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 $[0, 1 )$ ?
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.