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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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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