In my functional analysis textbook, the space $L^1$ is introduced. It is then explained that $|| f || = \int |f(x)|\ \text{d} x$ is not a norm (there are non-identity functions for which this integral is zero). Even if it were, $L^1$ would not be complete: There exist sequences $f_k \rightarrow f$ for which every $f_k$ is Riemann integrable, but $f$ is not. Then, Lebesgue integrals are introduced (after quite some measure theory).

I think this approach is correct, but can't this be done in an easier way?

Concretely, I want to understand why we need all the measure theory to define the Lebesgue integral. For example, I can think of setting up a 'weak Riemann integral' as follows:

Definition: A function $f : D \supset [a, b] \rightarrow \mathbb{R}$ is weakly Riemann integrable on $[a, b]$ if there exists a Riemann integrable function $g : [a, b] \rightarrow \mathbb{R}$, and a countable set $S \subset \mathbb{R}$ such that $f = g$ on $[a, b] \setminus S$. Then we define the weak Riemann integral over $[a, b]$ of $f$ as $\int_a^b f\ \text{d}\nu := \int_a^b g(x)\ \text{d}x$, where the right integral is a Riemann integral.

The example of a Lebesgue integrable function that is not Riemann integrable is a function that assumes another value on the rationals than on the irrationals. This definition works for that case, since the rationals are countable. I couldn't find cases for which this doesn't work (but I think there should be, or someone would have thought of such a simpler definition).

Why wouldn't this work? Is it not rigorous to define an integral like this? Is there a function that is Lebesgue integrable 'weakly Riemann integrable'?

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    $\begingroup$ The real success of the Lebesgue integral is the ability to integrate over arbitrary spaces. It's not built on the real line entirely, unlike the Riemann integral. There are many alternatives to the Riemann integral on the real line, but most do not generalize well to general Euclidean space or even more abstract spaces. $\endgroup$ May 19, 2017 at 13:55
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    $\begingroup$ I always liked this version: en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral $\endgroup$
    – user404127
    May 19, 2017 at 13:56
  • $\begingroup$ ^ That is one such generalization I alluded to in my first comment. $\endgroup$ May 19, 2017 at 13:57
  • $\begingroup$ The characteristic function of a fat Cantor set is Lebesgue integrable, but not "weakly Riemann integrable". $\endgroup$ May 19, 2017 at 13:59
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    $\begingroup$ @DanielFischer that seems right: a fat Cantor set has positive measure, is uncountable but does not contain any intervals. If you make it an answer, I'll accept it. $\endgroup$
    – Ruben
    May 19, 2017 at 14:04

1 Answer 1


The idea to generalise the Riemann integral by allowing redefinition of a function on a countable set is good, but it's "not good enough". A nice property of the Lebesgue spaces is that they are complete, and this extension of the Riemann integral doesn't achieve completeness of the space of integrable functions. Since the space of Riemann integrable functions is dense in $L^1([a,b])$, that is - up to isometric isomorphism - the only completion of the space of Riemann integrable functions. But this extension makes only functions whose set of discontinuities is of measure $0$ after redefinition on a countable set integrable, and there are Lebesgue integrable functions which don't have that property, e.g. the characteristic function of a fat Cantor set. Thus the space of weakly Riemann integrable functions lies strictly between the space of Riemann integrable functions and the space of Lebesgue integrable functions.


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