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](https://en.wikipedia.org/wiki/Fubini%27s_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.



Why isn't integral defined as the area under the graph of function?
Actually, in the following book the Lebesgue integral is defined the way you suggested:


**Pugh, C. C.** [*Real mathematical analysis*](https://link.springer.com/book/10.1007%2F978-3-319-17771-7). 
 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 $$ \mathcal{U}f=\{(x,y)\in\mathbb{R}\times [0,\infty):0\leq y<f(x)\}. $$ 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.


