Conditional expectation as a random variable new We  consider  the  probability  space  $([0,1),  F  ,  \lambda  )$  , where  $F$  denotes  the  Borel-$\sigma$-field  on  $[0,1)$  and  $\lambda$  is  the  Lebesgue-measure.  We  define  for  each  $n\in \mathbb N$  :
$$F_n := \sigma \left({\left[\frac{k}{2^n} , \frac{k+1}{2^n}\right) : 0\le k \le 2^n-1}\right)$$
Let  $X : [0,1) \to \mathbb R$  be  an  integrable  random  variable. 
(a)  Calculate  the  random  variable  $E[X|F_n]$  for  all  $n\in \mathbb N$  .
(b) Additionally  assume  that  the  mapping  $X$  is  continuous.  To  which  random variable  does  the  limit  $ \lim \limits_{n\to \infty } E[X|F_n]$  converge  pointwise  in  $[0, 1  )$  ?     
 A: According to the definition of the conditional expectation we need a random variable $E[X\mid F_n](\omega)$ for which
$$\int_A E[X\mid F_n](\omega)\ dP=\int_A X(\omega)\ dP$$
for all $A\in F_n$and $E[X\mid F_n]$ is constant over the sets $[\frac k{2^n},\frac{k+1}{2n})$ because $E[X\mid F_n]$ has to be $F_n$ measurable. For a $k$, let $E[X\mid F_n](\omega)=c_k$ if $\omega\in[\frac k{2^n},\frac{k+1}{2n})$, then
$$\int_{[\frac k{2^n},\frac{k+1}{2n})}c_k\ dP=c_kP\left(\big[\frac k{2^n},\frac{k+1}{2n}\big)\right)=\frac{c_k}{\ \ 2^n}=\int_{[\frac k{2^n},\frac{k+1}{2n})}X(\omega)\ dP$$ that is$$c_k=2^n\int_{[\frac k{2^n},\frac{k+1}{2^n})}X(\omega)\ dP.$$
If $X$ is continuous and finite then if $\omega\in [\frac k{2^n},\frac{k+1}{2^n})$
$$\lim_{n\to\infty}E[X\mid F_n](\omega)=\lim_{n\to\infty}2^n\int_{[\frac k{2^n},\frac{k+1}{2^n})}X(\omega)\ dP.$$
If $n$ is large enough then
$$2^n\int_{[\frac k{2^n},\frac{k+1}{2^n})}X(\omega)\ dP\approx 2^nX(\omega)\frac1{2^n}=X(\omega).$$
So, the limit of $E[X\mid F_n]$ is $X$ if $n$ tends to the infinity.
