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 σfield that makes X measurable.
Does it follow that there exists some Borelmeasurable function f: R → R such that Y=f(X)?
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 σfield that makes X measurable. Does it follow that there exists some Borelmeasurable function f: R → R such that Y=f(X)? 


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). 


If R is the reals, yes (as already explained). But if R is just some other Borel space, then perhaps not. 


It is trivially true but maybe worth noting that the converse is also true  if there exists such an f, then Y is σ(X)measurable. This and the question asked are theorem 20.1(ii) in Billingsley's Probability and Measure, 3rd edition. 

