TITLE: Is this a good argument on what makes the Lebesgue integral more general than the Riemann integral?
QUESTION [6 upvotes]: Ever since I learned about the Lebesgue integral I have tried to understand what exactly makes it more general than the Riemann integral. By this I mean what part of its definition makes it more general. Below I will only talk about positive functions defined on some interval $[a, b]\subset \mathbb{R}$.
Here is what I've come up with: when defining the Riemann integral, we use step functions, which are linear combinations of characteristic functions of intervals; when defining the Lebesgue integral, we use measurable simple functions, which are linear combinations of characteristic functions of measurable sets. I think that this is the point of the theory that makes the Lebesgue integral more general. In particular, I think that this is what makes the function $$f(x)=\begin{cases} 1 & x\in [0, 1]\cap \mathbb{Q} \\ 0 & x\in [0, 1]\cap (\mathbb{R}\setminus \mathbb{Q}) \end{cases}$$ Lebesgue integrable and not Riemann integrable.
Furthermore, I think that the reason why for the Riemann integral we use characteristic functions of intervals is because these are the only Jordan measurable subsets of $\mathbb{R}$ that are "interesting" (I can't really think of a Jordan measurable subset of $\mathbb{R}$ that is not an interval except some weird looking examples that end up having Jordan measure zero) and for the Riemann integral we wish to have a good connection with the Jordan measure. So, I think that we may also say that the reason why the Lebesgue integral is more general than the Riemann integrable is that the Lebesgue measure is more general than the Jordan measure.
Are the above arguments correct? I have really thought about this a lot and I really want to make sure that I know what part of the definition of the Lebesgue integral grants it its generality.

REPLY [2 votes]: Good question.  One reason why one might ponder about this is to recall Lebesgue's original defintion, developing measure theory first, then defining the integral for all bounded measurable functions, then extending to unbounded "summable"
functions.   What has this to do with Riemann's definition?   Why on earth is every Riemann integrable function a bounded measurable function? And (in the other direction) why on earth is there   a bounded measurable function that is not Riemann integrable?   If you don't ask such questions you weren't paying attention.
Lebegue doesn't help.  His definition is so remote from Riemann's that it takes a long time in his thesis to prove the connection.  That remoteness also alienated many mathematicians of his time.
The short historical answer (Maybe not satisfying for you).

Volterra in the 1880s gave an example of a function $F:[0,1]\to\mathbb R$
with $f=F'$ bounded but not Riemann integrable.

Lebesgue, motivated by that example, designed an integral guaranteed to include all Riemann integrable functions but also integrate all bounded derivatives.  [Indeed he explicitly claimed that as goal.]



Preliminaries:  First let's clarify what we are talking about.  Take an interval $[a,b]$ and write the following:
${\cal C}[a,b]$ --- all continuous functions on $[a,b]$
${\cal D}[a,b]$ --- all derivatives on $[a,b]$, i.e., $F'=f$ for all points for some function $F$.
$b{\cal D}[a,b]$ --- all bounded derivatives on $[a,b]$
${\cal R}[a,b]$ --- all Riemann integrable functions on $[a,b]$
${\cal IR}[a,b]$ --- all improperly Riemann integrable functions on $[a,b]$
${\cal L}[a,b]$ --- all Lebesgue integrable functions on $[a,b]$
${\cal IL}[a,b]$ --- all improperly Lebesgue integrable functions on $[a,b]$
If you understand your integration theory (and you should before considering such questions as this) you should know these facts intimately:

${\cal C}[a,b] \subset b{\cal D}[a,b] \subset b{\cal L}[a,b] 
\subset  {\cal L}[a,b] \subset {\cal IL}[a,b]$  and  all inclusions proper.

While $ b{\cal D}[a,b] \subset b{\cal L}[a,b]$  it is the case that
$b{\cal D}[a,b] \not\subset  {\cal R}[a,b] $.   This is the original motivation Lebesgue gave for his integral.

$  {\cal IR}[a,b] \not \subset {\cal L}[a,b]$
and $  {\cal L}[a,b] \not \subset {\cal IR}[a,b]$  although they are compatible, i.e., if a function is integrable in both senses the integrals agree.

But $  {\cal D}[a,b] \not \subset  {\cal L}[a,b]$  and  $  {\cal D}[a,b] \not \subset  {\cal IL}[a,b]$  which is the reason why integration theory did not stop with Lebesgue. [To elaborate a bit: Lebesgue's integral handles all bounded derivatives but only some unbounded derivatives.  He did not like this fact!]



Q.  Back to your question:  What part of the definition
of ${\cal L}[a,b]$ explains why $b{\cal L}[a,b]\supset {\cal R}[a,b]$ and
$b{\cal L}[a,b] \not= {\cal R}[a,b]$?
A1.  Well this depends on which of the very many definitions of the Lebesgue integral you are using.   If you use the measure-theoretic definition for
${\cal L}[a,b]$ and the orginal definition for ${\cal R}[a,b]$ this is a puzzle.  But you are correct:  if you use instead the Peano-Jordan measure definition of ${\cal R}[a,b]$ (as you suggested) then the explanation is quite clear.  Just check that all PJ-measurable sets are measurable and give one simple example of a measurable set that is not  PJ-measurable.  Your thinking here is fine!
A2.  But you might have seen a different definition for ${\cal L}[a,b]$ as
the completion of the space ${\cal C}[a,b]$ using the norm
$\|f\| = \int_a^b |f(x)|\,dx$.   In that case your explanation  will need to show that this completion contains ${\cal R}[a,b]$ but that there are bounded functions in that completion not contained in ${\cal R}[a,b]$.
A3.  Perhaps you are a fan of defining ${\cal L}[a,b]$ using Riemann sums monitored by a gauge (as developed in the 1960s).  Then it is immediate and trivial that ${\cal L}[a,b]\supset{\cal R}[a,b]$  and, again, one bounded counterexample will suffice.
A4.  I doubt you want any more definitions for ${\cal L}[a,b]$.  So bye.