# What exactly do I need to show before using Fubini-Tonelli

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 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. – 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.