Note that L^2(-) defines a contravariant functor from probability spaces to Hilbert spaces and bounded maps. This functor carries the limit \Omega _\infty = \lim _i \Omega _i to a colimit. To see this, first note that it carries each map to an embedding (since every measure-preserving map is onto up to a set of measure zero), so it suffices to show that every L^2 measurable function f on the limit is a limit of functions f_i which factor over \Omega _i.

Let such an f be given. Fix K so that the L^2-norm of the part of f outside [-K,K] is small. Partition [-K,K] into intervals of width 1/i. Each of the sets f^{-1}([ k/i, k+1/i )) is measurable. Hence there is some N_i so that each of these intervals is approximated up to probability \epsilon / i by some \Omega _{N_i}-measurable set. Define f_i to be equal to k/i on these approximating sets and zero outside of that. Then f_i is within 2\epsilon of f in L^2.

Now since forming adjoints is a self-duality on Hilbert spaces, this immediately implies that L^2(\Omega _\infty ) is also the inverse limit of the adjoint diagram, where the map L^2(\Omega _{i+1}) \to L^2(\Omega _i) is given by conditional expectation. It is apparent that an element of this inverse limit is precisely a martingale. Hence every martingale is the conditional expectation of a unique L^2-function on \Omega _\infty . It's clear that if a martingale converges, it is the conditional expectation of its limit, so we only need to show that given f \in L^2(\Omega _\infty ), E[f \mid \Omega _i] \to f.

Since the union of the inclusions L^2(\Omega _i) \hookrightarrow L^2(\Omega _\infty ) is dense, this is straightforward.