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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 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.

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 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.

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.

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 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.

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.

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.