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 $$ \mathcal{U}f=\{(x,y)\in\mathbb{R}\times [0,\infty):0\leq y<f(x)\}. $$ 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 intoroduction 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, 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.