Let $A\subset\mathbb{R}^n$ be a Borel measurable set of finite Lebesgue measure $\lambda (A)\in(0,\infty)$ and let $X$ be uniformly distributed on $A$. Let $B\subset A$ be measurable with $\lambda(B)>0$. Show that the conditional distribution of $X$ given $\{X\in B\}$ is the uniform distribution on $B$.
Let $A\in\mathscr{B}(\mathbb{R}^n)$ be a measurable set with the n dimensional Lebesgue measure. Then we can define a probability measure on $\mathscr{B}(\mathbb{R}^n)|_A$ by:
$\mu(B)=\frac{\lambda^n(B)}{\lambda^n(A)}\:\:\forall B\in\mathscr{B}(\mathbb{R}^n)\text{such that} B\subset A$.
The measure $\mu$ is the uniform distribution on $A$.
I was thinking of the following definition of conditional distribution $f_{Y|X}=\frac{f(x,y)}{f_X(x)}$
I think $X=(X_1,X_2,...X_n)$, right? However I have no idea on how to deal with this problem? If I apply the definition above I would not be able to compute the conditional distribution.
Question:
How should I prove this assertion?
Thanks in advance!
It is given that $P(X^{-1} (E))=\frac {\lambda (E)} {\lambda (A)}$ for Borel sets $E \subset A$. So $P(X^{-1} (E) |X^{-1} (B))=\frac {P(X^{-1} (E\cap B)} {P(X^{-1} (B))}=\frac {\lambda (E \cap B)} {\lambda (A)} /(\frac {\lambda (B)} {\lambda (A)})$. This is same as $\frac {\lambda (E\cap B)} {\lambda (B)}$. So the conditional distribution is uniform on $B$.