Suppose $P_n=\mathscr{U}([0,1])$ and $Q_n=\mathscr{U}([0,1+\frac{1}{n}])$ on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$.
How can I show if the probability measures are contiguous or not to each other? And does the same hold for $\hat{P}_n=\mathscr{U}([0,1])^n$ and $\hat{Q}_n=\mathscr{U}\left([0,1+\frac{1}{n}]\right)^n$ on $(\mathbb{R}^n, \mathscr{B}(\mathbb{R}^n))$?
I know the definition of contiguity. But I not so sure how to prove a measure is contiguous to another. And to prove a measure is not contiguous to another measure, do I need to find a measurable set $A_n$ such that $P_n(A_n)=0$ does not imply $Q_n(A_n)=0$?
Edit: For $P_n=\mathscr{U}([0,1])$ and $Q_n=\mathscr{U}\left([0,1+\frac{1}{n}]\right)$ on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$, I found that $\lim_{n \rightarrow \infty}Q_n(x)=\frac{1}{1-0}\mathbf{1}_{[0,1]}(x)=\lim_{n \rightarrow \infty}P_n(x)$. Therefore for any $A_n \in \mathscr{B}(\mathbb{R})$, $\lim_{n \rightarrow \infty}Q_n[A_n]=0$ iff $\lim_{n \rightarrow \infty}P_n[A_n]=0$, hence $P_n$ and $Q_n$ are contiguous to each other.
Now I need to show whether $\hat{P}_n=\mathscr{U}([0,1])^n$ and $\hat{Q}_n=\mathscr{U}\left([0,1+\frac{1}{n}]\right)^n$ on $(\mathbb{R}^n, \mathscr{B}(\mathbb{R}^n))$ are contiguous to each other. But my calculations show that $\lim_{n \rightarrow \infty}\hat{Q}_n(x)=\lim_{n \rightarrow \infty}\hat{P}_n(x)$ is no longer true in this scenario. Is there another way I can show whether the two measures are contiguous to each other or not? (Le Cam's first lemma, counter example, etc.?)
Sorry about the confusion.
$\widehat P$ and $\widehat Q$ are not (mutually) contiguous; set $A_n = [0, 1 + 1/n]^n \setminus [0,1]^n$. Then, clearly $\widehat P_n(A_n) = 0$, since $A_n \cap [0, 1]^n = \emptyset$, but $\widehat Q_n(A_n) = \frac{(1 + 1/n)^n - 1}{(1 + 1/n)^n} \to \frac{e-1}{e} \neq 0$. So $\widehat P$ is not contiguous to $\widehat Q$. However, $\widehat Q$ is contiguous to $\widehat P$, since $$\widehat Q_n(A_n) \to 0 \implies 0 \leq \mu_n(A_n \cap [0,1]^n) \leq \mu_n(A_n \cap [0,1+1/n]^n) \to 0 \implies \widehat P_n(A_n) \to 0.$$