Does “every element of a set $A$ is $m^*$-measurable” imply that “every element of the $\sigma$-algebra generated by $A$ is $m^*$-measurable”?

Question

Let $A$ is a subset of the ground set $X$ such that every element in $A$ is $m^*$-measurable. Then, can I say that every element of the $\sigma$-algebra $B$, generated by $A$, is $m^*$-measurable? Here, by “a set $A$ is $m^*$-measurable” I mean that, for all $E$ in $X$, $m^*(E) = m^*(E \cap A) + m^*(E - A)$.