Conditional expectation on a filtered space Let $\mathcal F_n$ be the $\sigma$-algebra generated by the
intervals
$((j - 1)2^{-n}; j2^{-n}], j = 1, 2, ... , 2^n$, on a probability space $(\Omega,\mathcal F, P)$, where $Ω$ is $[0, 1]$, $\mathcal F$ the Borel $\sigma$-algebra, and $P$ Lebesgue measure. Let $X$ be a bounded continuous function on $[0, 1]$.
Show that the sequence ${\{Y_n}\}={\{E(X|\mathcal F_n)}\}, n = 1, 2, ..., $ converges.
 A: Fix $w$ such that it defines a unique sequence $\{w_j\}$  and consider behaviour of $Y_n(w)$ for this $w$ as $n\to\infty$. Recall that
$$Y_n(w)=\sum_{j=1}^{2^n} c_j I_{\left(\frac{j-1}{2^{n}}, \frac{j}{2^{n}}\right]}(w).$$
The value 
$$
w=\frac{w_1}{2}+\frac{w_2}{2^2}+\ldots+\frac{w_n}{2^n}+\sum_{j=n+1}^\infty \frac{w_j}{2^j}
$$
belongs to the interval $\left(\frac{j_n-1}{2^{n}}, \frac{j_n}{2^{n}}\right]$ iff 
$$\frac{j_n-1}{2^{n}} = \frac{w_1}{2}+\frac{w_2}{2^2}+\ldots+\frac{w_n}{2^n}=\frac{\lfloor{2^n w}\rfloor}{2^n}\; \text{ and } \; \frac{j_n}{2^{n}} =\frac{\lceil{2^n w}\rceil}{2^n}.$$
Then 
$$Y_n(w)= c_{j_n} = 2^n \int_{\tfrac{\lfloor{2^n w}\rfloor}{2^n}}^{\tfrac{\lceil{2^n w}\rceil}{2^n}} {X(s)ds}.$$
The Mean Value theorem implies that the r.h.s. of the above equality equals to $X(w_{n,w}),$ where $w_{n,w}$ is some point in the interval  $\left({\tfrac{\lfloor{2^n w}\rfloor}{2^n}},\;{\tfrac{\lceil{2^n w}\rceil}{2^n}}\right]$: 
$$
Y_n(w)=X(w_{n,w}), \;\; \frac{\lfloor{2^n w}\rfloor}{2^n}<w_{n,w}\leq \frac{\lceil{2^n w}\rceil}{2^n}.
$$
Since $|w_{n,w}-w|\leq 2^{-n}$, the continuity of $X(s)$  implies that $Y_n(w)=X(w_{n,w})$ converges to $X(w)$ as $n\to\infty$. 
