Show that a $\sigma$-algebra is independent of itself iff $\mathbb
P(A)\in\{0,1\}$ for each set in the $\sigma$-algebra.
My try:
For the first direction, suppose that the $\sigma$-algebra is independent of itself. Then
$$\mathbb P(A_j\cap A_k)=\mathbb P(A_j)\cdot \mathbb P(A_k)$$
for each $j\neq k$. (Is this what is meant by independent of itself?) Since a $\sigma$-algebra is closed under complementation, then it must be the case that
$$\mathbb P(A_j\cap A_j^c)=\mathbb P(A_j)\cdot \mathbb P(A_j^c)$$
for each $j$. But since $A_j\cap A_j^c=\emptyset$ and $\mathbb P(\emptyset)=0$ then it must be the case that $\mathbb P(A_j)\in\{0,1\}\Leftrightarrow \mathbb P(A_j^c)\in\{0,1\}$.
For the other direction, suppose $\mathbb P(A_j)\in\{0,1\}$ for each $j$. Then
$$\mathbb P(A_j)\cdot \mathbb P(A_k)\in\{0,1\}$$
for all $j\neq k$. However, it's not clear to me why this would imply
$$\mathbb P(A_j\cap A_k)=\mathbb P(A_j)\cdot \mathbb P(A_k)$$
Am I on the right track?
Best Answer
Not an axiomatic proof, but I hope it helps for the only if part.
Using the property that $A \subset B \implies P(A) \leq P(B)$, we get the conditions $$\left\{ \begin{array}{ll} A_j \cap A_k \subset A_j \implies P(A_j\cap A_k) \leq P(A_j)\\ A_j \cap A_k \subset A_k \implies P(A_j\cap A_k) \leq P(A_k) \end{array} \right.$$ Then if either $P(A_j) = 0$ or $P(A_k)=0$, $P(A_j\cap A_k) \in \{0,1\} \implies P(A_j\cap A_k)=0$. On the other hand, if both $P(A_j) = 1$ and $P(A_k) = 1$, then $\underbrace{P(A_j\cup A_k)}_{=1} = \underbrace{P(A_j)}_{=1}+\underbrace{P(A_k)}_{=1}-P(A_j \cap A_k) \implies P(A)=1$. So in any case, $P(A_j\cap A_k) = P(A_j)\cdot P(A_k)$.