Does this proof for the MCT hold for the extended real valued functions.

Question

Here is a proof for the MCT, but it says that it is for the real numbers, not the extended real numbers. If we allow the function f to take the value infinity does the proof still hold? I can not see how it would not hold, but I would like your opinion.

PS: I assume the simple functions can not take the value infinity.

Answer 1

The proof still works, as long as the integral is indeed defined as (or known to be equal to) the supremum of $\{\int s d\lambda \mid 0 \leq s \leq f\}$, where the $s$ are simple functions that are not allowed to take the value $\infty$ (as you say they are).

One simply has to take care in the proof to check that $\bigcup E_n = \mathcal{R}$ still holds in this case, but it does, because there are three cases:

1. $f(x) = 0$. Then $s(x) = 0$, so that $\alpha s(x) \leq f_n(x)$ is trivial because of $f_n (x) \geq 0$.
2. $0 < f(x) < \infty$: As before.
3. $f(x) = \infty$. Then $\alpha \cdot s(x)$ is some finite value and $f_n(x) \to \infty$, which implies $f_n (x) \geq \alpha s(x)$ for $n$ large.