# Definite Integral with unequal partitions

I am trying to understand how definite integral works when the partitions of the function are of unequal length. I found this link and am stuck here: $$I = \int_a^b f(x)dx = \lim_{mesh(P) \to 0} R(f,P,T)$$

I understand that $\int_a^b f(x)dx = \text{Area under the curve}$ but not how that is equal to $\lim\limits_{mesh(P) \to 0} R(f,P,T)$.

The way I see this, in the case of equal partitions, the mesh is $\Delta x_{i} = \frac{1}{n}(b-a)$ so a 0 mesh means that there will be infinitely many thin partitions so that the sum of the area of each partition equals the area under the curve.

In the case of unequal partitions, since the mesh is $\Delta x_{i} = x_i - x_{i-1}$, if each mesh goes to 0, it appears that the number of partitions $n$ remains fixed so that the sum of the area of each partition definitely won't cover the area under the curve. Am I visualizing this correctly? If the number of partitions increase, how is $n$ related to $\Delta x_{i}$?

-
The number of partitions increases in this case, too, and the width of each $\Delta x_i$ goes to zero. It is just not required that all the $\Delta x_i$ go to zero at the same rate. – Ross Millikan Feb 26 '11 at 15:38
The link does not work for me... – Fabian Feb 26 '11 at 15:38
No, n also goes to infinity. – Qiaochu Yuan Feb 26 '11 at 16:03
Part of the definition of "Partition" is that it must cover all of [a,b]. So in order for the mesh size to get smaller, the number of elements in the partition must get bigger (so if $|P| \to 0$, then the number of elements must go to infinity). – Brian Feb 26 '11 at 18:10
This may not help, since it just glosses over the point of arbitrary partitions, but you might want to take a look at math.stackexchange.com/questions/15294/… – Arturo Magidin Feb 26 '11 at 20:13

The "convergence" of partitions, associated with defining a Riemann integral through a limiting process, is a bit trickier than convergence of a sequence.

We want two notions: that one partition "refines" another partition, and that a partition of an interval [a,b] produces subintervals of width at most h > 0. You seem to have a good grasp on the second of these, namely that even if the partition is not constructed to have n equal length subintervals, the subintervals may still have some "small" bound on their maximum width. We can specify that "mesh size" is small in this way, setting the stage for proving something about the limit as mesh size tends to zero.

It is the first concept, that of one partition refining another, which typically causes more difficulty. The notes you linked to do not seem to treat that topic. Here is a PDF on the Riemann integral which does.

For a given function f on interval [a,b] we define the lower Riemann integral as the sup over all partitions of their lower Riemann sums, and the upper Riemann integral as the inf over all partitions of their upper Riemann sums. If these agree, then we say the function f is Riemann integrable on [a,b], and the common value of the upper and lower Riemann integrals is the value of the Riemann integral.

A partition $Q$ of [a,b] is said to refine partition $P$ when the endpoints of $P$ are also endpoints of $Q$. The theory of Riemann integration uses a construction that given two partitions $P_1$ and $P_2$ produces a common refinement of both $Q$ by taking the union of the two sets of endpoints. Just having a smaller mesh size is not enough to guarantee that one partition refines another.

See the notes of the link above for more details of how the theory of Riemann integration is developed using the concept of partition refinement.

-
The limit in the definition of the integral is not related to the refinement order (it can be done that way, but that is unnecessarily complicated if you are trying to explain it to someone who's learning...) – Mariano Suárez-Alvarez Feb 26 '11 at 17:30
I mean: the limit written in the question means: «for all $\varepsilon>0$ there exists $\delta>0$ such that if $P$ is a partition of mesh $<\delta$ then...» and refinement does not appear there. – Mariano Suárez-Alvarez Feb 26 '11 at 17:32
@Mariano Suárez-Alvarez: You are right that the definition of a Riemann integral can be stated without the "complicated" notion of refinement (though inf/sup are needed, see above), and that some theory can be developed treating only a sequence of partitions whose mesh size tends to zero. If someone wishes to "visualize" the number of subintervals of a partition increasing, the notion of refinement may have a pedagogical value, as does the notion of the mesh size tending to zero. In any case an arbitrary increase in the number of subintervals doesn't get us to convergence. – hardmath Feb 26 '11 at 19:07
There is no need for any inf/sup, either, really: in my comment above, the ellipsis was meant to stand for «for any choice $\xi$ of intermediate points in $P$, we have $|s_{P,\xi}(f)-I|<\varepsilon$, where $s_{P,\xi}(f)$ is the Riemann sum corresponding to $P$, $\xi$ and $f$. – Mariano Suárez-Alvarez Feb 26 '11 at 21:44
This link was also useful in explaining partition refinement: – Sara Mar 5 '11 at 2:58