# Integral defined as a limit using regular partitions

Definition. Given a function $f$ defined on $[a,b]$, let $$\xi_k \in [x_{k-1},x_k],\quad k=1,\ldots,n$$ where $$x_k=a+k\frac{b-a}n, \quad k=0,\ldots,n \; .$$ One says that $f$ is integrable on $[a,b]$ if the limit $$\lim_{n\to\infty}\frac{b-a}n\sum_{k=1}^n f(\xi_k)$$ exists and is independent of the $\xi_k$.

I seek a proof of the:

Theorem. If $a<c<b$ and $f$ is integrable on $[a,c]$ and $[c,b]$, then $f$ is integrable on $[a,b].$

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please try it first –  trying Jun 4 '13 at 14:46
@Anindya Ghatak: I did but ... –  Tony Piccolo Jun 4 '13 at 14:49

HINT:

Take Two cases:

1.When $c$ is a tag of a sub-interval $[x_{k},x_{k+1}]$ of $\dot{P}$,where $\dot{P}$ is your tagged partition ${(I_{i},t_{i})}_{i=1}^{n}$,such that $I_{i}=[x_{i},x_{i+1}]$

2.When $c$ is an end-point of a sub-interval of $\dot{P}$.

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Can you develop one of the two cases ? –  Tony Piccolo Jun 5 '13 at 6:51

Your version of integrability is equivalent to the standard one but (to me at least) seems rather awkward to work with. Thus I would begin by verifying that your definition implies that $f$ is Darboux Integrable: use Darboux's Integrability Criterion (Theorem 8.7 of these notes) and notice that by suitable choices of $\xi_k$'s you can make your sums arbitrarily close to the upper and lower sums.

Your theorem is one of the basic properties of the Darboux integral, so is proved in all the standard texts (see e.g. Theorem 8.8 of loc. cit.).

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