Does $f$ measurable iff there is a measurable $G$ s.t. $f=1_G$?

Question

Let $(X,\mathcal M,\mu)$ a measurable space. I know that if $M\in \mathcal M$, then $\boldsymbol 1_M:X\longrightarrow \mathbb R$ will be measurable. But if a function $f$ is measurable, will we have that $f=\boldsymbol 1_G$ for a $G\in \mathcal M$. I ask this question, because I want to show that $$\mathbb E[\alpha X+\beta Y\mid \mathcal G]=\alpha \mathbb E[X\mid \mathcal G]+\beta \mathbb E[Y\mid \mathcal G],$$ and to do this, my teacher took $G\in \mathcal G$ and he checked that $$\mathbb E\Big[\mathbb E[\alpha X+\beta Y\mid \mathcal G]\boldsymbol 1_G\Big]=\mathbb E\Big[(\alpha \mathbb E[X\mid \mathcal G]+\beta \mathbb E[Y\mid \mathcal G])\boldsymbol 1_G\Big],$$ where as in my definition, we want this result not for $\boldsymbol 1_G$ only but for all $\mathcal G-$measurable. But still it work ?

Answer 1

There are different definitions of conditional expectation, but your teacher's is the standard one. That is, if $\mathbb E|X|<\infty$, we have $Y=\mathbb E[X|\mathcal G]$ if and only if $$ \mathbb E[X\mathbf 1_G]=\mathbb E[Y\mathbf 1_G] $$ for all $G\in\mathcal G$. The existence and uniqueness (up to sets of measure zero) of this random variable is given by the Radon-Nikodym theorem. You may then show, by approximating by simple functions and using the monotone (or dominated) convergence theorem, that $$ \mathbb E[XZ]=\mathbb E[YZ] $$ for any bounded $\mathcal G$-measurable function $Z$. However if we know this is true for all $\mathcal G$-measurable $Z$, it clearly holds for the case where $Z=\mathbf1_G$, so it in fact could be taken as definition of conditional expectation.