Is the estimator $\hat \theta =\max{\{X_1,X_2,\ldots,X_n\}}$ consistent? This is a follow up question to this: Conceptual question about estimators
I am still stuck at showing consistency for the second part.
A random number generator produces uniformly distributed random numbers on the intervall $[0,a]$ where $a>0$ is unknown. We can draw $n$ independent $\mathcal U_{[0,a]}$ random numbers $X_1,...,X_n$ for the estimation. Is:
$$\begin{equation*}
    \hat \theta =\max{\{X_1,X_2,\ldots,X_n\}}
\end{equation*} $$
a consistent estimator? I know that I need to show:
$$\lim_{n \to \infty} \Pr\left(\vert \hat{\theta}-a\vert > \varepsilon\right)=0$$
But I don't really know how. I can show that:
$$E[\hat \theta]=\frac{n}{n+1}a$$
and therefore $E[\hat \theta]-a=\frac{n}{n+1}a-a \not= 0 \implies \text{biased}$
But I am not sure if that helps me show consistency.
 A: Clearly $\hat{\theta}$ depends on $n$, so I denote it as $\hat{\theta}_{n}$.
Observe that $\hat{\theta}_{n}\leq a,$ so $\left|\hat{\theta}_{n}-a\right|>\varepsilon$
iff $\hat{\theta}_{n}<a-\varepsilon$ iff $X_{k}<a-\varepsilon$ for
$k=1,\ldots,n$. It follows that
\begin{eqnarray*}
 &  & \{\omega\in\Omega\mid\left|\hat{\theta}_{n}(\omega)-a\right|>\varepsilon\}\\
 & = & \cap_{k=1}^{n}\{\omega\in\Omega\mid X_{k}(\omega)<a-\varepsilon\}.
\end{eqnarray*}
Since $X_{1},X_{2},\ldots,X_{n}$ are i.i.d., we have
\begin{eqnarray*}
 &  & P\left(\{\omega\in\Omega\mid\left|\hat{\theta}_{n}(\omega)-a\right|>\varepsilon\}\right)\\
 & = & \prod_{k=1}^{n}P\left(\{\omega\in\Omega\mid X_{k}(\omega)<a-\varepsilon\}\right)\\
 & = & \prod_{k=1}^{n}\frac{a-\varepsilon}{a}\\
 & = & \left(\frac{a-\varepsilon}{a}\right)^{n}\\
 & \rightarrow & 0
\end{eqnarray*}
as $n\rightarrow\infty$.
Therefore $\hat{\theta}$ is a consistent estimator for $a$.

Remark:
Actually, $\hat{\theta}_{n}\rightarrow a$ pointwisely a.e.. For,
we have already proved that $\hat{\theta}_{n}\rightarrow a$ in probability.
By a result due to Vitali, there exists a subsequence $\left(\hat{\theta}_{n_{k}}\right)_{k}$
such that $\hat{\theta}_{n_{k}}\rightarrow a$ pointwisely a.e. as
$k\rightarrow\infty$. Let $\Omega_{0}=\{\omega\in\Omega\mid\hat{\theta}_{n_{k}}(\omega)\rightarrow a\}$,
then $P(\Omega_{0})=1$. Let $\omega\in\Omega_{0}$. Observe that
$(\hat{\theta}_{n}(\omega))_{n}$ is monotonic increasing and bounded
above by $a$, so $\lim_{n\rightarrow\infty}\hat{\theta}_{n}(\omega)$
exists. In particular, $\lim_{n\rightarrow\infty}\hat{\theta}_{n}(\omega)=\lim_{k\rightarrow\infty}\hat{\theta}_{n_{k}}(\omega)=a$.
This shows that $\hat{\theta}_{n}\rightarrow a$ a.e..
