If you are looking for a good reason to prefer the Lebesgue integral over the Riemann integral and remember something like “Lebesgue goes better with limits” but could not memorize a specific example, then you should have a look on this stunning sequence of functions (which I found in Brenner and Scotts “Mathematical Theory of Finite Element Methods”):

Let ${(r_n)_{n\in\mathbb{N}}}$ be a dense subset of ${[0,1]}$ (e.g. an enumeration of the rational numbers) and build the functions

$\displaystyle f_n(x) = \sum_{k=0}^n 2^{-k}\log(|x-r_k|).$

The functions ${f_n}$ accumulate singularities at the places ${r_k}$ as illustrated vaguely in this picture:

However, each function ${f_n}$ is (improper) Riemann-integrable (note that ${\log(|x|)}$ is Riemann-integrable over any compact interval) Moreover, even the integrals ${|\int_0^1 f_n(x){\mathrm d}{x}|}$ stay bounded for ${n\rightarrow\infty}$. Of course, the functions are also Lebesgue measurable.

To see why this sequence of functions is problematic for Riemann integration, consider the potential limit function

$\displaystyle f(x) = \sum_{k=0}^\infty 2^{-k}\log(|x-r_k|)$

which has “the value ${-\infty}$ on a dense subset”. Hence, the Riemann integral can not be finite.

However, it is clear that ${\|f - f_n\|_p\rightarrow 0}$ for ${1\leq p<\infty}$ (due to exponentially decaying weights) and since the ${L^p}$-spaces are Banach spaces, ${f\in L^p}$.

Advertisements