## 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.

Integration I think is one of the more fun parts of analysis because it has so many variations that are intimately related with each other. Have you heard of gauge integrals? An almost imperceptible generalization of the Riemann integral that turns out to subsume almost all of the other definitions.

Comment by david karapetyan | December 3, 2009 |

Denjoy’s integral does manage to handle some particularly troubling singularities that Lebesgue doesn’t, but it’s significantly more complicated and not really worth the time in this pass.

Comment by John Armstrong | December 3, 2009 |

[...] We’ve got Riemann’s condition, which is necessary and sufficient to say that a given function is integrable on a given interval. [...]

Pingback by Jordan Content « The Unapologetic Mathematician | December 3, 2009 |

[...] We wanted a better necessary and sufficient condition for integrability than Riemann’s condition, and now we can give a result halfway to our goal. We let be a bounded function defined on the [...]

Pingback by Jordan Content Integrability Condition « The Unapologetic Mathematician | December 9, 2009 |

[...] this end, we have five assertions relating the upper and lower single and double integrals involving a function which is defined and bounded on the rectangle above. [...]

Pingback by Iterated Integrals I « The Unapologetic Mathematician | December 16, 2009 |

[...] the last assertion is a consequence of the first four. Indeed, if the integral over exists, then the upper and lower integrals are equal, which collapses all of the inequalities into [...]

Pingback by Iterated Integrals II « The Unapologetic Mathematician | December 17, 2009 |

[...] now when we try to set up the upper and lower integrals to check Riemann’s condition we find that every subinterval of any partition must contain [...]

Pingback by Iterated Integrals III « The Unapologetic Mathematician | December 18, 2009 |

nice web

Comment by Omveer Singh | September 28, 2011 |

nice

Comment by Omveer Singh | September 28, 2011 |