Conditional Expectation on similar sigma algebras

Question

I'm trying to prove the following, (or to find a counterexample): Let on a probability space. $Y$ be a Bernoulli variable, $X\in L^1$ be another random variable, let $\mathcal{G}$ be some sub-sigma-algebra. If we set $\mathcal{F}:=\sigma(\mathcal{G},\sigma(Y))$ does it hold true that, $$\mathbb{E}[Y\cdot X|\mathcal{F}]=Y\cdot \mathbb{E}[Y\cdot X|\mathcal{G}]$$

Answer 1

No.

Let $A$ and $B$ denote two events such that $B\subset A$, $\mathcal G=\{\varnothing,\Omega\}$, $Y=\mathbf 1_A$ and $X=\mathbf 1_B$. Then the LHS is $E[\mathbf 1_B\mid\sigma(A)]$ and the RHS is $P[B]\cdot\mathbf 1_A$.

But $E[\mathbf 1_B\mid\sigma(A)]=b\mathbf 1_A$ for some $b$ which can be identified by integrating this once again, since $P[B]=E[E[\mathbf 1_B\mid\sigma(A)]]=E[b\mathbf 1_A]=bP[A]$, that is, $b=P[B]/P[A]$.

If $P[A]\ne1$ and $P[B]\ne0$, then $b\ne P[B]$ hence the LHS and the RHS are not always equal.

Which reasoning led you to suspect they might coincide?