# A converse version of Fubini's Theorem

Assume two $$\sigma-$$finite measure spaces $$(X,\mathcal{M},\mu)$$ and $$(Y,\mathcal{N},\nu)$$. Consider a $$\mathcal{M}\times\mathcal{N}$$-measurable function $$f$$, and we are interested in computing $$\int f d\left(\mu \times \nu\right)$$ where $$\mu \times \nu$$ is the product measure. I know the results of Fubini's theorem. Lets assume that one first observes that $$f(x,y)\in L^{1}(\mu)$$ for fixed $$y$$ and $$f(x,y)\in L^{1}(\nu)$$ for fixed $$x$$. Then, he naively computes $$\int \left[ \int f(x,y) d\mu(x) \right]d\nu(y)$$ and $$\int \left[ \int f(x,y) d\nu(y) \right]d\mu(x)$$ . and, he notices that the value of these two integral exist and are equal.

I have two questions:

1) Whether we can find a case as I described above?

2) Provided that we can find such an example, can we conclude that $$f \in L^{1}(\mu \times \nu)$$, and value of $$\int f d\left(\mu \times \nu\right)$$ equal to the $$\int \left[ \int f(x,y) d\nu(y) \right]d\mu(x)=\int \left[ \int f(x,y) d\mu(x) \right]d\nu(y)$$.

EDIT: Answer: Assume the following function

$$f=\begin{cases} 0 & (x,y)=(0,0)\\\frac{xy}{(x^2+y^2)^2} & \mbox{else}\end{cases}$$ It is exactly an example which satisfies the above conditions.

• No. It is not sufficient that the iterated integrals exist and are equal. See e.g. here: math.stackexchange.com/questions/1812264/… Nov 8 '19 at 23:17
• Thank you very much. It helps a lot! Nov 8 '19 at 23:23