# Cardinalic flaw of Riemann integral

I have learnt that integral is the Riemann sum of infinite rectangle, that:
Ʃ$^{n=1}_{∞}$f(xi)Δxi = ∫$^{b}_{a}$f(x)dx
However, I think that (a,b) is the continuous interval, so the number of rectangle should be c instead of $\aleph$0 (cardinality of natural number N).
So I wonder whether there are some problem that this definition is not valid anymore.

Bacle2
How so? The oo you're using is the countable infinity. An uncountable sum will

necessarily diverge , unless only countably-many are non-zero. Still, good

question.

Edit: after reading SteveL's comment, I guess I should be more precise:

The limit in the sum you describe is a limit as you approach countable infinity;

so you are selecting one point x_i* in each subinterval , and , as N-->oo (countable

infinity) there is a bijection between the number of rectangles and the x_i* you choose.

Since the x_i* are indexed by countable infinity, so are the rectangles.

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mathman
Each Δxi is a continuum - there is no contradiction.

I have learnt that integral is the Riemann sum of infinite rectangle, that:
Ʃ$^{n=1}_{∞}$f(xi)Δxi = ∫$^{b}_{a}$f(x)dx

I'm a little confused about this definition. Typically the Riemann integral is the limit of Riemann sums, each one of which is a finite sum over a partition of the interval. Each partition is a finite set of subintervals.

There is no infinite sum such as you've notated. Is this a definition you saw in class or in a book?

Thanks for explanation, I have understood.
And I mean it's the limit of finite sum, but I am a bit lazy so I remove the limit part for convenience.

HallsofIvy
Ʃ$^{n=1}_{∞}$f(xi)Δxi = ∫$^{b}_{a}$f(x)dx
However, I think that (a,b) is the continuous interval, so the number of rectangle should be c instead of $\aleph$0 (cardinality of natural number N).