A few days ago I had my last discussion session on probability theory as a TA. In the end I asked students to ask me questions as this is the last class. One of the student asked me about the (real) definition of expectation; he said he is confused by "simple facts" like $$ E(c)=c, c\in \mathbb{R} $$ and why the expectation is linear.

Embarrassed that I did not explain this well enough earlier, I tried to explain to him expectation is a form of weighted sum, and for continuous case we are using some kind of probability measure coming from the density function. So the expectation is really some kind of integration. The student seemed to follow at this point. Then I tried to explain that the integration involved this way is not the same as the Riemann integration he learned in calculus classes, which is partly why the expectation of Cauchy distribution does not exist. I drew a graph and showed the Lebesgue integral can be viewed as a kind of "horizontal"-decomposition of the integration area. He asked me a very good question:

"What is the benefit of using horizontal instead of vertical? Isn't that the same thing?"

I did not really know how to answer this appropriately in a short time. I showed him that the horizontal decomposition would involve general (measurable) sets, not rectangles. And I also intuitively defined the outer measure of a set using open boxes. As a practice I showed that $\mathbb{Q}\cap [0,1]$ has zero measure using this definition, and as a result the Dirichlet function has zero integral on the unit interval, while it is impossible to define the Riemann integral rigorously. However, I noticed that by this point he was more or less lost when I showed $m(\mathbb{Q}\cap [0,1])=0$. I told him that to understand it properly he needed to take a year of real analysis classes, and that he should consult a professor I know in my department.

I want to ask what is a good way to explain the ideas of the Lebesgue integral to a student like him next time without making him/her confused. I later learned that the student did not have a proper proof writing background (like he did not know what it means to be injective and surjective). Since these students constitute the majority of my my classes, I feel obliged to find a way to explain myself better without forcing them to pay a visit to my professor or read a serious textbook.

  • $\begingroup$ Not a duplicate but some of the information could be helpful matheducators.stackexchange.com/questions/135/… $\endgroup$
    – quid
    Dec 7 '14 at 11:44
  • $\begingroup$ @quid: I took a look on that, but most of the "motivation" are too high-brow for low level students I have. Most of them probably never heard of Fourier series or convergence of functions. $\endgroup$ Dec 7 '14 at 12:15
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    $\begingroup$ I think I understand the issue. However, if you use instead of convergence of functions the fact that the measure is/should be sigma-addivity this might work. And, arguably, a main reasons why one uses Lebesgue integral in that context is just to have sigma-addivity. I think additivity is really intuitive as something one wants to have and then to generalize to countable unions should also be alright. I would try the message: we use this integral since in this way we get this nice property sigma-additive, while otherwise we do not. (Might expand to answer later.) $\endgroup$
    – quid
    Dec 7 '14 at 12:40
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    $\begingroup$ Why go to Lebesgue integration at this point? There are plenty of Riemann integrable probability distributions (there are plenty of finite probability distributions!) and the question can be answered in terms of them. $\endgroup$ Dec 7 '14 at 15:20
  • $\begingroup$ @DavidSpeyer: I agree. Perhaps I should not have mentioned a concept I cannot explain very well. $\endgroup$ Dec 7 '14 at 16:23

As you told the student, the easiest way is to regard the Lebesgue integral as beginning with a partition of the range, rather than the domain. Perhaps a more refined way to view this is that the partition, rather than the "heights" of the rectangles, can be used to encode the "shape" of function being integrated.

The way to encode the function in the partition is to build the parts using the natural sets $S_{\lambda}=\{t\in \mathbb{R}:f(t)> \lambda\}$. These sets behave well with respect to limits. (Of course to be honest you must talk about sigma algebras…but this is a trailer for the film!)

Since you are talking to a non-major, you could stop here and just say that following up with this we obtain natural and desirable properties that Riemann integrability is too strong to satisfy, like the Monotone Convergence Theorem. Also, you might note that the question of when a function is Riemann integrable is naturally answered using the resulting class of sets (f on [a,b] is Riemann integral if and only if it is bounded and continuous almost everywhere in terms of the Lebesgue measure). All of this is discussed on the Wikipedia page…so then you could point the student there!

Also, see the excellent book by David Bressoud for an extremely detailed account of the history and motivation.


To see the reason why Lebesgue integral is preferred in probability theory on must go beyond the setting of real functions $f \colon \mathbb{R} \mapsto \mathbb{R}$. In this setting both the Riemann and the Lebesgue integral can be defined, and reasons for choosing one of them are quite subtle.

But in probability the interest is in random variables, which are functions $X(\omega) \colon \Omega \mapsto \mathbb{R}$, say. Here $\Omega$ is a probability space, which is an abstract space which can be pretty much anything (with a $\sigma$-algebra included). Generally we cannot assume that this space carries a structure necessary to define a Riemann integral! To define the Riemann integral we would need to be able to define intervals and partitions of them in $\Omega$, and we cannot.

But the Lebesgue integral do not need such things, as we use the structure of the value space (here $\mathbb{R}$) to define it. That way it can be used in the great generality we need in probability theory.

One can in part get away with defining expectation in a bits-and-pieces manner, like one integral when the random variable $X$ has a density, a sum when $X$ has a probability mass function, some weird beast when it is a mixture of those two cases, ... But using the Lebesgue integral we can get one, unifying definition. I think that is the real reason for preferring the Lebesgue integral, not the subtleties with the convergence theorems (which is an added premium).

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    $\begingroup$ Do you use non-standard Borel probability spaces often? If not, then all of your probability spaces just look like (are Borel isomorphic) the unit interval union finitely many atoms. I love this motivation you point to, but the Borel isomorphism here gives me pause, as most probability spaces any non-major would care about are certainly standard, and saying that these spaces are terribly more exotic than the real line may not be really telling the truth! $\endgroup$
    – Jon Bannon
    Jan 27 '16 at 17:59
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    $\begingroup$ @Jon Bannon: You might be right, but how do you, in practice, define a Riemann integral on such spaces? While the Borel isomorphism is there, probability theory does not really use the probability space as if it was a unit interval. $\endgroup$ Jan 27 '16 at 18:07
  • $\begingroup$ Indeed you are right! Anyhow, nice answer! $\endgroup$
    – Jon Bannon
    Jan 27 '16 at 19:14

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