I have been attempting to calculate the following Lebesgue integral, but I'm not very confident that my method is correct, so I would appreciate it if someone could take a look and let me know if this is right. $$ \int_{[-\pi/2,\pi]\times[0,\pi]}\sin{(x+y)}\hspace{1mm} dm(x,y) $$
We begin by showing that the following integral is finite: $$ \int_{[-\pi/2,\pi]\times[0,\pi]}|\sin{(x+y)}|\hspace{1mm} dm(x,y) $$
But $\sin{(x+y)} \leq 1$ so by monotonicity of the Lebesgue integral we have: \begin{align*} \int_{[-\pi/2,\pi]\times[0,\pi]}|\sin{(x+y)}|\hspace{1mm} dm(x,y) & \le \int_{[-\pi/2,\pi]\times[0,\pi]}1\hspace{1mm} dm(x,y) \\ & = \int_{-\pi/2}^{\pi}\int_{0}^{\pi}1\hspace{1mm} dy dx =\int_{-\pi/2}^{\pi} \pi \hspace{1mm} dx = \frac{3\pi^{2}}{2} < \infty \end{align*}
So we can calculate our integral as an iterated integral: \begin{align*} \int_{[-\pi/2,\pi]\times[0,\pi]}\sin{(x+y)}\hspace{1mm} dm(x,y) & =\int_{[-\pi/2,\pi]}\int_{[0,\pi]}\sin{(x+y)}\hspace{1mm} dm(y)dm(x) \\ & =\int_{[-\pi/2,\pi]}[-\cos(x+y)]^{\pi}_{0}\hspace{1mm} dm(x) \\ & =\int_{[-\pi/2,\pi]}[\cos(x)-\cos(x+\pi)]\hspace{1mm} dm(x) \\ & = [\sin(x)-\sin(x+\pi)]^{\pi}_{-\pi/2} =2. \end{align*}