Explicit computation of a conditional expectation 
Let $\Omega=[0,1], \mathcal{F}=\mathcal{B}([0,1]), \mathbb{P}=\lambda$ on $[0,1]$. Let $\mathcal{G}$ be the smallest $\sigma$-algebra containing the Borel subsets of $[0,\frac{1}{2}]$. Compute for $X\in L^1$ the conditional expectation $\mathbb{E}[X|\mathcal{G}]$.

My attempt: we can 'take out what is known', so it makes sense to rewrite $X=X\chi_{[0,\frac{1}{2}]}+X\chi_{(\frac{1}{2},1]}$. Then, using linearity and the 'taking out what is known'-property, we have $\mathbb{E}[X|\mathcal{G}]=X\chi_{[0,\frac{1}{2}]}+\mathbb{E}[X\chi_{(\frac{1}{2},1]}|\mathcal{G}]$. The answer file, however, says 
$$\mathbb{E}[X\chi_{(\frac{1}{2},1]}|\mathcal{G}]=2\mathbb{E}[X\chi_{(\frac{1}{2},1]}]\chi_{(\frac{1}{2},1]}$$
Could anyone explain me how to derive the last equality?
EDIT:
After a 2nd thought I got: $$\mathbb{E}[X\chi_{(\frac{1}{2},1]}|\mathcal{G}]=\frac{1}{\lambda(\frac{1}{2},1]} \int_{(\frac{1}{2},1]} X d\mathbb{\lambda}=2\mathbb{E}[\chi_{(\frac{1}{2},1]}X]$$
Which I guess kinda solves the problem, yet I don't understand why we should multiply it with $\chi_{(\frac{1}{2},1]}$

Recall that $\mathbb E(X|\mathcal G)=Y$ a.s., where $Y$ is a r.v. on $(\Omega, \mathcal G)$. You find that for $\omega\in[0,\frac12]$ 
$$Y(\omega)=X(\omega)\text{ a.s.}$$
And the second summand is the (constant) value of $Y$ for $\omega\in(\frac12,1]$: 
$$Y(\omega) \stackrel{a.s.}{=} 2\mathbb E[X\chi_{(\frac12,1]}]:=C $$
So, 
$$\mathbb E(X|\mathcal G) \neq X(\omega)\chi_{[0,\frac12]}+C,$$
but
$$\mathbb E(X|\mathcal G) \stackrel{a.s.}{=}\begin{cases} X(\omega), & \omega \in [0,\frac12]\cr C, & \omega\in (\frac12,1]\end{cases} = X\chi_{[0,\frac12]} + C\chi_{(\frac{1}{2},1]}$$
