Before I prove this, let's show that f is measurable. In some sense this does not matter. The integral of a nonnegative function is well defined, whether that function is measurable or not; and monotone convergence holds, whether f is measurable or not; but if a reasonable topology is placed on the reals, then f is indeed measurable. Let's prove this first. For this result, the functions could be negative, but they must still be measurable, and increasing.
Let the open intervals act as a base for σ on the reals. This is equivalent to the open rays in both directions. If the preimage of an open ray, up or down, belongs to σ(X), then the same holds for their ongoing unions and intersections, the preimage of every measurable set is measurable, and f is measurable.
Start with an open ray pointing up. The preimage of this ray, under fn, belongs to σ. This because fn is measurable. Take the next step, and the preimage of the same ray under fn+1 is a (possibly larger) set that also belongs to σ. The union of all these sets is measurable, and it belongs to the preimage of the ray under f. Suppose some x in X is in the preimage of the ray under f, but is not in any of these sets. Let f(x) = c, with c > e, where e is the endpoint of the ray. By the definition of limit, there is some n such that fn(x) lies strinctly between e and c. This puts x in one of our ascending preimages after all. Therefore, the preimage of our open ray, under f, is measurable.
A similar proof holds for an open ray pointing down. The countable intersection of measurable sets is measurable, hence the preimage under f is measurable. That completes the proof.
Now return to the monotone convergence theorem, i.e. the integral of the limit is the limit of the integrals. By monotonicity, the integral of f is at least the integral of each fn. Thus the integral of f is at least the limit of the integrals of fn.
If the integral of any fn is infinite, then we're done. (The integral of the limit and the limit of the integrals are both infinite.) Similarly, if the integrals are finite, but they increase without bound, then we're done. So assume the integral of each fn is finite, and the sequence of integrals is bounded, with b as a least upper bound. The integral of f is at least b. We only need show it is not greater than b.
Consider a real number α, strictly between 0 and 1. Then consider a simple function u between 0 and f. Let An be the set of x in X such that α×u(x) ≤ fn(x).
Now you've probably noticed that A depends on n, u, and α, and you're right. Well u is fixed, and α is fixed, at least for the next couple paragraphs, so I'll just call it An.
Assume the reals have the usual topology. Thus α×u-fn is a measurable function. The reals ≤ 0 belong to σ, and their preimage, An, also belongs to σ. In other words, An is a measurable set.
Since fn is increasing, An is expanding. Also, the union over An is all of X. Why? Suppose some point x is left out. Thus αu(x) exceeds fn(x) for each n. Push u(x) up to f(x); this only makes the left side larger. Still, with α < 1, the left side is bounded below f(x). The right side approaches f(x), and for some n, fn(x) exceeds αf(x). This is a contradiction; hence the union of An is all of X.
Remember that αu ≤ fn, and take the integral over the region An. By monotonicity, the integral of αu ≤ the integral of fn. Extend the right side to the integral over all of X. This is of course ≤ b.
∫ αu { over the region An } ≤ b
Since integration and scaling commute, write it this way.
α × ∫ u { over the region An } ≤ b
∫ u { over the region An } ≤ b / α
Now it's time to be more precise about A.
∫ u { over the region Aα,n } ≤ b / α
Hold n fixed, and take the limit as α approaches 1 from below. The right side becomes b. On the left, Aα,n is a decreasing sequence of regions whose intersection is A1,n. (Restrict α to rational numbers approaching 1, so that the sequence is countable.) Now the left side is a reverse indefinite integral, and is equal to the integral of u over A1,n.
Henceforth, I shall not worry about α, as it will be equal to 1. Our relationship can be written as follows.
∫ u { over the region An } ≤ b
For notational convenience, let v(An) be the integral of u over An. If σ is the measurable sets of X, v is a map from σ into the nonnegative reals. Note that v is well defined for every measurable set, but when the set is drawn from An, v attains a value between 0 and b. By the properties of concatenation, v() is a measure.
Since An is an expanding sequence, apply the expansion theorem, and v(x) is the supremum of v(An). The former is the integral of u over X, and the latter is no larger than b.
Take the supremum over all simple functions u, and the integral of f is no larger than b. That completes the proof.