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Let $f(x, y)$ be a Lebesgue integrable function on $\mathbb{R}^{n} = \mathbb{R}^{n_{1}} \times \mathbb{R}^{n_{2}}$. By $f^{y}(x)$, we mean the function of $x$ obtained by fixing $y$. Furthermore, suppose that $f^{y}$ is integrable on $\mathbb{R}^{n_{1}}$ and $\int_{\mathbb{R}^{n_{1}}}f^{y}(x)\, dx$ is integrable on $\mathbb{R}^{n_{2}}$. Why does there exist a set $A$ in $\mathbb{R}^{n_{2}}$ of measure 0 such that $f^{y}(x)$ is Lebesgue integrable on $\mathbb{R}^{n_{1}}$ whenever $y \not\in A$?

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up vote 2 down vote accepted

You have to suppose f measurable and integrable, that is assured bu Fubini's Theorem, whose proof uses an approach with simple function and taking limits!

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Sorry, I have edited the question to clarify what I was asking about. – Shayla Mar 12 '12 at 12:20
What he wrote is still an answer. If you change everything to Borel measurable, then the cuts are actually all measurable. The issue with Lebesgue integration is that when you complete the sigma algebra in the product space you get a bunch of sets of measure zero crossed with non-measurable sets. When you project those non-measurable sets to the lower space, issues can pop up. The thing is though the function you started with is measurable, so you cannot have a set of positive measure where the cuts are non-measurable and similarly for integrability. – Chris Janjigian Mar 12 '12 at 13:06

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