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Say we have $f(x)=1$ for $x\in \Re $

$\sum_{x\in [0,1]} f(x)=\infty$ but $\int_{[0,1]}f(x)dx=1$

I thought it was intuitive to think that integrals are just representations of infinite sums but I'm starting to think that is not the case. My initial hunch it has something to do with the integral being a limit sum of countable partitions but doesn't the integral cover the entire [0,1] area?

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    $\begingroup$ If we're thinking intuitively, note that on the left what you're summing is just $f(x)$, but on the right it's $f(x) dx$ - so yes we're summing 'infinitely' many items, but each of those items is 'infinitesimally small'. $\endgroup$ – AakashM Apr 16 '13 at 14:34
  • $\begingroup$ You used "uncountable" in your title, so I think you might find this interesting to add to your intuition: Every uncountable sum of positive real numbers diverges. That gives some indication of what it is only useful to consider countable and finite sums. To prove this, just notice that given a list of uncountably many numbers, there has to be an $n$ such that infinitely many of the numbers are greater than $1/n$. $\endgroup$ – rschwieb Apr 16 '13 at 14:45
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The Riemann integral is a limit of sums over finite partitions. Uncountable sums aren't well-defined in general, nor even countably infinite sums.

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