= Solution
Put $Q_n(\theta)=n^{-1}\sum_{i=1}^n\log f(Y_i;\theta)$ and $Q(\theta)=\mathbb E_{\theta_0}\log f(Y_1;\theta)$. The integrable absolute-log envelope makes these expectations finite. Compactness and continuity of the likelihood ensure that a maximum exists, and maximizing it is equivalent almost surely to maximizing $Q_n$.
First identify the unique maximizer of $Q$. Let $r(y)=f(y;\theta)/f(y;\theta_0)$ on the support of the true density. Since $\log r\leq r-1$,
$$
Q(\theta)-Q(\theta_0)=\mathbb E_{\theta_0}\log r(Y_1)\leq\mathbb E_{\theta_0}r(Y_1)-1=\int_{\{f(y;\theta_0)>0\}}f(y;\theta)dy-1\leq0.
$$
Equality requires $r=1$ almost surely under the true distribution and no mass outside its support, so the two probability distributions coincide. By <identifiability>, this happens only for $\theta=\theta_0$. Equivalently $Q(\theta_0)-Q(\theta)$ is the <Kullback-Leibler divergence> from the true distribution to the candidate.
Fix $\varepsilon>0$ and let $K_\varepsilon=\{\theta\in\Theta:\|\theta-\theta_0\|\geq\varepsilon\}$. If this set is empty there is nothing to prove. Otherwise it is compact, so continuity and the strict maximum imply
$$
\delta_\varepsilon=Q(\theta_0)-\max_{\theta\in K_\varepsilon}Q(\theta)>0.
$$
On the event $\sup_{\Theta}|Q_n-Q|<\delta_\varepsilon/3$, every point of $K_\varepsilon$ has $Q_n(\theta)<Q(\theta_0)-2\delta_\varepsilon/3$, whereas $Q_n(\theta_0)>Q(\theta_0)-\delta_\varepsilon/3$. Hence no maximizer can belong to $K_\varepsilon$. The supplied <uniform law of large numbers> therefore proves <compact-parameter consistency of maximum likelihood>:
$$
\mathbb P_{\theta_0}(\|\widehat\theta_n-\theta_0\|\geq\varepsilon)\leq\mathbb P_{\theta_0}\!\left(\sup_\Theta|Q_n-Q|\geq\frac{\delta_\varepsilon}{3}\right)\longrightarrow0,
\qquad\boxed{\widehat\theta_n\xrightarrow{p}\theta_0.}
$$
Under the usual regularity conditions, with nonsingular one-observation <Fisher information matrix> $I(\theta_0)$, the <asymptotic normality of a maximum likelihood estimator> is
$$
\boxed{\sqrt n(\widehat\theta_n-\theta_0)\xrightarrow{d}N_d\bigl(0,I(\theta_0)^{-1}\bigr),}
$$
where $I(\theta_0)=\mathbb E_{\theta_0}[S_{\theta_0}(Y_1)S_{\theta_0}(Y_1)^T]$ and $S_\theta=\nabla_\theta\log f(\cdot;\theta)$ is the <score function>.
For the uniform model, let $M_n=\max_iY_i$. For a sample in its support, the likelihood is $\theta^{-n}\mathbf1_{\{\theta\geq M_n\}}$, decreasing on its nonzero region. Thus the <uniform endpoint maximum-likelihood estimator> is
$$
\boxed{\widehat\theta_n=M_n.}
$$
The all-zero sample is a probability-zero exception and may be assigned an arbitrary positive estimate. Independence gives $\mathbb P_\theta(M_n\leq y)=(y/\theta)^n$ on $[0,\theta]$. For $0<\varepsilon<\theta$,
$$
\mathbb P_\theta(|M_n-\theta|>\varepsilon)=\left(1-\frac\varepsilon\theta\right)^n\longrightarrow0;
$$
for $\varepsilon\geq\theta$ this probability is zero. This proves consistency directly.
Let $T_n=n(\theta-M_n)/\theta$. Its exact <cumulative distribution function> is
$$
\mathbb P_\theta(T_n\leq t)=\begin{cases}0&t<0,\\1-(1-t/n)^n&0\leq t\leq n,\\1&t>n.\end{cases}
$$
For fixed $t\geq0$, this converges to $1-e^{-t}$, so $T_n\xrightarrow{d}\operatorname{Exp}(1)$. In particular the errors are of order $1/n$. Explicitly, for every fixed $\eta>0$ and sufficiently large $n$,
$$
\mathbb P_\theta\bigl(\sqrt n|M_n-\theta|>\eta\bigr)=\left(1-\frac\eta{\theta\sqrt n}\right)^n\leq e^{-\eta\sqrt n/\theta}\longrightarrow0.
$$
Consequently
$$
\boxed{\frac{n(\theta-\widehat\theta_n)}\theta\xrightarrow{d}\operatorname{Exp}(1),\qquad\widehat\theta_n=\theta+o_p(n^{-1/2}).}
$$
One violated regularity condition is parameter-independent support: the density is supported on $[0,\theta]$. Differentiation under the density integral is not valid in the usual regular-model form. Indeed the interior score equals $-1/\theta$, whose expectation is not zero; the changing endpoint contributes the omitted boundary term. Thus the uniform model is handled by its exact maximum distribution, rather than by the regular asymptotic normality theorem.
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