Given a random vector $\mathbb{X}$ with joint density $f$, and a set $A=\{\mathbb{X}\in B\}$ with $B\in \mathscr{B}(\mathbb{R})$ Prove that:\ $$f_{(x|A)}=\frac{f(x)}{\mathbb{P}(A)}\text{ if } x\in B $$
I tried to prove this using the next theorem:\ Given X a random variable. Let $B\in\mathscr{F}$ a fixed event. Then there is a measurable function $g(x)$ such that\ $$\mathbb{P}(\{X\in A\}\cap B)=\int_{A}g(x)d\mathbb{P}_{x}\text{ For every }A\in\mathscr{B}(\mathbb{R})$$ And $g$ is denoted ad $\mathbb{P}(B|X=x)\\$ But i think that my proof is wrong... Let's consider some $D\in\mathscr{B}(\mathbb{R}^{n})$. Then, the event $A\cap\{X\in D\}$ is equivalent to the event $\{X\in (B\cap D)\}$ Then\ $$\mathbb{P}(A\cap \{X\in D\})=\int_{B\cap D}f(x)dx_{1},\dots,dx_{n}$$ $$=\int_{D}\frac{f(x)}{\mathbb{P}(A)}\mathbb{P}(A)1^{\mathbb{X}}_{B} dx_{1},\dots,dx_{n}$$ $$=\mathbb{P}(A)\int_{D}\frac{f(x)}{\mathbb{P}(A)}1^{\mathbb{X}}_{B} dx_{1},\dots,dx_{n}$$ Then it satisfies the theorem. I'm almost sure that this is wrong but I don't know how to prove it correctly...
$$F_X(x|A)=P(X\leq x|A)=E(1_{\{X\leq x\}}|A)\overset{(1)}{=}\frac{E(1_{\{X\leq x\}} 1_{\{A\}})}{P(A)}\overset{(2)}{=}\frac{P(\{X\leq x\}\cap A)}{P(A)}$$
If $x\in B$
$$=\frac{P(\{X\leq x\})}{P(A)}=\frac{F_X(x)}{P(A)}$$
(1) $E(Z|A)=\frac{E(Z1_A)}{p(A)}$ Conditional_expectation#Conditional_expectation_with_respect_to_an_event
(2) $1_B 1_A=1_{B\cap A}$