Suppose that $Z\triangleq \frac{dQ}{dP},X \in L^2_{\mathbb{P}}(\Omega;\mathcal{F})$ and $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$.
How can I prove that: $E_P [ Z X| \mathcal{G}]= E_{Q} [X | \mathcal{G}]E_P [ Z | \mathcal{G}]$, only using $L^2_{\mathbb{P}}(\Omega;\mathcal{F})$ arguments?
Note: The $L^1_{\mathbb{P}}(\Omega;\mathcal{F})$-proof is easy to find and can be found for example here.However, I don't want to prove it in $L^1_{\mathbb{P}}(\Omega;\mathcal{F})$ and restrict to $L^2_{\mathbb{P}}(\Omega;\mathcal{F})$, but I'm really looking for a direct argument. I was thinking of some sort of Cauchy-Schwartz type of thing?