[proofplan] Disjointness forces the probability of the intersection $A\cap B$ to be zero. Independence would force that same probability to equal the product $\mathbb P(A)\mathbb P(B)$. Since both factors are strictly positive, the product is strictly positive, giving a contradiction. [/proofplan] [step:Compare the zero intersection probability with the positive product probability] Let $\varnothing$ denote the empty event in $\mathcal F$. Since $A$ and $B$ are disjoint, we have \begin{align*} A\cap B=\varnothing. \end{align*} Because $\mathbb P$ is a probability measure on $(\Omega,\mathcal F)$, the empty event has probability zero, so \begin{align*} \mathbb P(A\cap B)=\mathbb P(\varnothing)=0. \end{align*} On the other hand, the hypotheses give $\mathbb P(A)>0$ and $\mathbb P(B)>0$, hence \begin{align*} \mathbb P(A)\mathbb P(B)>0. \end{align*} Therefore \begin{align*} \mathbb P(A\cap B)=0<\mathbb P(A)\mathbb P(B), \end{align*} so $\mathbb P(A\cap B)\neq \mathbb P(A)\mathbb P(B)$. By the definition of [independence of events](/page/Independence), this means that $A$ and $B$ are not independent. [guided] Let $\varnothing$ denote the empty event in $\mathcal F$. The point of the argument is that the same quantity, $\mathbb P(A\cap B)$, is forced to have two incompatible values if $A$ and $B$ were independent. First use disjointness. Since $A$ and $B$ are disjoint events, their intersection is the empty event: \begin{align*} A\cap B=\varnothing. \end{align*} Because $\mathbb P$ is a probability measure on the measurable space $(\Omega,\mathcal F)$, it assigns probability zero to the empty event. Therefore \begin{align*} \mathbb P(A\cap B)=\mathbb P(\varnothing)=0. \end{align*} Now compare this with what independence would require. By the definition of [independence of events](/page/Independence), the events $A$ and $B$ are independent exactly when \begin{align*} \mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B). \end{align*} But the hypotheses say that both $\mathbb P(A)$ and $\mathbb P(B)$ are strictly positive real numbers. The product of two strictly positive real numbers is strictly positive, so \begin{align*} \mathbb P(A)\mathbb P(B)>0. \end{align*} Combining the two computations gives \begin{align*} \mathbb P(A\cap B)=0<\mathbb P(A)\mathbb P(B). \end{align*} Thus \begin{align*} \mathbb P(A\cap B)\neq \mathbb P(A)\mathbb P(B). \end{align*} By the definition of independence of events, $A$ and $B$ are not independent. [/guided] [/step]