We have written in our text book:
Let $(X_{1},\mathcal{A}_{1},\mu_{1})$ and $(X_{2},\mathcal{A}_{2},\mu_{2})$ be two $\sigma-$finite measures. Define $(X,\mathcal{A},\mu):=(X_{1}\times X_{2},\mathcal{A_{1}} \otimes \mathcal{A_{2}},\mu_{1}\otimes\mu_{2})$
Let $f: X \to \bar{\mathbb R}$ be $\mathcal{A}-$measurable
Then for $g \in \{f_{-},f_{+}\}$:
$X_{1}\to [0,\infty],x_{1}\mapsto\int_{X_{2}}g(x_{1},x_{2})d\mu_{2}(x_{2})$ is $\mathcal{A_{1}}-$measurable
and
$X_{2}\to [0,\infty],x_{2}\mapsto\int_{X_{1}}g(x_{1},x_{2})d\mu_{1}(x_{1})$ is $\mathcal{A_{2}}-$measurable
Then we can use Tonelli for $f \geq 0$ a.e. as well as Fubini.
Problem:
I have a case, let's say $f(x,t):=e^{-xt}\sin{x}$ and would like to use Fubini-Tonelli for $R> 0$ on
$\int_{[0,R]}\int_{[0,\infty[}e^{-xt}\sin{x}d\lambda(x)d\lambda(t)$
Part of the solution simply states:
$\int_{[0,R]}\int_{[0,\infty[}|e^{-xt}\sin{x}|d\lambda(t)d\lambda(x)<\infty$ (drastically shortened)
Which intuitively makes sense in order to use Fubini.
However, where in this solution is shown that $f:[0,R]\times[0,\infty[\to\bar{\mathbb R}$ is indeed measurable.
Does it suffice to simply state $f$ is continuous on $[0,R]$ as well as on $[0,\infty[$
and therefore it (i.e. the function as a whole) is measurable?
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show $$\int_{[0,R]}\int_{[0,\infty[}|e^{-xt}\sin{x}|d\lambda(t)d\lambda(x)<\infty.$$