I'm confused about the generality of a result from our text book. The result reads:
Let $(\mathcal{X},\Bbb{E}, \mu)$ be a measure space and let $f$ be non-negative and $\Bbb{E}-\Bbb{B}$-measurable from $\mathcal{X}$ to $\Bbb{R}$ (written as $f\in \mathcal{M}^+(\mathcal{X}, \Bbb{E}) $). The set function $\nu : \Bbb{E} \rightarrow [0,\infty]$ defined by $$\nu(A)= \int_A f \;d\mu, \;\;\; \text{with} \; A\in\Bbb{E} $$
is a measure on $(\mathcal{X}, \Bbb{E})$
How general is this? Does this hold e.g., if $\mathcal{X}=\Bbb{R} \otimes \Bbb{R}$, $\Bbb{E}=\Bbb{B}\otimes \Bbb{B}$ and $\mu=m \otimes m$, where m is the Lebesgue measure. Or for even higher dimensions?
If $\{A_i\}$ is a sequence of disjoint sets in $\mathbb E$ then $$ \nu\left(\bigcup_{i=1}^\infty A_i\right) = \int_{\bigcup_{i=1}^\infty A_i} f\ \mathsf d\mu = \int_{\mathcal X} f\sum_{i=1}^\infty \mathsf 1_{A_i}\ \mathsf d\mu. $$ Let $f_n = f\sum_{i=1}^n \mathsf 1_{A_i}$. Then $f_n(x)\geqslant 0$ and $f_n(x)\leqslant f_{n+1}(x)$ for all $x\in\mathcal X$, so by the monotone convergence theorem we have \begin{align} \int_{\mathcal X} f\sum_{i=1}^\infty \mathsf 1_{A_i}\ \mathsf d\mu &= \int_{\mathcal X} \lim_{n\to\infty}f_n\ \mathsf d\mu\\ &= \lim_{n\to\infty} \int_{\mathcal X} f_n\ \mathsf d\mu\\ &=\lim_{n\to\infty} \int_{\bigcup_{i=1}^n A_i} f\ \mathsf d\mu\\ &=\lim_{n\to\infty} \sum_{i=1}^n \nu(A_i)\\ &= \sum_{i=1}^\infty \nu(A_i). \end{align} It follows that $\nu$ is a measure on $(\mathcal X,\mathbb E)$.