In a book that I am reading, they give the following proposition:

Let (X, A) be a measure space, and let B inside X that belongs to A. Let f be a [0, infinity]-valued measurable function on A. Then there is a sequence {f_n} of [0, infinity)-valued simple measurable functions on A that satisfy f_1(x) <= f_2(x) <= ... and f_n(x) -> f(x) as n -> infinity.

The proof given is:

Let A(n, k) = {x in B : (k - 1)/2^n <= f(x) <= k/2^n} for each k = 1, 2, .., n2^n.
Define f_n(x) be (k - 1)/2^n if x in A(n, k) and 0 if x not in union from n = 1 to n2^n of A(n, k).

Apparently it is easy to see that f_1(x) <= f_2(x) <= ... and f_n(x) -> f(x) as n -> infinity hold, but I just cannot see this. Can anyone explain?