Monotone Convergence Theorem One simple case of the monotone convergence theorem for integration is:
Let $E \subset \mathbb{R}^n$ and suppose that $f_k : E \rightarrow \mathbb{R}$ is a sequence of non-negative measurable functions which increases monotonically to a limit $f$. Then $f : E \rightarrow \mathbb{R}$ is measurable and
\begin{equation*} \lim_k \int_E f_k = \int_E f \end{equation*}
where here we mean the Lesbegue integral on $\mathbb{R}^n$.
I know one proof of this theorem which is very measure-theoretic, and I was wondering is there a non-measure theoretic proof for the theorem if we only assume that the $f_k$ are Riemann integrable?
 A: There are proofs of the monotone and bounded convergence theorems for Riemann integrable functions that do not use measure theory, going back to Arzelà in 1885, at least for the case where $E=[a,b]\subset\mathbb R$.  For the reason t.b. indicated in a comment, you have to assume that the limit function is Riemann integrable.  A reference is W.A.J. Luxemburg's "Arzelà's Dominated Convergence Theorem for the Riemann Integral," accessible through JSTOR.  If you don't have access to JSTOR, the same proofs are given in Kaczor and Nowak's Problems in mathematical analysis (which cites Luxemburg's article as the source).  
In the spirit of a comment by Dylan Moreland, I'll mention that I found the article by Googling "monotone convergence" "riemann integrable", which brings up many other apparently helpful sources.
A: If you restrict your attention to regulated functions (= uniform limits of step functions, see e.g. Bourbaki), then you can get a nice 'structural proof' of monotone convergence by using Dini's theorem, which states that if a sequence of continuous functions converges pointwisely to 0 on a compact metric space, then it must also converge uniformly. If you use Dini's result by approximating step functions in a smart way with continuous functions, then you arrive at monotone convergence, at least for regulated functions on a compact interval.
A: It is precisely uniform convergence that gives us the Riemann integrability of $f$.  Monotonicity is not sufficient, as t.b.'s comment shows.  Assuming $f$ is Riemann integrable to begin with is sort of cheating, although it may be useful in specific examples.
