# For independent $A,B$, is it possible to have "$B$ is a subset of $A^c$

One of my probability homework consists of the following theorem:

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. If $A,B \in \mathcal{F}$ are independent events, then $B \subset A^c.$

I have an objection that if this theorem is true , then $A \cap B= \emptyset$ since $B \subset A^c$ is equivalent that intersection is empty. But my earlier exposition tells me that if two events are independent then they can intersect. How do we go about this?

My objection is that the theorem above is wrong, since you can find independent events and still their intersection is not empty.

• I think that a better statement would be : if $A,B$ are independent, then $B\subset A^c\iff(\mathbb{P}(B)=0\text{ or }\mathbb{P}(A)=0)$ ? – Balloon Sep 24 '17 at 10:21
• so, how do you prove the 'if, then ' part? – Quantes Sep 24 '17 at 10:27
• Are you given any other properties of $A, B$ or is that it? – Graham Kemp Sep 24 '17 at 10:32
• @Quantes: What's your definition of $\,$"$A,B$ are independent"? – quasi Sep 24 '17 at 10:33
• Yes, of course -- it's blatantly wrong. – quasi Sep 24 '17 at 10:37

Let me rectify my first comment : if $A,B\in\mathcal{F}$ are independant and disjoint, then $(\mathbb{P}(A)=0$ or $\mathbb{P}(B)=0$), which can be easily stated since $$0=\mathbb{P}(\varnothing)=\mathbb{P}(A\cap B)=\mathbb{P}(A)\cdot\mathbb{P}(B).$$ Hence, any example of independent events with the probabilities of twe two being non-zero are a counter-example of the theorem.
Exemple : let $(\Omega,\mathcal{F},\mathbb{P})$ be $(\{1,\dots,6\},\mathcal{P}(\{1,\dots,6\}),\mathbb{P})$, where $\mathbb{P}$ is the uniform probability on $\{1,\dots,6\}$. Then for $A=\{1,2,3,4\}$ and $B=\{2,4,6\}$ we have $$\mathbb{P}(A)=\frac{2}{3},\mathbb{P}(B)=\frac{1}{2},\mathbb{P}(A\cap B)=\frac{1}{3} \text{ and }\mathbb{P}(A)\cdot\mathbb{P}(B)=\frac{1}{3},$$ which says that throwing a dice and get a number $\leq 4$ or even are two compatible and independent events of probability non-zero.
Why did I want to rectify my (false) comment : take $(\{1,2,3,4\},\mathcal{P}(\{1,2,3,4\}),\mathbb{P})$ with $\mathbb{P}(1)=\mathbb{P}(2)=0$, $\mathbb{P}(3)=\mathbb{P}(4)=\frac{1}{2}$. Then, with $A=\{1,2\}$ and $B=\{2,3\}$, we have that $\mathbb{P}(A\cap B)=\mathbb{P}(2)=0=\mathbb{P}(A)\cdot\mathbb{P}(B)$, so the two events are independent, $(\mathbb{P}(A)=0\text{ or }\mathbb{P}(B)=0)$ is checked since $\mathbb{P}(A)$ is zero, but $A\cap B=\{2\}\neq\varnothing$.