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


$X_{2}\to [0,\infty],x_{2}\mapsto\int_{X_{1}}g(x_{1},x_{2})d\mu_{1}(x_{1})$ is $\mathcal{A_{2}}-$measurable

Then we can use Tonelli for $f \geq 0$ a.e. as well as Fubini.


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


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?

  • $\begingroup$ Yes, it is because $f$ continuous on $[0,R]$ and on $[0,\infty[$ implies $f$ continuous on $[0,R] \times [0,\infty[$ (as a whole) and this implies that $f$ is measurable. $\endgroup$ – mathcourse Jan 13 at 11:30

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 $$\int_{[0,R]}\int_{[0,\infty[}|e^{-xt}\sin{x}|d\lambda(t)d\lambda(x)<\infty.$$


You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,\infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] \times[0,\infty)$ to $\mathbb R$ is Borel measurable.


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