Fubini' theorem says

Suppose $A$ and $B$ are complete measure spaces with measures $\mu$ and $\nu$. Suppose $f(x,y)$ is $A \times B$ measurable. If $$ \int_{A\times B} |f(x,y)|\,\text{d}(\mu \times \nu)<\infty, $$ where the integral is taken with respect to a product measure on the space over $A \times B$, then $$ \int_A\left(\int_B f(x,y)\,\text{d}\nu\right)\,\text{d}\mu=\int_B\left(\int_A f(x,y)\,\text{d}\mu\right)\,\text{d}\nu=\int_{A\times B} f(x,y)\,\text{d}(\mu \times \nu), $$

I was wondering if

  • $f(x,) \in L^1(B), \forall x \in A$?
  • $\int_B f(x,y) d\nu \in L^1(A)$ ?
  • the reason $A$ and $B$ are complete measure spaces?

Thanks and regards!


See this discussion on MathOverflow for a discussion on sigma-finiteness and completeness of the measure space.

In general you can't say that $f(x,\cdot)\in \mathcal{L}^1(B)$ for all $x\in A$ but only that $$ \{x\in A\mid f(x,\cdot)\notin\mathcal{L}^1(B)\} $$ is a measurable (with respect to the sigma-algebra on $A$) null-set. As to the second item: $$ \int_A\left|\int_B f(x,y)\nu(\mathrm dy)\right|\mu(\mathrm dx)\leq \int_A\int_B|f(x,y)\nu(\mathrm dy)|\,\mu(\mathrm dx)=\int_{A\times B}|f(x,y)|\,\mathrm d(\mu\times \nu)<\infty. $$

  • $\begingroup$ Thanks! is the set "measurable null set" $\{x∈A∣ f(x,⋅)∈ L^1(B)\}$ or $\{x∈A∣ f(x,⋅)\notin L^1(B)\}$? $\endgroup$ – Tim Feb 22 '13 at 14:16
  • $\begingroup$ It's the last one - edited accordingly :) $\endgroup$ – Stefan Hansen Feb 22 '13 at 15:25
  • $\begingroup$ And, of course, proof of the fact that this null set is measurable requires completeness of the measure $\mu$... $\endgroup$ – GEdgar Feb 22 '13 at 15:40
  • $\begingroup$ @GEdgar: It does? $\endgroup$ – Stefan Hansen Feb 22 '13 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.