What does (Conditional probability = 1) imply

Question

What does $p(A|B)=1$ (or equivalently $p(A∩B)=p(B)$) imply?

I guess it leads to $p(B⊆A)=1$ but I couldn't prove it. Is it right? If not, give a counter example.

Answer 1

The expression $$ p(B⊆A) $$ does not make sense, so you can't prove it.

Probability is a function whose domain consists of subsets of the sample space, extended to the expressions that define conditional probability, not relations between those subsets. So all you can say is the equivalent statement $$ p(A∩B)=p(B) \ . $$

The idea you are trying to express is correct. Knowing that an event in $B$ occurred, you know that an event in $A$ occurred. That means that $B$ must be a subset of $A$, up to probability. In a finite sample space, it must be an actual subset (assuming individual elements have nonzero probability). In the continuous case it can happen that $A - A\cap B$ is nonempty but has probability $0$.

Answer 2

We have $$1=\Pr(A|B)=\frac{\Pr(A\cap B}{\Pr(A\cap B)+\Pr(B\setminus A)}$$ so that $\Pr(B\setminus A)=0$.

As a concrete example, let the probability space be $[0,1]$ with the uniform distribution, and let $A=(0,1],\ B=[0,1)$.