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

Lierre gave a very helpful insight at answer 5 on "Clean" approach to integrals 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!

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Not sure if this is what you are looking for, but $L^1 \mathbb{R}$ is the completion of $C_c(\mathbb{R})$ with respect to the (Riemann) integral norm. –  copper.hat May 16 '12 at 14:54
Yes it is possible and you do not miss anything. But keep in mind that the usual construction that decomposes a function into positive and negative parts naturally offers also the meaning of an integral being plus or minus infinite, which you have to define by "hand" if you use the completion approach. –  timur May 16 '12 at 19:53

As copper.hat pointed out $C_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, so we had to define such an integral for continuous functions with compact support.