I imagine that you mean that $(a_n)_{n \ge 1}$ are independent random variables taking values on a finite subset $B$, and that $\nu_l(b) \le P_n(b):=P[a_n = b] \le \nu_u(B)$ for every $b \in B$ ?
If yes, then for each $b \in B$, the formula $$M_n(b) := \sum_{k=1}^n\frac{1}{k}\big(1_{[a_k=b]}-P_k(b)\big)$$ defines a square-integrable martingale. This martingale has orthogonal increments and is bounded in $L^2(P)$, since $$E\Big[\frac{1}{k^2}\big(1_{[a_k=b]}-P_k(b)\big)^2\Big] \le \frac{1}{4k^2}.$$ Hence it converges almost surely and in $L^2$.
We deduce that the averages
$$\frac{S_n(b)}{n} := \frac{1}{n}\sum_{k=1}^n \big(1_{[a_k=b]}-P_k(b)\big)$$
converge almost surely to $0$, by Cesàro lemma since
$$\frac{S_n(b)}{n} = \frac{1}{n}\sum_{k=1}^n k(M_n(b)-M_{n-1}(b))
= \frac{1}{n}\Big(nM_n(b) - \sum_{k=0}^{n-1}M_k(b)\Big).$$
As a result, the averages $\frac{S_n(b)}{n}$ and $\frac{1}{n}\sum_{k=1}^n P_k(b)$ have the same limit points as $n \to +\infty$, which belong to $[\nu_l(b),\nu_u(b)]$