I really have some problems in proving the following result.
Let $X,Y$ be two integrable random variables defined in $(\Omega, F, \Bbb{P})$. Assume that $X$ and $Y$ are independent. Then show that for any $f,g:\Bbb{R}\rightarrow \Bbb{R}$ bounded we have $$\Bbb{E}(f(X)g(Y)|X)=f(X)\Bbb{E}(g(Y))$$
I am really confused because I don't see where to start. So let me tell you what I have. In my opinion we work here in a continuous situation, i.e. we do not have the formulas for discrete conditioning. Therefore I have the following definition/theorem for the conditioning:
Let $B\subset F$ be a sub-$\sigma$-algebra. Let $X\in L^1$ be a random variable. Then there exists a unique random variable $\xi\in L^1$ such that $\xi$ is $B$-measurable and for all $B$-measurable and bounded random variables $Z$ we have $$\Bbb{E}(XZ)=\Bbb{E}(\xi Z)~~~~~~~~~~~~(1)$$Then $\xi$ is called the conditional expectation of $X$ given $B$ and we write $\xi=\Bbb{E}(X|B)$
But now in my opinion this would mean that it is enough to show that $\xi=f(X)\Bbb{E}(g(Y))$ satisfy $(1)$ or am I wrong?
But the problem is that I am not really comfortable with the whole topic and notations thus I wanted to ask if someone could help me.
Thanks a lot
Since $f(X),g(Y)$ are independent and $f(X)$ is $\sigma(X)$ measurable, it is enough to prove that if $X$ is (bounded, due to $f$ being bounded) $\mathcal G$ measurable and $Y$ is (bounded, due to $g$ being bounded) independent of $\mathcal G$, then $\mathbb E[XY|\mathcal G] = X\mathbb E[Y]$ (then, we use this fact with $\mathcal G = \sigma(X)$, $X \to f(X), Y \to g(Y)$)
We'll do it in two steps, firstly we'll prove that if $X$ is (bounded) $\mathcal G$ measurable, then $\mathbb E[XY|\mathcal G] = X \mathbb E[Y|\mathcal G]$, i.e. we can pull out $\mathcal G$ measurable stuff out of the conditional expectation (this is the part that Kurt. G refers to) and then we'll prove that if $Y$ is (bounded) independent of $\mathcal G$, then $\mathbb E[Y|\mathcal G] = \mathbb E[Y]$.
Take any bounded $\mathcal G$ measurable random variable $Z$. We want to show (1), i.e $\mathbb E[X YZ] = \mathbb E[\xi Z]$, where $\xi = X \mathbb E[Y|\mathcal G]$. But $$ \mathbb E[\xi Z] = \mathbb E[XZ \mathbb E[Y|\mathcal G]] = \mathbb E[XZ Y],$$ where at the last step we used (1) (with bounded, $\mathcal G$ measurable random variable $XZ$, and $\xi = \mathbb E[Y|\mathcal G]$. In other words, we used (1) not for $\mathbb E[XY|\mathcal G]$, but for $\mathbb E[Y|\mathcal G]$). This proves that $\mathbb E[XY|\mathcal G] = X\mathbb E[Y|\mathcal G]$ for bounded $X$ (note that assumption of boundedness of $X$ and $Y$ isn't necessary. It works for integrable $X,Y$ such that $XY \in L_1$ via some sort of dominated convergence theorem).
Next, we need to show that for $Y$ bounded and independent of $\mathcal G$, we have $\mathbb E[Y|\mathcal G] =\mathbb E[Y]$. We proceed similarly, take any bounded, $\mathcal G$ measurable random variable $Z$. We need to prove $\mathbb E[YZ] = \mathbb E[Z \mathbb E[Y]]= \mathbb E[Z]\mathbb E[Y]$. But since $Z$ is $\mathcal G$ measurable and $Y$ is independent of $\mathcal G$, then $Z,Y$ are independent, so in particular uncorrelated, hence $\mathbb E[YZ] = \mathbb E[Y]\mathbb E[Z]$.
Putting those facts together, we can conclude.