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Let $(\Omega,\mathcal F,\mathbb P)$ a probability space and $\mathcal G\subset \mathcal F$ a subfield. To prove that

$$\mathbb E[\alpha X+\beta Y\mid \mathcal G]=\alpha \mathbb E[X\mid \mathcal G]+\beta \mathbb E[Y\mid \mathcal G]\ \ a.s.$$

I proves that $$\forall G\in \mathcal G, \ \ \mathbb E\Big[\mathbb E[\alpha X+\beta Y\mid \mathcal G]\boldsymbol 1_G\Big]=\mathbb E\Big[\left(\alpha \mathbb E[X\mid \mathcal G]+\beta \mathbb E[Y\mid \mathcal G]\right)\boldsymbol 1_G\Big],$$ but why is it enough to conclude that $\mathbb E[\alpha X+\beta Y\mid \mathcal G]=\alpha \mathbb E[X\mid \mathcal G]+\beta \mathbb E[Y\mid \mathcal G]$ a.s. ?

What I thought is that if for all $G\in \mathcal G$, $$\mathbb E[X\boldsymbol 1_G]=\mathbb E[Y\boldsymbol 1_G]\implies \mathbb E[(X-Y)\boldsymbol 1_G]=0\implies (X-Y)=0\ a.s.\implies X=Y\ \ a.s.$$ but as I could ask before here, it looks wrong. Any explanation ?

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    [Here](http://math.stackexchange.com/questions/2095990/if-mathbb-ex-boldsymbol-1-g-0-for-all-g-in-mathcal-g-does-x-0) – 2017-01-13
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    Remember that $\mathbb{E}[(X-Y)\mathbb{1}_G]$ doesn't mean that $X-Y = 0$. – 2017-01-13

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From $\def\E{\mathbf E}\E[A\def\G{\mathcal G}1_G] = \E[B1_G]$ for all $G\in \G$, for $\G$-measurable (that is the difference to your other question!) $A,B$ you can infer $$ A-B = 0 \def\P{\mathbf P}\qquad\text{$\P|_\G$-a. s.} $$ by choosing for $G$ the sets $\{A-B > \frac 1n\}$ and $\{B-A < -\frac 1n\}$, which are elements of $\G$ due to the $\G$-measurability of $A$ and $B$.\par Now apply this for the $\G$-measurable functions $A := \E[\alpha X + \beta Y\mid \G]$ and $B := \alpha\E[X\mid \G]+\beta\E[Y\mid \G]$.