Let X,Y to be mappings from the sample space Ω to R and suppose Y is measurable with respect to σ(X), the smallest σ-field that makes X measurable. 


Does it follow that there exists some Borel-measurable function f: R → R such that 
Y=f(X)? 



question on sigma-fields
The answer is yes. The proof is quite standard.
**1.** If Y=1\_A, where A is in \simga(X), then by the definition of \sigma(X) there exists a Borel set B such that A = X^{-1}(B) and therefore


Y(\omega) = 1\_A(\omega) = 1\_B(X(\omega)) = f(X(\omega)),


where we put f:= 1\_B (of course f is now a Borel function).


**2.** If now Y is a simple r.v. i.e. it can be written in form Y = \sum\_{i=1}^n c\_i 1\_{A\_i}, where A\_i are sets in \sigma(X), then using the previous point we can find Borel functions f\_i such that 1\_{A\_i} = f\_i(X) and obviously in this case f = \sum\_{i=1}^n c\_i f\_i.


**3.** Finally, any r.v. Y measurable w.r. to \sigma(X) can be approximated by a sequence of simple r.v. Y\_n measurable w.r. to \sigma(X) i.e. Y\_n -> Y almost surely. By the previous point there exist f\_n such that Y\_n = f\_n(X). Now we can define
f(x) = \lim\_n f\_n(x) if the limit exists and put f(x)=0 otherwise. It is easy to check that f is a Borel function (basically it is a limit of Borel functions), and that Y = f(X).


