Conditional expectation on sigma algebra containing Borel subsets Let $\Omega=[0,1]$, $\mathcal{F}=\mathcal{B}([0,1])$ and $\mathbb{P}$ the Lebesgue measure on $[0,1]$. Let $\mathcal{G}$ be the smallest $\sigma$-algebra on $[0,1]$ containing the Borel subsets of $[0,\frac{1}{2}]$.
Now, assuming that $X\in L^1$, I want to compute $\mathbb{E}(X|\mathcal{G})$. 
For me it is not clear why we want to split the interval $[0,1]$ into $[0,\frac{1}{2}]$ and $(\frac{1}{2},1]$? How is this related to $\mathcal{G}$ containing the Borel subsets of $[0,\frac{1}{2}]$?
 A: $\mathcal{G}=\mathcal{B}\left[0,\frac{1}{2}\right]\cup\left\{ B\cup\left(\frac{1}{2},1\right]\mid B\in\mathcal{B}\left[0,\frac{1}{2}\right]\right\} $. 
$Y:\Omega\to\mathbb{R}$ is measurable with respect to $\mathcal{G}$
if it is $\mathcal{B}\left[0,1\right]$-measurable and is
constant on $\left(\frac{1}{2},1\right]$.
To achieve $Y=\mathbb{E}\left(X\mid\mathcal{G}\right)$ we need $\int_{A}X\left(\omega\right)\mathbb{P}\left(d\omega\right)=\int_{A}Y\left(\omega\right)\mathbb{P}\left(d\omega\right)$
for each $A\in\mathcal{G}$.
Prescribing $Y$ by $\omega\mapsto X\left(\omega\right)$ if $\omega\in\left[0,\frac{1}{2}\right]$
yields this for $A\in\mathcal{B}\left[0,\frac{1}{2}\right]$.
Let it be that $Y\left(\omega\right)=c$ for $\omega\in\left(\frac{1}{2},1\right]$.
Then we need: $$\int_{\left(\frac{1}{2},1\right]}X\left(\omega\right)\mathbb{P}\left(d\omega\right)=\int_{\left(\frac{1}{2},1\right]}Y\left(\omega\right)\mathbb{P}\left(d\omega\right)=\frac{1}{2}c$$
showing that $c=2\mathbb{E}X1_{\left(\frac{1}{2},1\right]}$
This together results in: $Y=X1_{\left[0,\frac{1}{2}\right]}+2\left[\mathbb{E}X1_{\left(\frac{1}{2},1\right]}\right]1_{\left(\frac{1}{2},1\right]}$.
