Finding a conditional expectation of a function over a sigma algebra I want to find a conditional expectation of a function on a sigma-field. 
I am given:


*

*$F_n = \sigma([\frac{i}{2^n}, \frac{i+1}{2^n}), i=0,1,2,...,2^n-1)$

*$E[|f(X)|]<\infty$

*X is a uniform random variable


I want to find $E[f|F_n]$.
In this question, what would it mean that X is an uniform r.v.?
I think I have to show the measurability condition and partial averaging condition somehow, but I do not know how to start. Any help would be much appreciated!
 A: To do this question, you need to start with the basics.
Your probability space is $[0,1]$ with the Lebesgue (or Borel) measure. The random variable $X$ is uniform, in the sense that $X(\omega) = \omega$ for all $\omega \in [0,1]$. This is obviously Borel measurable.
Now, you have $F_n$. The key thing to realize about $F_n$ is that it in fact has finitely many sets. The way to see this is that the sets $[\frac i{2^n}, \frac{i+1}{2^n})$ , along with the singleton $\{1\}$ , are disjoint and their union is $[0,1]$. So, $F_n$ is the sigma algebra generated by a finite partition of $[0,1]$. I leave you to see that $F_n$ has only finitely many sets.
Suppose you do not see the above : no problem. We want to find out when a function is $F_n$ measurable. Let $g : [0,1] \to \mathbb R$ be $F_n$ measurable. Let $y \in [0,1)$. By definition, $g^{-1}(\{g(y)\})$ is a set that must belong in $F_n$. Therefore, it must, in particular, contain $[\frac i{2^n}, \frac {i+1}{2^n})$, where $i$ is the unique such index that $y$ belongs to this set.
In particular, $g$ is constant on on each $[\frac i{2^n} , \frac {i+1}{2^n})$.
You can show the converse as well : if $g$ is a function that is constant on each of these intervals then $g$ is $F_n$ measurable!
Now, the other condition is that $E[1_AE[X|F_n]] = E[1_AX]$. By monotone class theorem , it is sufficient to check that this holds for $A = [\frac i{2^n}, \frac{i+1}{2^n})$.
Now, $E[X|F_n]$ is $F_n$ measurable, so it is a constant $C_i$ on any such interval (constant could depend on $i$). The RHS for any such interval is $\frac {2i+1}{2^{n+1}}$ (the midpoint of $A$, easily calculated from the definition of $E[1_AX] = \int_A X$), while in the LHS, the constant $C_i$ comes out to become $C_iP(A)= C_i \frac 1{2^n}$. Thus, $C_i = \frac{2i+1}{2}$.
Finally, the answer is that the conditional expectation takes the value $C_i$ on the interval $[\frac{i}{2^n}, \frac{i+1}{2^n})$. Note that $\{1\}$ is a set of measure zero, so it does not matter what value is taken here.
Note that $E[f | F_n]$ for any integrable function $f$, is then equal to $\frac 1{2^n} \sum f(C_i)$, because the conditional expectation is a step function, so you can just use that definition and get through.
