Is the Lebesgue integral equal to the Lebesgue measure of the enclosed region?

Question

Let $u:\mathbb{R}\to[0,+\infty]$ be Lebesgue integrable, and let

$$A:=\{(x,y)\in\mathbb{R}^2 \; | \; 0\leqslant y\leqslant u(x)\}.$$

Thus $A$ is the set of all points enclosed between the graph of the function $u(x)$ and the $x$-axis.

Question: $$\int u(x)\,\lambda^1(dx) \overset{?}{=} \lambda^2(A). $$

The left side is the Lebesgue integral of $u(x)$, whereas the right side is the Lebesgue measure (on $\mathbb{R}^2$) of the set of points enclosed by the graph and the $x$-axis. Of course, one would expect these to be equal, at least on well-behaved functions.

Answer 1

$\lambda^{2}(A)=\int_{\{(x,y): 0\leq y \leq u(x)\}} dxdy=\int \int I_{\{0 \leq y \leq u(x)\}}dy dx=\int u(x)dx$ because $\int_0^{u(x)} 1dy=u(x)$.

Justification: Fubini/Tonelli.