Last month, I mentioned to a friend that I was going to re-read the novel Slaughterhouse-Five; I had read it in high school, but I wanted to re-read it as a mature adult. This prompted him to quote the saying that it's not youth that's wasted on the young, it's literature that's wasted on the young. (I did re-read the novel, which retains its power and interest after 50 years.)

But while doing some work to build a new course, I was looking at various texts on Fourier analysis, and on real analysis. This is mathematics I learned, or was supposed to have learned, when I was in college or graduate school. But undoubtedly my understanding was rather rough when I first learned these subjects. At one point last week, I was reading an online excerpt of a introductory (senior/first-year graduate) level real analysis text by Elias Stein. Stein was a leading analyst, and as I quickly realized, a lucid writer. I realized as I read the free excerpt of his text that I would certainly benefit, thirty-some years after completing graduate school, from studying again this level of mathematics. It occurred to me that perhaps mathematics can be wasted on the young, who are too inexperienced to really understand what's going on.

One lovely bit of mathematics in Stein's text in particular caught my attention: the definition of an outer measure in the theory of Lebesgue integration. This is as follows, for a set \(E\) in the real line: \[ m_*(E) = \inf \sum_{j=1}^\infty|I_j|,\] where the infimum is taken over all countable coverings of \(E\) by open intervals \(I_j\) (where \(|I_j|\) is the length of the interval \(I_j\)). Stein points out that this is different than the Jordan outer content defined by \[ J_*(E) = \inf \sum_{j=1}^n|I_j|,\] where the infimum is taken over all finite coverings of \(E\). I was surprised that that these would be different; that to make Lebesgue measure and integration work properly, it is necessary to to use countable covers. Stein refers to Exercise 14 in that chapter for an explanation: This exercise is to show that there is a countable set \(E\subset[0,1]\) such that \(m_*(E)=0\) but \(J_*(E)=1\). The exercise gives as a hint to show that \(J_*(E)=J_*(\overline{E})\), where \(\overline{E}\) is the closure of the set \(E\). The solution to the exercise is to let \(E=\mathbb Q\cap[0,1]\). Any cover of \(E\) by finitely many open intervals will cover all of \([0,1]\), forcing \(J_*(E)=1\). But if we enumerate \(E\) as \(\{r_1,r_2,r_3,\ldots\}\), we can cover \(E\) by a countable collection of open intervals of arbitrarily small total length: Let \(I_j=\big(-\epsilon/2^{j+1},\epsilon/2^{j+1}\big)\), so that \[\sum_{j=1}^\infty|I_j| = \sum_{j=1}^\infty \frac\epsilon{2^{j+1}}=\epsilon,\] forcing \(m_*(E)=0\).