The Unapologetic Mathematician

Mathematics for the interested outsider

Jordan Content Integrability Condition

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 f:[a,b]\rightarrow\mathbb{R} be a bounded function defined on the n-dimensional interval [a,b].

The condition hinges on defining a certain collection of sets. For every \epsilon>0 we define the set

\displaystyle J_\epsilon=\left\{x\in[a,b]\vert\omega_f(x)\geq\epsilon\right\}

of points where the oscillation exceeds the threshold value \epsilon. The first thing to note about J_\epsilon is that it’s a closed set. That is, it should contain all its accumulation points. So let x be such an accumulation point and assume that is isn’t in J_\epsilon, so \omega_f(x)<\epsilon. So there must exist a neighborhood N of x so that \Omega_f(N)<\epsilon and this means that \omega_f(y)<\epsilon for any point y\in N, and so none of these points can be in J_\epsilon. But if this is the case, then x can’t be an accumulation point of J_\epsilon, and so we must have x\in J_\epsilon.

And now for our condition. The function f is integrable if and only if the Jordan content c(J_\epsilon)=0 for every \epsilon>0.

We start by assuming that \overline{c}(J_\epsilon)\neq0 for some \epsilon>0, and we’ll show that Riemann’s condition can’t hold. Given a partition P of [a,b] we calculate the difference between the upper and lower sums

\displaystyle\begin{aligned}U_P(f)-L_P(f)&=\sum\limits_{i_1=1}^{m_1}\dots\sum\limits_{i_n=1}^{m_n}\left(\max\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}-\min\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}\right)\mathrm{vol}(I_{i_1\dots i_n})\\&=A_1+A_2\geq A_1\end{aligned}

where A_1 is the part of the sum involving subintervals which contain points of J_\epsilon and A_2 is the rest. The intervals in A_1 have total length \overline{J}(P,J_\epsilon)\geq\overline{c}(J_\epsilon), and in these intervals we must have

\displaystyle\max\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}-\min\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}\geq\epsilon

because if the difference were less then the subinterval would be a neighborhood with oscillation less than \epsilon and thus couldn’t contain any points in J_\epsilon. Thus we conclude that A_1\geq\epsilon\overline{c}(J_\epsilon), and the difference between the upper and lower sums is at least as big that. This happens no matter what partition we pick, and so the upper and lower integrals must also differ by at least this much, violating Riemann’s condition. Thus if the function is integrable, we must have \overline{c}(J_\epsilon)=0.

Conversely, take an arbitrary \epsilon>0 and assume that \overline{c}(J_\epsilon)=0. For this to hold, there must exist a partition P_0 so that \overline{J}(P_0,J_\epsilon)<\epsilon. In each of the subintervals not containing points of J_\epsilon, we have \omega_f(x)<\epsilon for all x in the subinterval. Then we know there exists a \delta so that we can subdivide the subinterval into smaller subintervals each with a diameter less than \delta, and the oscillation on each of these subintervals will be less than \epsilon. We will call this refined partition P_\epsilon.

Now, if P is finer than P_\epsilon, we can again write

\displaystyle\begin{aligned}U_P(f)-L_P(f)&=\sum\limits_{i_1=1}^{m_1}\dots\sum\limits_{i_n=1}^{m_n}\left(\max\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}-\min\limits_{t\in I_{i_1\dots i_n}}\{f(t)\}\right)\mathrm{vol}(I_{i_1\dots i_n})\\&=A_1+A_2\end{aligned}

where again A_1 contains the terms from subintervals containing points in J_\epsilon and A_2 is the remainder. In all of these latter subintervals we know the difference between the maximum and minimum values of f is less than \epsilon, and so

\displaystyle A_2<\epsilon\mathrm{vol}([a,b])

For A_1, on the other hand, we let M and m be the supremum and infimum of f on all of [a,b], and we find

\displaystyle A_1\leq\overline{J}(P_0,J_\epsilon)(M-m)<\epsilon(M-m)

Thus we conclude that


Since this inequality is valid for any \epsilon>0, we see that Riemann’s condition must hold.

December 9, 2009 Posted by | Analysis, Calculus | 4 Comments