Let $\mu$ be an absolutely continuous (w.r.t. the Lebeasgue measure $\lambda$) Borel measure on $(X, \mathcal{X})$, where $X \subset \mathbb{R}^p$, for some $p \geq 1$. Then we know that $\mu$ is continuous from above/below: i.e. if $(A_n)_{n\geq 1} $ is a sequence of measurable sets then if $A_n \subset A_{n+1}$ (or $A_{n+1} \subset A_n$) we have $ \lim_{n \to \infty}\mu_n (A_n)=\mu(\cup_{n\geq 1} A_n) $ (or $\lim_{n \to \infty}\mu_n (A_n)=\mu(\cap_{n\geq 1} A_n)$). However, it is legitimate to wonder whether other forms of continuity hold true.
In particular, for a sequence of vectors $x_n=(x_{n,1}, \ldots, x_{n,d})$ with positive components, one may denote $$ x_n \cdot B= \{x \cdot x_n: \, \in B\}, \quad B \in \mathcal{X}, $$ where $x_n \cdot x=(x_{n,1}x_1, \ldots, x_{n,p}x_p)$ denotes the componentwise product, and wonder whether $$ \lim_{n\to \infty}{x_{n,j}}=1, \forall j \in \{1, \ldots,p\} \implies \lim_{n \to \infty}\mu(x_n \cdot B)= \mu (B). $$ CASE 1: For bounded sets $B$, it should be true that the condition on the left-hand side implies $\lambda(x_n \cdot B \, \triangle \, B)\to 0$ and hence $\mu(x_n \cdot B \, \triangle \, B)\to 0$, since $\mu$ is assumed absolutely continuous (see Continuity for sets of asymptotically null Lebesgue measure); herein $A \, \triangle \, B$ denotes the symmetric set difference. Is this correct?
CASE 2: If $B$ is unbounded, we still have that $\mu$ is assumed finite, how can we proceed in this case?
EDIT 1 In both cases, if we have that, denoting $B_n:=x_n \cdot B$, $$ B=\limsup_{n\to \infty}B_n=\liminf_{n\to \infty}B_n=\lim_{n\to \infty} B_n, $$ then we can apply property (M12) on page 48 of Vestrup's book - The Theory of Measures and Integration, 2003, Wiley - and conclude that $\lim_{n\to \infty}\mu(B_n)=\mu(B)$.
EDIT 2 Resorting to the existence of $\lim_{n \to \infty}B_n$ is not necessary if $B$ is a closed set. Notice that $\mu(B_n \, \triangle \, B)\to 0 \implies \mu(B_n)\to \mu(B)$. Also, notice that we can always construct a monotone nonincreasing positive sequence $r_n$ such that $1>r_n \downarrow 0$ and, ultimately, $$ \max_{1\leq j \leq d}|x_{n,j}-1|<r_n, $$ see e.g. Limits and existence of decay rates. Consequently, defining $$ \bar{B}_n^{(1)}=\{x \in X: \, \exists x' \in B\, \text{ s.t. } x'_j(1-r_n)\leq x_j \leq x_j'(1+r_n)\}, $$ $$ \bar{B}_n^{(2)}=\{x \in X: \, \exists x' \in B^c\, \text{ s.t. } x'_j(1-r_n)< x_j < x_j'(1+r_n)\}, $$ we have $\bar{B}_{n+1}^{(j)} \subset \bar{B}_n^{(j)}$ with $j=1,2$. Ultimately $B_n \subset \bar{B}_n^{(1)}$ and $B_n^c= x_n \cdot B^c \subset \bar{B}_n^{(2)}$, since we can define $T_n(x)=(x_{n,1}^{-1}, \ldots, x_{n,d}^{-1}) \cdot x$, $\forall x \in X$, and deduce $$ B_n^c=(T_n^{-1}(B))^c=T_n^{-1}(B^c)=x_n \cdot B^c, $$ see also Preimage of the complement. Moreover, $\lim_{n\to \infty}\bar{B}_n^{(1)}=B$ and $\lim_{n\to \infty}\bar{B}_n^{(2)}=B^c$, thus $$ \lim_{n\to \infty}\bar{B}_n^{(1)}\cap B^c=\emptyset, \quad \lim_{n\to \infty}\bar{B}_n^{(2)}\cap B=\emptyset. $$ Ultimately, $$ \mu(B_n \cap B^c)\leq \mu(\bar{B}_n^{(1)}\cap B^c) \to 0, \quad \mu(B_n^c \cap B)\leq \mu(\bar{B}_n^{(2)}\cap B) \to 0. $$ It now follows that $\mu(B_n \, \triangle \, B)\to 0$. Is this reasoning correct?
