# Why isn't integral defined as the area under the graph of function?

In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way?

The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).

What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?

Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by Fubini's theorem, so the two integrals shouldn't behave very differently.

EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points $$(x,y) \in E \times \mathbb{R}$$ where $$E$$ is a space with measure and $$y \leq f(x)$$. I assume that $$f$$ is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.

The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.

• You do realize that the point of defining the integral is to come up with a notion of "area under the graph". This notion does not exist a priori. In reality, we can't geometrically compute areas except of very specific figures, mainly rectilinear ones. As to why we have to define measure theory, perhaps look at the answer on this question: math.stackexchange.com/questions/7436/lebesgue-integral-basics – Arturo Magidin Jan 28 '19 at 19:22
• I think the question is why develop integration theory in addition to measure theory. Once you have measure theory, just define the integral to be the measure of the region under the graph. Right? – Nik Weaver Jan 28 '19 at 19:29
• Is your question why Lebesgue integral is defined as an increasing limit of areas of pluri-rectangles (with possibly contable pieces with measurable "base") instead of directly as the measure of the subgraph? It's basically the same thing, once you've proved the theorem in Lebesgue measure theory that $\mu(E)=\lim_n \mu(E_n)$ when good $E$ is approximated by good $E_n$ from the inside ($E$ increasing union of $E_n$). Take $E$ as the subgraph and $E_n$ the pluri-rectangles. – Qfwfq Jan 28 '19 at 20:10
• The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers. – Todd Trimble Jan 29 '19 at 22:01
• Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space. From the usual definitions via simple functions, this is fairly straightforward, and one can prove the standard theorems (dominated convergence, etc) without too much trouble. But if you want to use a definition as "area under the graph", you have to take a detour and construct Lebesgue measure on $\mathbb{R}$, which is a fair amount of work. – Nate Eldredge Jan 30 '19 at 5:15

Actually, in the following book the Lebesgue integral is defined the way you suggested:

Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.

First we define the planar Lebesgue measure $$m_2$$. Then we define the Lebesgue integral as follows:

Definition. The undergraph of $$f:\mathbb{R}\to[0,\infty)$$ is $$\mathcal{U}f=\{(x,y)\in\mathbb{R}\times [0,\infty):0\leq y The function $$f$$ is Lebesgue measurable if $$\mathcal{U}f$$ is Lebesgue measurable with respect to the planar Lebesgue measure and then we define $$\int_{\mathbb{R}} f=m_2(\mathcal{U}f).$$

I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:

You get the monotone convergence theorem for free: it is a straightforward consequence of the fact that the measure of the union of an increasing sequence of sets is the limit of measures.

As pointed out by Nik Weaver, the equality $$\int(f+g)=\int f+\int g$$ is not obvious, but it can be proved quickly with the following trick: $$T_f:(x,y)\mapsto (x,f(x)+y)$$ maps the set $$\mathcal{U}g$$ to a set disjoint from $$\mathcal{U}f$$, $$\mathcal{U}(f+g)=\mathcal{U}f \sqcup T_f(\mathcal{Ug})$$ and then

$$\int_{\mathbb{R}} f+g= \int_{\mathbb{R}} f +\int_{\mathbb{R}} g$$

follows immediately once you prove that the sets $$\mathcal{U}(g)$$ and $$T_f(\mathcal{U}g)$$ have the same measure. Pugh proves it on one page.

• Wow! I'm going to have to look that up. – Nik Weaver Jan 28 '19 at 20:04
• @user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+. – Nik Weaver Jan 28 '19 at 20:14
• Well, on that last point the issue is that you have to prove that $\mathcal{U}g$ and $T_f(\mathcal{U}g)$ have the same measure ... – Nik Weaver Jan 28 '19 at 20:50
• You said you first need to define the planar Lebesgue measure $m_2$. I don't remember, but doesn't defining such a measure properly take a good amount of time? And then wouldn't defining the Lebesgue integral traditionally be close already? – Basj Jan 29 '19 at 10:38
• @Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly.. – user57888 Jan 29 '19 at 11:41

If $$f: \mathbb{R} \to [0,\infty)$$ is Borel (or Lebesgue) measurable, then for each rational $$a > 0$$ define $$X_a = f^{-1}([a,\infty)) \times [0,a)$$. Then each $$X_a$$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.

I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $$\int f + \int g = \int (f + g)$$ aren't obvious if you take "measure of the region under the graph" as the definition.

• There are uncountably many of $X_a$, so I'm not sure i follow your argument. – user57888 Jan 28 '19 at 19:41
• I said rational $a > 0$. – Nik Weaver Jan 28 '19 at 19:41
• Right, right, right. Ok, sorry for my mistake. – user57888 Jan 28 '19 at 19:42
• On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument. – user57888 Jan 28 '19 at 21:08
• Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT. – Nik Weaver Jan 28 '19 at 21:50

The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).

(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $$f:E \rightarrow [0, +\infty],$$ where $$E \subseteq {\mathbb R}^n$$ is measurable] is based on the notion that the integral of a nonnegative $$f$$ should represent the volume of the region under the graph of $$f.$$

I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.

Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $$(X,\mu)$$. From the usual definitions via simple functions, this is fairly straightforward, and one can prove the standard theorems (dominated convergence, etc) without too much trouble. But if you want to use a definition as "area under the graph", you have to take a detour and construct Lebesgue measure on $$\mathbb{R}$$, which is a fair amount of work.

Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $$\mathbb{R}$$. But for finding the product measure of the region under the graph of a function $$f : X \to \mathbb{R}$$ via a Fubini-like approach, the only subsets of $$\mathbb{R}$$ you really need to measure are intervals, of the form $$[0,f(x)]$$. We don't need any fancy measure theory to tell us the length of such sets.

Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $$\mathbb{R}$$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.

• That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way. – user57888 Jan 30 '19 at 8:50

As others have pointed out, the Lebesgue integral is still computing the area under the graph. I'd just like to point out how it computes that area in a different way than the Riemann integral. For the sake of example let $$f: I \to \mathbb{R}_{\geq 0}$$ be a non-negative function on the unit interval $$I := [0,1]$$.

The Riemann-sum recipe for the integral is: take a "partition" $$I = \bigcup_{i=0}^{N-1} [t_i, t_i+1]$$ of the domain $$I$$, where $$0 = t_0 < t_1 < t_2 < \dots < t_N = 1$$. Then on each sub-interval $$[t_i, t_{i+1}]$$ of the partition, approximate $$f$$ by a constant $$c_i$$ (usually either $$\inf_{t \in [t_i, t_{i+1}]} f(t)$$ for the "lower sum" or $$\sup_{t \in [t_i, t_{i+1}]} f(t)$$ for the "upper sum"). The resulting approximation of the integral is $$\sum_i c_i (t_{i+1} - t_i)$$. Finally we take limits over finer and finer partitions of $$I$$.

On the other hand, the "Lebesgue-sum" recipe is: take a partition of the co-domain $$\mathbb{R}$$, for instance by choosing $$N \gg 0$$ and writing $$\mathbb{R} = \bigcup_{n \in \mathbb{Z}} [\frac{n}{2^N}, \frac{n+1}{2^N})$$. Now on each of the pre-images $$f^{-1}([\frac{n}{2^N}, \frac{n+1}{2^N}]) \subset I$$ approximate $$f$$ by a constant. In this case there is a fixed choice (presumably made by Lebesgue): we use $$\frac{n}{2^N}$$. Note that the $$f^{-1}([\frac{n}{2^N}, \frac{n+1}{2^N}))$$ partition the interval $$I$$, but since there's no reason for these sets to be intervals we can't calculate their length via subtraction, like we did with "$$(t_{i+1} - t_i)$$". Hence the whole concept of a measure. If $$\mu$$ is the Lebesgue measure on $$I$$, our approximation of the integral is $$\sum_n \frac{n}{2^N} \mu(f^{-1}([\frac{n}{2^N}, \frac{n+1}{2^N})))$$. Lastly we take limits over finer and finer partitions of $$\mathbb{R}$$ (e.g. let $$N \to \infty$$).

Vastly oversimplifying things:

• Riemann integral: vertical rectangles.
• Lebesgue integral: horizontal (unions of) rectangles.

There are several references to various books defining the Lebesgue integral this way in the answers, but the first person to define it this way is ... Lebesgue. In his thesis "Intégrale, Longueur, Aire" (1902) he first discusses the Darboux treatment of Riemann's integral, and then defines his own integrals geometrically, and only then goes on to define them analytically. Curiously, it seems to me that he proves that his analytic definition agrees with the geometric definition for Riemann integrable functions, but he does not really say that it's also the same for all (Lebesgue) measurable functions. Here are some quotes (I think "mesurable (J)" means "measurable in the sense of Jordan"):

1. Pour interpréter géométriquement ces nombres, attachons à toute fonction $$f$$ positive définie dans $$(a, b)$$ l'ensemble $$E$$ des points dont les coordonnées vérifient à la fois les deux inégalités

$$a\leq x\leq b \hspace{5mm} 0\leq y \leq f(x)$$

[...]

Si la fonction $$f$$ est de signe quelconque, nous lui faisons correspondre l'ensemble $$E$$ des points dont les coordonnées vérifient les trois inégalités $$a\leq x\leq b \hspace{5mm} yf(x)\geq 0 \hspace{5mm} 0\leq y^2 \leq f(x)^2.$$

L'ensemble $$E$$ est somme de deux ensembles $$E_1$$ et $$E_2$$ formés des points à ordonnées positives pour $$E_1$$, et négatives pour $$E_2$$(*). L'intégrale par défaut est l'étendue intérieure de $$E_1$$ moins l'étendue extérieure de $$E_2$$ l'intégrale par excès est l'étendue extérieure de $$E_2$$ moins l'étendue intérieure de $$E_1$$. Si $$E$$ est mesurable (J) (auquel cas $$E_1$$ et $$E_2$$ le sont) la fonction est intégrable, l'intélgrale étant $$m(E_1) - m (E_2)$$

1. Ces résultats suggèrent immédiatement la généralisation suivante: si l'ensemble $$E$$ est mesurable, (auquel cas $$E_1$$, et $$E_2$$, le sont) nous appellerons intégrale définie de $$f$$, prise entre $$a$$ et $$b$$, la quantité $$m(E_1)-m(E_2).$$ Les fonctions $$f$$ correspondantes seront dites sommables.

Relativement aux fonctions non sommables, s'il en existe, nous définirons les intégrales inférieure et supérieure comme égales à $$m_i (E_1) - m_e (E_2) \hspace{5mm} m_e (E_1) - m_i (E_2).$$

Ces deux nombres sont compris entre les intégrales par défaut et par excès.

1. Nous allons définir analytiquement les fonctions sommables.