How I can show that $(\Omega,\mathbb{F},P_{B})$ is a probability space given that $(\Omega,\mathbb{F},P)$ is a probability space and $B \in \mathbb{F}$ with $P(B)>0.$ setting $P_{B} : \mathbb{F} \to [ 0,\infty): P_{B} = \frac{P(A\cap B)}{P(B)}$ ?
And also that $(B,\mathbb{F}_{B},P_{B})$ is a probability space with $\mathbb{F}_{B} = \{ A \subset B : A \in \mathbb{F} \}$
My effort so far is that :
The restriction that $B \in \mathbb{F}$ is defined as: $$ \mathbb{F}_{B} := \{ A \cap B \mid A \in \mathbb{F} \} = \{ A \in \mathbb{F} \mid A \subset B \} $$
And also we know that $P(B)>0.$
The probability $P_{B}$ in the measurable space $(B,\mathbb{F}_{B} ,P_{B})$ is the restriction of this probability measure in the $\sigma$-algebra $\mathbb{F}_{B}$.
This restriction simply ignores all the sets $A$ for which $A \cap B = \oslash$ provided that is specified in the same $\sigma$-algebra $\mathbb{F}_{B}$.
Of course these both restrictions give us the same values in the sets $A \in \mathbb{F}$ which are subsets of $B.$
According to the definition of conditional probability we obtain: $$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(\oslash)}{P(B)}= 0 $$ because $\mu(A \cap B)$ for every $A \in \mathbb{F}$ assign the value $0$ for $A \in \mathbb{F}$.
And because if we take the sequence $A_{i} \in \mathbb{F},i \in \mathbb{N}$ which are pairwise disjoint we know: $$ \mu_{B} ( \cup_{i=1}^{\infty} A_{i}) = \sum_{i=1}^{\infty} \frac{\mu(A_{i} \cup B )}{ \mu(B)} = \sum_{i=1}^{\infty} \mu(A_{i} \mid B) = \sum_{i=1}^{\infty} \mu_{B}(A_{i}) $$
Dividing with $\mu(B)$ both the measures in the question are probability measures and according to the previous we have:
$$ P( \bigcup_{n=1}^{\infty} A_{i}) = \sum_{n=1}^\infty P(A_{i}) $$ the countable additivity,which also implies for $B_{i}$ for every $B_{i} \in \mathbb{F}_{B},i \in \mathbb{N}.$