(Lebesgue integral) Can simple functions take on infinity?

Question

Let (X, M, u) be a measure space. The definition of the Lebesgue integral over X for a function defined on X and with values in [0, oo] is given in terms of simple functions, that is,
Int f du = supremum {Int phi du | phi is a measurable simple function satisfying 0 <= phi <= f}. Some authors require phi to take on only finite values, but others allow phi to take on oo. It's not hard to show that in both cases the value of Int f du is the same. But if we allow simple functions to take on oo, then the standard proof of the Monotone Convergence Theorem, the one due to H. Levy, is not valid, Because, for c in (0, 1) and phi a measurable simple function such that 0 <= phi <= f, we define X_n = {x in X | f_n(x) >= c phi(x)}, n=1,2 ,3 ...f_n being a monotone increasing sequence of measurable functions on X. If we require phi to have only finite values, then X = Union X_n. But if we allow phi to take on oo, then this is not true. That's the reason for my question.

Answer 1

Yes, simple functions can take on infinity as a value. But if they are going to be integrable, the measure of the set where they are infinite must be zero. So, if you let A={x:phi(x)=infty}, either u(A)=0 or
int phi du=infty.
In the first case, you can write X as the union of the X_n and A.
In the second case, you just have to show that the integrals of f_n over A go to infinity.

In either case, the Monotone Convergence Theorem still holds for functions taking on infinite values.

Answer 2

But your assumption was that phi <= f, if phi becomes unbounded then surely Int f du becomes infinite and the supremum{Int phi du | phi} = infinity, so in that sense it's still true. That's what I'd think, but it might be better to consult a text such as Mathematical Analysis by Rudin. Unfortunately I'm strictly an amateur at this sort of thing...