilon$ (and this probability goes to $0$ as $i \to \infty$), why isn't it the case that the probability that infinitely many $X_i$ are greater than $\epsilon$ zero (or at least very small)? The answer basically stems from the fact that infinitely many samples is very many, and small probabilities don't really matter if they aren't too small. You can of course try to update your intuition regarding this by thinking about your example with intervals. Alternatively, if a coin has a $\frac{1}{10^{10}}$ probability of landing on heads and if you flip it infinitely many times, there will be infinitely many heads, and this is still true if you replace $\frac{1}{10^{10}}$ by $\frac{1}{n}$ for the $n^{th}$ toss. The reason I said "if they aren't too small" is that if the probability of heads on the $n^{th}$ toss is $\frac{1}{n^2}$, then you will, with probability $1$, only get heads finitely many times. And indeed, it's actually critical for your example with intervals that the sum of the lengths of the intervals is infinite.  
