[proofplan] We verify the probability measure axioms directly from the definition of conditional probability. First, $\mathbb P_B$ is well-defined and non-negative because $A \cap B \in \mathcal F$ and $\mathbb P(B)>0$. The total mass is computed by evaluating at $\Omega$. Finally, countable additivity follows by intersecting a disjoint family with $B$, using countable additivity of $\mathbb P$, and dividing by the fixed positive number $\mathbb P(B)$. [/proofplan] [step:Verify that $\mathbb P_B$ is well-defined and non-negative] Let $A \in \mathcal F$. Since $A \in \mathcal F$ and $B \in \mathcal F$, closure of the $\sigma$-algebra $\mathcal F$ under finite intersections gives $A \cap B \in \mathcal F$. Therefore $\mathbb P(A \cap B)$ is defined. Since $\mathbb P(B)>0$, the quotient \begin{align*} \mathbb P_B(A)=\frac{\mathbb P(A \cap B)}{\mathbb P(B)} \end{align*} is defined. Because $\mathbb P$ is non-negative on $\mathcal F$, we have $\mathbb P(A \cap B)\ge 0$. Dividing by the positive number $\mathbb P(B)$ gives \begin{align*} \mathbb P_B(A) \ge 0. \end{align*} Also, since $A \cap B \subset B$, the set $B \setminus A$ belongs to $\mathcal F$ and the union \begin{align*} B = (A \cap B) \cup (B \setminus A) \end{align*} is disjoint. Finite additivity of $\mathbb P$ gives \begin{align*} \mathbb P(B)=\mathbb P(A \cap B)+\mathbb P(B \setminus A) \ge \mathbb P(A \cap B), \end{align*} so \begin{align*} 0 \le \mathbb P_B(A) \le 1. \end{align*} Thus $\mathbb P_B:\mathcal F \to [0,1]$ is well-defined. [/step] [step:Compute the total mass of $\Omega$] Using $\Omega \cap B = B$, we obtain \begin{align*} \mathbb P_B(\Omega) = \frac{\mathbb P(\Omega \cap B)}{\mathbb P(B)} = \frac{\mathbb P(B)}{\mathbb P(B)} = 1. \end{align*} Thus $\mathbb P_B$ assigns total mass $1$ to $\Omega$. [/step] [step:Prove countable additivity on disjoint measurable families] Let $(A_n)_{n=1}^{\infty}$ be a sequence of pairwise disjoint sets in $\mathcal F$. Since $\mathcal F$ is closed under countable unions, the set $\bigcup_{n=1}^{\infty} A_n$ belongs to $\mathcal F$, so $\mathbb P_B\left(\bigcup_{n=1}^{\infty} A_n\right)$ is defined. Define \begin{align*} C_n := A_n \cap B \end{align*} for each positive integer $n$. Since $A_n,B \in \mathcal F$, each $C_n$ belongs to $\mathcal F$. The sets $(C_n)_{n=1}^{\infty}$ are pairwise disjoint: if $m \ne n$, then \begin{align*} C_m \cap C_n = (A_m \cap B) \cap (A_n \cap B) = (A_m \cap A_n) \cap B = \varnothing \cap B = \varnothing. \end{align*} Moreover, distributivity of intersection over countable unions gives \begin{align*} \left(\bigcup_{n=1}^{\infty} A_n\right)\cap B = \bigcup_{n=1}^{\infty} (A_n \cap B) = \bigcup_{n=1}^{\infty} C_n. \end{align*} Using the countable additivity axiom for the probability measure $\mathbb P$ on the pairwise disjoint family $(C_n)_{n=1}^{\infty}$, the series $\sum_{n=1}^{\infty}\mathbb P(C_n)$ converges to $\mathbb P\left(\bigcup_{n=1}^{\infty} C_n\right)$. Since $\mathbb P(B)>0$, multiplying this convergent non-negative series by the constant $1/\mathbb P(B)$ is valid. Hence \begin{align*} \mathbb P_B\left(\bigcup_{n=1}^{\infty} A_n\right) &= \frac{\mathbb P\left(\left(\bigcup_{n=1}^{\infty} A_n\right)\cap B\right)}{\mathbb P(B)} \\ &= \frac{\mathbb P\left(\bigcup_{n=1}^{\infty} C_n\right)}{\mathbb P(B)} \\ &= \frac{\sum_{n=1}^{\infty} \mathbb P(C_n)}{\mathbb P(B)} \\ &= \sum_{n=1}^{\infty} \frac{\mathbb P(A_n \cap B)}{\mathbb P(B)} \\ &= \sum_{n=1}^{\infty} \mathbb P_B(A_n). \end{align*} Therefore $\mathbb P_B$ is countably additive on $\mathcal F$. [/step] [step:Conclude that $\mathbb P_B$ is a probability measure] We have shown that $\mathbb P_B:\mathcal F \to [0,1]$ is well-defined, non-negative, satisfies $\mathbb P_B(\Omega)=1$, and is countably additive on pairwise disjoint countable families in $\mathcal F$. These are exactly the axioms for a probability measure on the measurable space $(\Omega,\mathcal F)$. Hence $(\Omega,\mathcal F,\mathbb P_B)$ is a probability space. [/step]