Results about conditional expectations

Question

$\theta$, $\phi$ are integrable random variables on a probability space $(\Omega,\mathcal{F},P)$ and $\mathcal{G}$ is $\sigma$-field on $\Omega$ contained in $\mathcal{F}$. Now we want to prove $E(\theta\mid\mathcal{G})=E(\theta)$ if $\theta$ is independent of $\mathcal{G}$. The proof is, for any $B\in \mathcal{G}$, by independence, $\theta$ and $1_{B}$ are independent. And,$$\int_{B}E(\theta)dP=E(\theta)E(1_{B})=E(\theta 1_{B})=\int_{B}\theta dP,$$ and the conclusion follows. I'm really confused with the first equality $\int_{B}E(\theta)dP=E(\theta)E(1_{B})$. Can anyone explain this to me? Thanks!

Answer 1

$$\int_{B}E[\theta]dP=E[E[\theta]I_{B}]=E[I_{B}]E[\theta]$$,that's all.

Answer 2

For every real number $a$, one has: $\int\limits_Ba\mathrm dP=a\int\limits_B\mathrm dP=aP[B]$. Furthermore, $P[B]=E[I_B]$. Apply these to $a=E[\theta]$.