Let C be some arbitrary subset of X disjoint from A, and pick some arbitrary subset D of A that is in M. Then if we call B the union of C and D, this statement would imply B is in M. We can then intersect B with the complement of D, and find that C is in M. For a counterexample, then, just pick two disjoint subsets of a space, one in the algebra, one out.

Let [;(X, \mathfrak{M});] be a measurable space, [;A,B \subset X  ;]. Does [;A,A \cap B \in  \mathfrak{M}  ;] imply that [;B \in \mathfrak{M}  ;]? I'm having a tough time with this. My intuition is telling me this isn't true in general, but I can't think of a counterexample. If anyone has any advice, whether it be the answer, a counterexample, or a hint, all is welcome!

[Measure Theory/Sigma Algebras] Quick inclusion question