Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. $X$ a random variable and $B\in \mathcal F$. How can I prove that $$\mathbb E[X|\mathbb 1_B]=\frac{1}{\mathbb P(B)}\mathbb E[X\mathbb 1_B] $$ using the definition ?
Attempts
I know that $$\mathbb E[X\mathbb 1_B]=\int_{\mathbb R}x\mathbb P(X\in dx, Y=1)$$ $$=\int_{\mathbb R}x\frac{\mathbb P(X\in dx|Y=1)}{\mathbb P(Y=1)}\mathbb P(Y=1)=\mathbb P(B)\mathbb E[X\mid \mathbb 1_B].$$
But can I prove it using $$\mathbb E[X\mathbb 1_F]=\mathbb E[\mathbb E[X|\mathbb 1_B]\mathbb 1_F]$$ for all $F\in \sigma (\mathbb 1_B)=\{\Omega ,B,B^c,\emptyset\}$ ?