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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?

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One problem is that the monotone limit $f$ need not be Riemann integrable. Indeed, enumerate the rationals in $[0,1]$ and let $f_k$ be the characteristic function of $\{q_0,\ldots,q_k\}$. Then $f_k$ converges monotonically to the characteristic function of $\mathbb{Q} \cap [0,1]$ but, as you know, this function isn't Riemann integrable. So you need to assume that you already know that $f$ is Riemann integrable. – t.b. Dec 16 '11 at 5:12
@t.b.: Yes, thats true and a good point. So lets assume that $f$ is known to be Riemann integrable. – Eric Haengel Dec 16 '11 at 5:31
up vote 3 down vote accepted

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.

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This note is a very nice and freely available exposition the bounded convergence theorem for Riemann integrable functions. – t.b. Dec 16 '11 at 5:43

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.

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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.

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If they are all dominated by a Riemann integrable function, then because Riemann integrable functions are bounded, this is the same as bounded convergence. If you allow them to be dominated by an arbitrary Lebesgue integrable function, then I would be surprised if measure theory could be avoided in a proof. – Jonas Meyer Dec 16 '11 at 5:53
Interesting. So, proof in Rudin on page 155 (where Rudin says 'would be quite troublesome with the equipment we have at hand so far'), would indeed have been possible from Arzela's DCT for Riemann? – Forever Mozart Dec 16 '11 at 6:24
I don't know which Rudin or have it on hand (whichever it is) to check what you're referring to. (My comment above was posted in response to a question that was edited in and then removed within the 5 minute window.) – Jonas Meyer Dec 16 '11 at 6:29
Its not true that uniform convergence is necessary and sufficient that a sequence $f_k$ of riemann integrable functions converges to a riemann integrable function. If you take $f_k$ to have the graph which is a triangle of width $1/k$ and height $k$ with left end-point at the origin, then $f_k \rightarrow 0$ is riemann integrable, but this convergence is not uniform. – Eric Haengel Feb 9 '12 at 18:51

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