Bayes Theorem with change of measure

Question

Tomas bjork- arbitrage theory in continuous time. Appendix B, proposition B41 says:

The proof is not clear to me.

Thanks to Gordon's comment below of $E^Q (X/G)$ being $G$ measurable, I think the part where Bjork seems to imply that

$E^Q (X/G) . E^P (L/G) = E^P[(L.E^Q(X/G))/G]$

is valid since $E(x.y/\tau) = yE(x/\tau)$ if $y$ is $\tau$ measurable.

However in the next step, Bjork seems to say

$E^P[(L.E^Q(X/G))/G] = L.E^Q(X/G)$

Why would this be valid?

Moreover the RHS seems to imply

$E^P[(L.X)/G] = L.X$

Why is this valid?

Answer 1

Nothing new in this answer - I have just consolidated what others have said in the answer and the comments, and put the explanation next to each step. I have to move the original equations so as to have one equation on each line:

Answer 2

The last one is valid since it is a defining relation of conditional expectation. Ane also we have $\frac{dQ}{dP} = Z$ and it implies that $dP \thinspace Z = dQ$. And this is the last equation.

Let's consider the first equation: $\mathbb{E}^P[L \thinspace \mathbb{E}^Q(X | G) \thinspace | \thinspace G] = L \thinspace \mathbb{E}^Q(X | G)$

As it was said before, $\mathbb{E}^Q(X | G)$ is G-measurable, so we can take this expression before the whole conditional expectation and again we use defining relation of the conditional expectation $\int_{G} \mathbb{E}(L | G) dP = \int_{G} L dP$