I want to check my solutions for this problem:
Let $F$ be a $\sigma$-algebra over $\Omega$, and $Q: F \rightarrow [0, 1]$ be a normalized, additive set function (i.e., $Q(\Omega) = 1$ and $Q(A \cup B) = Q(A) + Q(B)$ for all disjoint $A, B \in F$).
Prove: (a) If for any $A_n \in F$ with $A_n \uparrow A$ (i.e., $A_n \subseteq A_{n+1}$ for all $n$ and $A = \cup_{n>1} A_n$), then $\lim_{n \to \infty} Q(A_n) = Q(A)$, then the same holds for any $B_n \in F$ with $B_n \downarrow B$ (i.e., $B_n \supseteq B_{n+1}$ for all $n$ and $B = \cap_{n>1} B_n)$, and $\lim_{n \to \infty} Q(B_n) = Q(B)$. In this case, $Q$ is called $\sigma$-continuous.
(b) $Q$ is a probability measure (i.e., $\sigma$-additive) if and only if $Q$ is $\sigma$-continuous.
My solutions:
(a) To prove that if $\lim_{n \to \infty} Q(A_n) = Q(A)$ for any increasing sequence $A_n \in F$ with $A_n \subseteq A_{n+1}$ for all $n$ and $A = \cup_{n>1} A_n$, then the same holds for any decreasing sequence $B_n \in F$ with $B_n \supseteq B_{n+1}$ for all $n$ and $B = \cap_{n>1} B_n$, and $\lim_{n \to \infty} Q(B_n) = Q(B)$, we can use the property of complements.
Let $C_n = \Omega \setminus B_n$. Then $C_n$ forms an increasing sequence, $C_n \subseteq C_{n+1}$ for all $n$, and $\lim_{n \to \infty} C_n = \Omega \setminus B$. Since $Q(\Omega) = 1$ and $Q$ is additive, we have $Q(\Omega \setminus B) = Q(\lim_{n \to \infty} C_n) = \lim_{n \to \infty} Q(C_n)$. Now, using the given assumption that $\lim_{n \to \infty} Q(A_n) = Q(A)$, we can write $Q(C_n) = 1 - Q(B_n)$. Taking the limit on both sides, we get $\lim_{n \to \infty} (1 - Q(B_n)) = 1 - Q(B)$, which implies $\lim_{n \to \infty} Q(B_n) = Q(B)$. Therefore, if $\lim_{n \to \infty} Q(A_n) = Q(A)$ for increasing sequences, then the same holds for decreasing sequences.
(b) To prove that $Q$ is a probability measure if and only if $Q$ is $\sigma$-continuous, we need to show both directions.
First, assume that $Q$ is a probability measure. We want to show that $\lim_{n \to \infty} Q(A_n) = Q(A)$ for any increasing sequence $A_n \in F$ with $A_n \subseteq A_{n+1}$ for all $n$ and $A = \cup_{n>1} A_n$. Since $Q$ is a probability measure, it satisfies $\lim_{n \to \infty} Q(A_n) = Q(\cup_{n>1} A_n)$. But by assumption, $Q(A) = Q(\cup_{n>1} A_n)$. Therefore, $\lim_{n \to \infty} Q(A_n) = Q(A)$, and $Q$ is $\sigma$-continuous.
Next, assume that $Q$ is $\sigma$-continuous. We want to show that $Q$ is a probability measure. Since $Q$ is $\sigma$-continuous, it satisfies $\lim_{n \to \infty} Q(A_n) = Q(A)$ for any increasing sequence $A_n \in F$ with $A_n \subseteq A_{n+1}$ for all $n$ and $A = \cup_{n>1} A_n$. Taking $A_n = \emptyset$ (the empty set) for all $n$, we have $\lim_{n \to \infty} Q(\emptyset) = Q(\cup_{n>1} \emptyset)$. Since $\cup_{n>1} \emptyset = \emptyset$, we have $\lim_{n \to \infty} Q(\emptyset) = Q(\emptyset)$. Since $Q(\Omega) = 1$, we must have $Q(\emptyset) = 0$. Therefore, $Q$ satisfies the property $Q(\emptyset) = 0$, which is one of the axioms for a probability measure. Additionally, $Q(\Omega) = 1$ is already given. Hence, $Q$ is a probability measure.
In conclusion, $Q$ is a probability measure if and only if $Q$ is $\sigma$-continuous.
Am I right or have I done any mistake? Thanks for the help