Independent but not identically distributed Bernoullis If $\{X_i\}_{i\geq 1}$ is a sequence of independent $\text{Bernoulli}(\theta_i)$ random variables, and there are constants $0\leq a<b\leq 1$ such that $a\leq\theta_i \leq b$, for every $i\geq 1$, is it the case that
$$
  P\!\left\{\omega:\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^n X_i(\omega)\geq a\right\} = 1 \quad\text{and}\quad P\!\left\{\omega:\limsup_{n\to\infty}\frac{1}{n}\sum_{i=1}^n X_i(\omega)\leq b\right\} = 1?
$$
Is there a known proof of this in the literature? If the assertions above are false, are there any simple counterexamples?
 A: For $i=1,2,\dots$
Let $B_{i}$ have Bernoulli distribution with parameter $b$.
Let $U_{i}$ have Bernoulli distribution with parameters $\frac{\theta_{i}}{b}$.
Let $V_{i}$ have Bernoulli distribution with parameters $\frac{a}{\theta_{i}}$.
Further let all these random variables be independent.
Then $U_{i}V_{i}B_{i}$ has Bernoulli distribution
with parameter $a$ and $U_{i}B_{i}$ has Bernoulli distribution with
parameter $\theta_{i}$.
This with $U_{i}V_{i}B_{i}\leq U_{i}B_{i}\leq B_{i}$ where $U_iB_i$ has the same distribution as $X_i$.
The strong law of large numbers assures us that:
$$P\left(\lim_{n\to\infty}\overline{U_{n}V_{n}B}_{n}=a\right)=1\text{ and }
P\left(\lim_{n\to\infty}\overline{B}_{n}=b\right)=1$$
This together with: $$P\left(\lim_{n\to\infty}\overline{U_{n}V_{n}B}_{n}\leq\liminf_{n\to\infty}\overline{U_{n}B}_{n}\leq\limsup_{n\to\infty}\overline{U_{n}B}_{n}\leq\lim_{n\to\infty}\overline{B}_{n}\right)=1$$
Taking $X_n=U_nB_n$ we conclude that: $$P\left(a\leq\liminf_{n\to\infty}\overline{X}_n\leq\limsup_{n\to\infty}\overline{X}_{n}\leq b\right)=1$$
