The Unapologetic Mathematician

Mathematics for the interested outsider

Some integrable functions

Okay, we know what it means for a function to be integrable (in either of the equivalent Riemann or Darboux senses), but we don’t yet know any functions to actually be integrable. I won’t give the whole story now, but a just a large enough part to work with for the moment.

The major theorem here is that a continuous function f on a closed interval \left[a,b\right] is integrable. Notice from the Heine-Cantor theorem that f is automatically uniformly continuous. That is, for any \epsilon there is some \delta so that for all x and y in \left[a,b\right] with |y-x|<\delta we have |f(y)-(x)|>\epsilon. Again, the important thing here is that we can choose our \delta independently of the point x, while continuity says \delta might depend on x.

So now we need to take our function and show that the upper and lower Darboux sums converge to the same value. Equivalently, we can show that their difference converges to zero. So given an \epsilon we want to show that there is some partition x so that the difference

\displaystyle U_x(f)-L_x(f)=\sum\limits_{i=1}^n(M_i-m_i)(x_i-x_{i-1})

is less than \epsilon, and the same is true of any refinement of this partition.

We’ll choose our partition with every slice having constant width \delta, so there are \frac{b-a}{\delta} of them. By the uniform continuity of f we can find a \delta so that for any points x and y with |y-x|<\delta we’ll have |f(y)-f(x)|<\frac{\epsilon}{b-a}. Then in particular the difference M_i-m_i will be less than \frac{\epsilon}{b-a}, while x_i-x_{i-1}=\delta and n=\frac{b-a}{\delta}. Thus the difference in the Darboux sums will be less than \epsilon, as we wanted.

What about refinements? Well, any refinement of a partition can only lower the upper Darboux sum and raise the lower one. This is because adding a point to a partition can’t raise the maximum in either of the new subintervals or lower the minimum, and in fact adding a point will usually lower the maximum and raise the minimum. So our partition has a small enough difference in the Darboux sums, and any refinement will make the difference even smaller, and thus we have the convergence we need.

Now, can we do better than continuous functions? Well, we can relax continuity at the endpoints a bit. If the function jumps to a different value at a or b than the limit seems to indicate, we can still get uniform continuity everywhere but that one point, and we’re still good. We still have problems with asymptotes where the function shoots off to infinity, like f(x)=\frac{1}{x} does at the left endpoint of \left[0,1\right].

What else? Well, we can allow a finite number of discontinuities, as long as none of them are asymptotes. If a discontinuity happens at c\in\left[a,b\right], we can choose c to be a partition point, and so on. Then a partition with these selected points is just the same as a partition on each of the continuous sections of the function in between the discontinuities, and we know that they’re all good.

Incidentally, we can use this same method of picking some of the partition points ahead of time to show another nice property of the integrals: we can break the integral in the middle of the interval and evaluate the two pieces separately, then add them together. That is, if c\in\left[a,b\right] then we have the equation

\displaystyle\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx

as long as both sides are integrable.

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February 1, 2008 - Posted by | Analysis, Calculus

2 Comments »

  1. [...] notice that this is just a Riemann sum for the function . Since is continuous, we know that it’s Riemann integrable, and so as gets larger and larger, these Riemann sums must converge to the Riemann integral. That [...]

    Pingback by Step Function Integrators « The Unapologetic Mathematician | March 14, 2009 | Reply

  2. [...] describes the area that the plane cuts out of this volume. It exists because because the integrand is continuous as a function of . In such classes, we make the reasonable assumption that as we vary this area [...]

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


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