TITLE: Why isn't integral defined as the area under the graph of function?
QUESTION [54 upvotes]: 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.

REPLY [5 votes]: 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.