In fact, it is equal to a, which is really included in B.

When A is really contained in B, A-B is an empty set with a probability of 0.

But the reverse is not necessarily true, even if the probability is 0, it is not necessarily an empty set.

For example, the probability of a point on a uniform distribution is 0.

A counterexample can be constructed as follows

Complete works U={x∈}, evenly distributed, A={x∈}, B={x belongs to (1, 3)}.

So A-B = {1 }≦? , P(A-B)=0, but A is not included in B.