Is the Lebesgue integral the completion of integrals on step functions?

Lierre gave a very helpful insight at answer 5 on about what Riemann integrals are. My question relates to whether this can be extended to Lebesgue integrals.

Lierre pointed out that Riemman integrals can be seen as the natural extension of the 'obvious' linear form on characteristic (or 'indicatrix') functions on real line intervals. There is a theorem which shows that any uniformly continuous function $f$ on a subset $V$ of a metric space $X$ into a complete metric space $Y$ can be uniquely extended to a u.c. function on the closure of $V$ into $Y$. If we take $X$ and $Y$ as $\mathbb{R}$, and $V$ as being the subspace of real functions on $\mathbb{R}$ spanned by the characteristic functions of closed finite intervals $[a,b]$, and $f$ as the function taking $[a,b]$ to $|a-b|$, it turns out that the Riemann integral is the extension of $f$ to closure of $V$.

My question is this: if we take countable unions of intervals $[a,b]$, with the 'obvious' $f$ defined as a countable sum (if it exists), and the subspace of functions on $\mathbb{R}$ spanned by these as our $V$, is it also true that the closure of $V$ consists of the Lebesgue integrable functions, and that $f$ extends to the Lebesgue integral on them? Also, how far if at all can one go in proving theorems such as the dominated convergence theorem based purely on the properties f must have as an extension to the closure, without actually constructing the integral itself?
Thanks!

5
2022-07-25 20:47:17
Source Share
Answers: 0