Let $\mathcal{U}$ denote the set of all open balls with rational centre and rational radius that are contained in $A_j$ for some $j$. This is a countable family with the same union as the original family $\{A_j : j\in J\}$. Now for $U\in\mathcal{U}$ we have $U\cap B\subseteq A_j\cap B$ for some $j$ so $U\cap B$ has measure zero. As $\mathcal{U}$ is countable and covers $B$ we conclude that $B$ has measure zero.

Let $\mathcal{U}$ denote the set of all open balls with rational centre and rational radius that are contained in $A_j$ for some $j$. This is a countable family with the same union as the original family $\{A_j : j\in J\}$. Now for $U\in\mathcal{U}$ we have $U\cap B\subseteq A_j\cap B$ for some $j$ so $U\cap B$ has measure zero. As $\mathcal{U}$ is countable and covers $B$ we conclude that $B$ has measure zero.

UPDATE: The answer above is correct if the sets $A_j$ are open, as in an earlier version of the question. The current version seems to be consistent with the following example in which $B$ does not have measure zero:

• $n=m=1$
• $B=[0,\infty)$
• $A_t=(-\infty,0]\cup\{t\}$ for $t>0$