Conditional probability with respect to a sub-algebra.

Question

Consider the space $([0,1]\times[0,1], B, \lambda_2)$ where $\lambda_2$ is the Lebesgue measure on the unit square let $G$ be the sub-algebra of the set of the form $A \times [0,1]$ where $A$ is a Borelian set of [0,1]. Let D be the region of $[0,1] \times [0,1]$ enclosed by the curves $y=x^{\frac{1}{2}}$ and $y=x^2$. I want to calculate $P(D|G)$ but I'm really stuck here... Any hint or help would be amazing.

Answer 1

The function $\nu$ define on $G$ by, for any $A \times [0,1] \in G$, $ \nu(A \times [0,1]) = \lambda_2(D \cap (A \times [0,1]) )$, is a finite measure and $\nu \ll \lambda_2|_G.$ The conditional probability of $D$ with respect to $G$ is the Radon-Nykodym derivative $\frac{\textrm{d}\nu}{\textrm{d}\lambda_2|_G}$.

Let $f$ from $[0,1]^2 $ to $[0,+\infty)$ be defined by, for all $(x,y)\in [0,1]^2 $, $f(x,y)= \sqrt{x} - x^2$. We have that $f$ is $G$-measurable and, for any $A \times [0,1] \in G$: $$ \nu(A \times [0,1] )=\lambda_2(D \cap (A \times [0,1]) )= \int_{D \cap (A \times [0,1])} d \lambda_2 = \int_{A \times [0,1]} \chi_{[x^2, \sqrt{x}]}(y) d \lambda_2(x,y)= \\ =\int_{A} \left ( \int_{[0,1]}\chi_{[x^2, \sqrt{x}]}(y) d\lambda(y) \right ) d\lambda(x)= \int_{A} (\sqrt{x}-x^2) d\lambda(x)=\int_{A \times [0,1]} f(x,y)\lambda_2(x,y)$$

where $\lambda$ is the Lebesgue measure on $[0,1]$. Since the Radon-Nykodym derivative is unique a.e.$[G]$ (with respect to G), we have that

$$\frac{\textrm{d}\nu}{\textrm{d}\lambda_2|_G}(x,y)= f(x,y) \;\;\;\;\textrm{ a.e.}[G]$$.