Let $(a_k)_{k \geq 1}$ be random variables taking values on a finite subset $B$. Assume that $$ (1) \quad \Pr\Big (\lim_{n\rightarrow +\infty}d(\frac{1}{n}\sum_{k=1}^n 1_{[a_k = b]}, [v_\ell(b,\theta_0),v_u(b,\theta_0)])= 0\Big)=1 \quad \forall b\in B $$ where $d$ stays for distance (i.e., absolute value difference); $v_\ell$ and $v_u$ are real-valued functions of $b\in B$ and $\theta \in \mathbb{R}$; $\theta_0\in \mathbb{R}$ is a specific value of $\theta$.
Define $$ \Theta_n\equiv \Big\{\theta \in \mathbb{R}: \frac{1}{n}\sum_{k=1}^n 1_{[a_k = b]}\in [v_\ell(b,\theta),v_u(b,\theta)] \forall b \in B\Big\} $$
Question: Is it possible to find an upper for $\Pr(\theta_0\in \Theta_n)$?
My attempt so far: We have that $E\Big(\frac{1}{n}\sum_{k=1}^n 1_{[a_k = b]}\Big)\leq v_u(b,\theta_0)$ (see here). By Markov inequality $$ \Pr \Big(\frac{1}{n}\sum_{k=1}^n 1_{[a_k = b]} \geq r\Big)\leq \frac{v_u(b,\theta_0)}{r} $$ for each $r>0$. From here?
As advised in the comment below, the problem as posed only provides trivial bounds. Therefore, in what follows, I add some additional structure. In particular, I show that $$ (2) \quad \Pr(a_k=b | a_{1},\dots, a_{k-1}) \in \big[\nu_{\ell}(b, \theta_0), \nu_u(b,\theta_0)\big] \quad \text{ for all } k, b $$ is a sufficient condition for (1), in the hope that this can suggest additional relevant details.
Proof of (1): From (2), we have that $$ \frac{1}{n}\sum_{k=1}^n \Pr(a_k=b | a_{1},\dots, a_{k-1})\in \big[\nu_{\ell}(b, \theta_0), \nu_u(b,\theta_0)\big] \text{ for all } n $$
Let $$ S_n(b)\equiv\sum_{k=1}^n \frac{1}{k} \left({1}_{[a_k=b]}- \Pr(a_k=b| a_1, \dots, a_{k-1})\right). $$ Observe that the sequence $(S_n(b))_{n\geq 1}$ is an $L^2$-bounded martingale. Therefore, $S_n(b)$ converges almost surely as $n\to +\infty$. Further, $\frac{1}{n}\sum_{k=1}^{n-1} S_k(b)$ converges to the same limit as $S_n(b)$.
Let $$ Q_n(b)\equiv \sum_{k=1}^n \left(1_{[a_k=b]}- \Pr(a_k=b| a_1, \dots, a_{k-1}\right). $$
Observe that $$ Q_n(b) = n S_n(b)-\sum_{k=1}^{n-1} S_{k}(b). $$
By the convergence of $S_n(b)$ and $\frac{1}{n}\sum_{k=1}^{n-1} S_k(b)$ to the same limit, $\frac{1}{n}{Q_n(b)}$ converges almost surely to 0 as $n\to +\infty$.
Hence, $$ \frac{1}{n}\sum_{k=1}^n 1_{[a_k=b]}-\frac{1}{n}\sum_{k=1}^n \Pr(a_k=b| a_1, \dots, a_{k-1})\overset{a.s.}{\to} 0 \text{ as $n\to +\infty$.} $$ Therefore, $\frac{1}{n}\sum_{k=1}^n 1_{[a_k=b]}$ and $\frac{1}{n}\sum_{k=1}^n \Pr(a_k=b| a_1, \dots, a_{k-1})$ have the same limit points which belong to $ \big[\nu_{\ell}(b, \theta_0), \nu_u(b,\theta_0)\big] $ and (1) holds.
