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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!

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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: $$ \int_A\left|\int_B f(x,y)\nu(\mathrm dy)\right|\mu(\mathrm dx)\leq \int_A\int_B|f(x,y)\nu(\mathrm dy)|\,\mu(\mathrm dx)=\int_{A\times B}|f(x,y)|\,\mathrm d(\mu\times \nu)<\infty. $$

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  • $\begingroup$ 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

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