Estimator for Uniform Random Variables Let $(U_n)$ be iid uniform random variables over $(0, \theta)$. Consider to estimators for $\theta$, namely, $\max\{2\bar U_n, 0\}$ and $\max\{U_n\}$. My question is how to show that both of these estimators are convergent almost surely to $\theta$. In addition, it seems to me that the former estimator is overly complicated since $2\bar U$ is always larger than $0$. Therefore, it is fine to simply write it as $2\bar U$. Right? More detailed answers are much appreciated. Thank you!
 A: First,
unless I'm missing something,
I agree with you that $\max\{2\bar{U}_n,0\}$ should be written as simply $2\bar{U}_n$.
Let's look at the two estimators you have:

$2\bar{U_n}$: This one is the easiest of the two.
In general,
if $X$ is a uniform random variable on the interval $(a,b)$,
then $X$ has an expected value of $\frac{b+a}{2}$.
In our case,
this means that the expected value of the $U_n$ is $\theta/2$.
Since uniform random variables have a finite variance,
we can apply the Strong Law of Large Numbers to conclude that $\bar{U}_n$ converges almost surely to the mean $\theta/2$,
from which the desired result immediately follows.

$\max\{U_n\}$: I'll first introduce some notation.
Let us use $(\Omega,\mathscr F,\Pr)$ to denote the probability space on which the $U_n$ are defined,
and for every $n\in\mathbb N$,
let $M_n=\max\{U_1,\ldots,U_n\}$.
Let $\omega\in\Omega$ be arbitrary.
Then,
the sequence $M_1(\omega),M_2(\omega),M_3(\omega),\ldots$ is monotonically increasing.
Indeed,
we have that
$$\{U_1(\omega)\}\subset\{U_1(\omega),U_2(\omega)\}\subset\cdots,$$
and $A\subset B$ necessarily implies that $\max A\leq\max B$.
Given that $\theta$ is an upper bound for the sequence $\{M_n(\omega):n\in\mathbb N\}$,
it must be the case that it converges to some value $\theta_0(\omega)\in[0,\theta]$.
Suppose that $\theta_0(\omega)<\theta$.
Then,
it clearly is the case that $U_n(\omega)<\theta-\epsilon(\omega)$ for every $n\in\mathbb N$,
where $\epsilon(\omega)>0$ is some constant independent of $n$.
Consequently,
we see that the subset $\Omega_0\subset\Omega$ on which $M_n(\omega)$ does not converge to $\theta$ is included in
\begin{align*}
\bigcup_{\omega\in\Omega_0}\left(\bigcap_{n\in\mathbb N}[U_n<\theta-\epsilon(\omega)]\right)&\subset
\bigcup_{q\in\mathbb Q,~0<q<\theta}\left(\bigcap_{n\in\mathbb N}[U_n<\theta-q]\right).
\end{align*}
At this point,
if you prove that
$$\Pr\left[\bigcap_{n\in\mathbb N}[U_n<\theta-q]\right]=0,\tag{1}$$
then the result will follow by countable sub-additivity.
(A hint for $(1)$: use Kolmogorov's $0$-$1$ law. Don't hesitate to comment if anything I've written is unclear.)
