Converting a definition to a limit form Set $S$ is the set of infinite sequences (each sequence of the form: $a_1,a_2,...$) where each term $a_i\in A$. Let $u_a^{T_0}$ represents number of times $a\in A$ occurs before index ${T_0}$ in a particular sequence. (The probability with which a particular $a$ occurs at any index is $p_a<1$) 
The statement is: 
"For every $u\geq0$ and $\delta>0,$ there exists $T_0\geq u$ such that
$$P(\forall a \in A: u_a^{T_0}>u)\geq 1-\delta$$
where $P(.)$ represents the sum of the probabilities of the event (inside the brackets) over all sequences." 
Note that this is just a statement, which may or may not be true. 
So is there any way this could be written as a limit of something? There seems to be a resemblance between the definition of limit to $\infty$ and this, but I cannot just work out the exact form.
 A: I read it as
$\quad$ "For each $u$ fixed, the probability of the event
$\quad$(each $a \in A$ appears at least $u$ times in $a_1,\ldots,a_{T_0}$)
$\quad$ tends to $1$ as $T_0\to ∞$."
Another way of writing this would be: $\forall u > 0$,
$$\mathbb P\left(\min_{a \in A}\#\{n \leq T_0\,:\, a_n = a\}\geq u\right) \to 1 \qquad \text{as}\qquad T_0 \to ∞.$$

A weaker statement than this is the following:

*

*"With probability 1, every element $a \in A$ appears infinitely many times in $(a_n)_{n=1}^∞$":

it is weaker than your statement because your's has a kind of "uniformity" built in — it is equivalent to the $p_a$ being uniformly bounded away from 0 (i.e. $\inf_{a\in A}p_a > 0$). A counterexample would be with any infinite $A$, e.g.
$$A = \mathbb N = \{1,2,3,\ldots\}, \qquad p_a = 2^{-a}:$$
even though every symbol will almost-surely appear infinitely often, there's no way that it can satisfy your statement. Same goes for $A$ finite with some of the $p_a$ equal to zero.
Otherwise, it should hold just fine! (Assuming the $p_a$ applies to all $n$ independently.) You may want to look up normal numbers and the proof that a number is normal with probability one (using the Borel-Cantelli Lemma).
It may yield good intuition why the above is true in this case.
