[proofplan]
We prove both directions directly from the [Radon--Nikodym derivative](/page/Radon-Nikodym%20Derivative) identity displayed in the first step. [Uniform integrability](/page/Uniform%20Integrability) lets us split the likelihood ratio into the bounded part $L_n\le M$ and the tail $L_n>M$, so small $P_n$-probability events have small $Q_n$-probability. Conversely, if uniform integrability fails, the likelihood-ratio tail sets themselves have vanishing $P_n$-probability but non-vanishing $Q_n$-probability, contradicting [contiguity](/page/Contiguity). The moment condition implies uniform integrability by a direct power estimate on the tail.
[/proofplan]
[step:Split the likelihood ratio into bounded and tail parts]
Let $A_n \in \mathcal{F}_n$ be a sequence of events such that $P_n(A_n)\to 0$. Fix $M>0$. Since $Q_n \ll P_n$ and $L_n=dQ_n/dP_n$, the [Radon--Nikodym derivative](/page/Radon-Nikodym%20Derivative) identity gives
\begin{align*}
Q_n(A_n)=\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\right].
\end{align*}
Splitting according to the disjoint events $\{L_n\le M\}$ and $\{L_n>M\}$ gives
\begin{align*}
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\right]
= \mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n\le M\}}\right]
+ \mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n>M\}}\right].
\end{align*}
On $\{L_n\le M\}$ we have $L_n\mathbb{1}_{A_n}\le M\mathbb{1}_{A_n}$, and on the tail we have
\begin{align*}
L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n>M\}}
\le L_n\mathbb{1}_{\{L_n>M\}}.
\end{align*}
Therefore
\begin{align*}
Q_n(A_n)
\le M P_n(A_n)+\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right].
\end{align*}
[guided]
The only available bridge between $Q_n$ and $P_n$ is the [Radon--Nikodym derivative](/page/Radon-Nikodym%20Derivative). For every event $A_n\in\mathcal{F}_n$, the identity $L_n=dQ_n/dP_n$ means
\begin{align*}
Q_n(A_n)=\int_{\Omega_n}\mathbb{1}_{A_n}(\omega)L_n(\omega)\,dP_n(\omega).
\end{align*}
Equivalently, in expectation notation,
\begin{align*}
Q_n(A_n)=\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\right].
\end{align*}
To compare this with $P_n(A_n)$, we separate the part where $L_n$ is bounded from the part where $L_n$ may be large. Fix $M>0$. Since
\begin{align*}
\Omega_n=\{L_n\le M\}\cup \{L_n>M\}
\end{align*}
as a disjoint union up to equality of indicators, we have
\begin{align*}
Q_n(A_n)
&= \mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n\le M\}}\right]
+ \mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n>M\}}\right].
\end{align*}
On $\{L_n\le M\}$, the likelihood ratio cannot amplify probabilities by more than the factor $M$, so
\begin{align*}
L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n\le M\}}
\le M\mathbb{1}_{A_n}.
\end{align*}
Taking $P_n$-expectations gives
\begin{align*}
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n\le M\}}\right]
\le M\mathbb{E}_{P_n}\!\left[\mathbb{1}_{A_n}\right]
= M P_n(A_n).
\end{align*}
On the complementary set $\{L_n>M\}$, we discard $\mathbb{1}_{A_n}\le 1$ and obtain
\begin{align*}
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{A_n}\mathbb{1}_{\{L_n>M\}}\right]
\le
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right].
\end{align*}
Combining the two estimates yields
\begin{align*}
Q_n(A_n)
\le M P_n(A_n)+\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right].
\end{align*}
This inequality is the core of the lemma: the bounded part is controlled by $P_n(A_n)$, and uniform integrability controls the tail uniformly in $n$.
[/guided]
[/step]
[step:Use uniform integrability to prove contiguity]
Assume
\begin{align*}
\lim_{M\to\infty}\sup_{n\in\mathbb{N}}\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]=0.
\end{align*}
Let $\varepsilon>0$. Choose $M>0$ such that
\begin{align*}
\sup_{n\in\mathbb{N}}\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]<\frac{\varepsilon}{2}.
\end{align*}
Since $P_n(A_n)\to 0$, choose $N\in\mathbb{N}$ such that for all $n\ge N$,
\begin{align*}
M P_n(A_n)<\frac{\varepsilon}{2}.
\end{align*}
The estimate from the previous step gives, for all $n\ge N$,
\begin{align*}
Q_n(A_n)
\le M P_n(A_n)+\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]
<\varepsilon.
\end{align*}
Thus $Q_n(A_n)\to 0$, so $Q_n\triangleleft P_n$.
[/step]
[step:Construct tail events when uniform integrability fails]
Assume $Q_n\triangleleft P_n$ and suppose, toward a contradiction, that $(L_n)$ is not uniformly integrable under $(P_n)$. Then there exists $\varepsilon>0$ such that for every $M>0$,
\begin{align*}
\sup_{n\in\mathbb{N}}\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]\ge \varepsilon.
\end{align*}
Choose an increasing sequence of thresholds $M_k\to\infty$. For each $k\in\mathbb{N}$, the definition of the supremum gives an index $n_k\in\mathbb{N}$ such that
\begin{align*}
\mathbb{E}_{P_{n_k}}\!\left[L_{n_k}\mathbb{1}_{\{L_{n_k}>M_k\}}\right]\ge \frac{\varepsilon}{2}.
\end{align*}
The indices may be chosen with $n_k\to\infty$: otherwise a finite set of indices would realize the lower bound $\varepsilon/2$ for arbitrarily large thresholds. For each fixed $n$, the random variables $L_n\mathbb{1}_{\{L_n>M\}}$ converge pointwise to $0$ as $M\to\infty$ and are dominated by the integrable [random variable](/page/Random%20Variable) $L_n$, because $L_n\ge 0$ and $\mathbb{E}_{P_n}[L_n]=Q_n(\Omega_n)=1$. The [Dominated Convergence Theorem](/page/Dominated%20Convergence%20Theorem) therefore gives
\begin{align*}
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]\to 0
\end{align*}
as $M\to\infty$, a contradiction. Passing to a subsequence if necessary, relabel the indices so that $n_k$ is strictly increasing.
Define events $B_k\in\mathcal{F}_{n_k}$ by
\begin{align*}
B_k:=\{L_{n_k}>M_k\}.
\end{align*}
Then
\begin{align*}
P_{n_k}(B_k)
=
P_{n_k}(L_{n_k}>M_k)
\le
\frac{1}{M_k}\mathbb{E}_{P_{n_k}}\!\left[L_{n_k}\right]
=
\frac{1}{M_k},
\end{align*}
because $L_{n_k}>M_k$ on $B_k$ and $\mathbb{E}_{P_{n_k}}[L_{n_k}]=Q_{n_k}(\Omega_{n_k})=1$. Hence $P_{n_k}(B_k)\to 0$. On the other hand,
\begin{align*}
Q_{n_k}(B_k)
=
\mathbb{E}_{P_{n_k}}\!\left[L_{n_k}\mathbb{1}_{B_k}\right]
=
\mathbb{E}_{P_{n_k}}\!\left[L_{n_k}\mathbb{1}_{\{L_{n_k}>M_k\}}\right]
\ge \frac{\varepsilon}{2}.
\end{align*}
[/step]
[step:Contradict contiguity using the tail events]
Define a sequence of events $A_n\in\mathcal{F}_n$ as follows. For $n=n_k$, set $A_n:=B_k$; for all other indices $n$, set $A_n:=\varnothing$. Since $n_k\to\infty$ and $P_{n_k}(B_k)\to 0$, we have $P_n(A_n)\to 0$. But along the strictly increasing subsequence $(n_k)$,
\begin{align*}
Q_{n_k}(A_{n_k})=Q_{n_k}(B_k)\ge \frac{\varepsilon}{2},
\end{align*}
so $Q_n(A_n)$ does not converge to $0$. This contradicts $Q_n\triangleleft P_n$. Therefore $(L_n)$ must be uniformly integrable under $(P_n)$.
[/step]
[step:Derive uniform integrability from the moment bound]
Assume there exist $\delta>0$ and $C<\infty$ such that
\begin{align*}
\sup_{n\in\mathbb{N}}\mathbb{E}_{P_n}\!\left[L_n^{1+\delta}\right]\le C.
\end{align*}
For every $M>0$ and every $n\in\mathbb{N}$, on the event $\{L_n>M\}$ we have
\begin{align*}
L_n \le M^{-\delta}L_n^{1+\delta}.
\end{align*}
Multiplying by $\mathbb{1}_{\{L_n>M\}}$ and taking $P_n$-expectations gives
\begin{align*}
\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]
\le
M^{-\delta}\mathbb{E}_{P_n}\!\left[L_n^{1+\delta}\right]
\le
C M^{-\delta}.
\end{align*}
Taking the supremum over $n$ yields
\begin{align*}
\sup_{n\in\mathbb{N}}\mathbb{E}_{P_n}\!\left[L_n\mathbb{1}_{\{L_n>M\}}\right]
\le C M^{-\delta}.
\end{align*}
Since $C M^{-\delta}\to 0$ as $M\to\infty$, the likelihood ratios are uniformly integrable under $(P_n)$. By the equivalence already proved, $Q_n\triangleleft P_n$.
[/step]
