Finding the "most" continuous representative of a class of functions equal almost everywhere. In measure theory, we consider functions to be basically the same if they are equal almost everywhere. It seems crazy, though, to choose any of these as the representative when doing calculations. Why wouldn't we choose the nicest one, if it exists?
So, given a function f, is there a clear way to find a function g such that g = f a.e. and if h = f a.e., g is continuous at every point h is? 
I tried g(x) = liminf e->0 int_{B_e(x)} f, but I couldn't prove much about it.
 A: If $(\Omega,\Sigma,\mu)$ is a probability space, it is possible to prove that one can choose representatives from elements of $L_\infty(\Omega,\Sigma,\mu)$ such that all finite algebraic operations are preserved. Such a choice of representaives is known as a Lifting. Liftings exist by the, terribly advanced, von Neumann-Maharam Lifting Theorem. But if we take the probability space to be $[0,1]$ with Lebesgue measure and the Borel $\sigma$-algebra, it is independent of the usual axioms of set theory that a lifting exists that chooses continuous functions as representatives when such a choice is possible. For $L_p$-spaces with $p<\infty$, not even liftings exist.
If this sounds confusing and complicated, I can assure you that this topic is confusing and complicated and belongs to the most advanced parts of measure theory.
By the way: I strongly disagree with the statement that we consider functions to be the same in measure theory when they agree outside a null set. There are many important issues in measure theoory where such an identification makes no sense.
