For the following : $\Omega = [0,1)$, $I_{k,n} = [k2^{-n},(k+1)2^{-n}]$ for $k=0$ to $k=2^n-1$, $F_n = \sigma(I_{k,n}, 0\leq k < 2^n)$
Let f be an integrable function on $\Omega$. I have already shown that
$g_n = \mathbb{E}[f|F_n] = \sum_{k=0}^{2^n-1} 2^n\mathbb{E}[f\mathbb{1}_{I_{k,n}}]\mathbb{1}_{I_{k,n}} $
Thus $g_n$ is a step function and I want to prove that it converges in $L_1$ toward f. Unfortunately I have some problems to finish the calculus.
$\mathbb{E} [ |\mathbb{E}[f|F_n] - f| ] = \int_0^1 |\sum_{k=0}^{2^n-1} 2^n\mathbb{E}[f\mathbb{1}_{I_{k,n}}]\mathbb{1}_{I_{k,n}}(x) - f(x)|dx $
$= \int_0^1 |\sum_{k=0}^{2^n-1} 2^n\int_{I_{k,n}}f(y)dy\mathbb{1}_{I_{k,n}}(x) - f(x)|dx$
After some steps I find
$\mathbb{E} [ |\mathbb{E}[f|F_n] - f| ] \leq \sum_{k=0}^{2^n-1} 2^n \int_{I_{k,n}^2} |f(x)-f(y)|dydx $ but I can't find a way to show that this term goes to zero as n goes to infinity. Maybe I have been to brutal on the majoration. Any idea ?