Upper and Lower Integrals and Riemann’s Condition
Yesterday we defined the Riemann integral of a multivariable function defined on an interval . We want to move towards some understanding of when a given function is integrable on a given interval .
First off, we remember how we set up Darboux sums. These were given by prescribing specific methods of tagging a given partition . In one, we always picked the point in the subinterval where attained its maximum within that subinterval, and in the other we always picked the point where attained its minimum. We even extended these to the Riemann-Stieltjes case and built up upper and lower integrals. And we can do the same thing again.
Given a partition of and a function defined on , we define the upper Riemann sum by
In each subinterval we pick a sample point which gives the largest possible sample function value in that subinterval. We similarly define a lower Riemann sum by
As before, any Riemann sum must fall between these upper and lower sums, since the value of the function on each subinterval is somewhere between its maximum and minimum.
Just like when we did this for single-variable Riemann-Stieltjes integrals, we find that these nets are monotonic. That is, if is a refinement of , then and . As we refine the partition, the upper sum can only get smaller and smaller, while the lower sum can only get larger and larger. And so we define
The upper integral is the infimum of the upper sums, while the lower integral is the supremum of the lower sums.
Again, as before we find that the upper integral is convex over its integrand, while the lower integral is concave
and if we break up an interval into a collection of nonoverlapping subintervals, the upper and lower integrals over the large interval are the sums of the upper and lower integrals over each of the subintervals, respectively.
And, finally, we have Riemann’s condition. The function satisfies Riemann’s condition on we can make upper and lower sums arbitrarily close. That is, if for every there is some partition so that . In this case, the upper and lower integrals will coincide, and we can show that is actually integrable over . The proof is almost exactly the same one we gave before, and so I’ll just refer you back there.