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Suppose $f : (0,\infty) \rightarrow \mathbb{R}$ is nonnegative and monotone. How might one find that the integral $\int_{[0,T]} f$ for $0 < T \leq \infty$ is a limit of averages in the sense that $\int_{[0,T]} f = \lim_{N\rightarrow \infty} \frac{1}{N} \sum_{0 < k < NT} f(k/N)$?

I've tried writing $f$ as a limit of a nondecreasing sequence of simple functions so that the monotone convergence theorem would apply, figuring that the monotone condition could allow the integrals to be calculated easily. I know that this is possible in general--I can write down such a sequence for an arbitrary nonnegative measurable function $f$--but I've not had luck isolating a specific sequence which would yield the above result. If this approach can be made profitable, I'd appreciate any hints on how to find a sequence that works in this case.

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Hint : a monotone function may not have more than countably many points of discontinuity (prove it !)

Then you're back to good old Riemann integrals.

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  • $\begingroup$ Thanks a lot for the wisely chosen hint. That's a much better method than what I was originally trying to force. $\endgroup$
    – Zach Conn
    Nov 23, 2011 at 1:39

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