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 $[0,\frac{1}{2}]$?

$\mathcal{G}=\mathcal{B}\left[0,\frac{1}{2}\right]\cup\left\{ B\cup\left(\frac{1}{2},1\right]\mid B\in\mathcal{B}\left[0,\frac{1}{2}\right]\right\}$.

$Y:\Omega\to\mathbb{R}$ 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: $Y=X1_{\left[0,\frac{1}{2}\right]}+2\left[\mathbb{E}X1_{\left(\frac{1}{2},1\right]}\right]1_{\left(\frac{1}{2},1\right]}$.

• Merci! Now, I am wondering how I can work from this prescribing mappings $Y$ towards the solution: $\mathbb{E}[X|\mathcal{G}]=X\mathbb{1}_{[0,\frac{1}{2}]} + 2 \mathbb{E}[\mathbb{1}_{(\frac{1}{2},1]}X]\mathbb{1}_{(\frac{1}{2},1]}$? Why is this factor 2 before the second expectation?
– iJup
Commented Feb 20, 2016 at 14:41
• I made a mistake (and repaired). Sorry for the confusion I caused. Commented Feb 20, 2016 at 21:23
• Why is Y measurable w.r.t. G if it B[0,1]-measurable and constant on (0.5,1]? Commented Feb 21, 2016 at 20:32
• @RoosJansen If $Y$ is $\mathcal B[0,1]$-measurable then for each $d$ we have $\{Y<d\}\in\mathcal B[0,1]$ so that $B:=\{Y<d\}\cap[0,\frac12]\in\mathcal B[0,\frac12]$. If secondly $Y$ is constant on $(\frac12,1]$ and takes value $c$ there, then $\{Y<d\}=B\cup(\frac12,1]$ if $c<d$ and $\{Y<d\}=B$ if $c\geq d$. In both cases $\{Y<d\}\in\mathcal G$. Commented Feb 21, 2016 at 20:52