Solution: x and y are independent, that is to say, the probability of Z=XY event has nothing to do with the preconditions of x and y. In this problem, because the distribution of y is (0, 1), it is relatively simple. That is, P(X.0≤z 乸 Y=0)=P(0≤z), P(X. 1≤z 乸 Y= 1)=P(X≤z). In this way, Fz(z) is only related to the distribution of X. However, x∈(-∞, +∞) and Z have different cases of z≥0 and Z ≥ X. The results show that their distribution functions are different. Only when z=0, the left and right limits of Fz(z) are not equal, so it is a breakpoint. For reference.