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Suppose $f:[0,1]\times[0,1]\to\Bbb R$ is R-integrable. Is it true that, for each $\alpha\in[0,1],$ at least one of the functions $x\mapsto f(x,\alpha)$ and $y\mapsto f(\alpha,y)$ has to be R-integrable?

I've seen an example of a non-R integrable function $f:[0,1]\times[0,1]\to\Bbb R,$ $$f(x,y)=\begin{cases}1,&x\in\Bbb Q\\ 2y,&x\notin\Bbb Q\end{cases}$$

In this example, $y\mapsto f(\alpha,y)$ is integrable, but $x\mapsto f(x,\alpha)$ fails to be $\forall\alpha\in[0,1]\setminus\left\{\frac12\right\}$ and so does $f$ over $[0,1]\times[0,1]$ as proven by RRL.

I tried building something from the answer in the thread, but I couldn't find any example of a function that isn't integrable "along any segment in $[0,1]\times[0,1].$ Does anyone have any suggestions?

Matcha Latte
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  • You can also show that if $f$ is Riemann integrable on any product of intervals $A \times B$. Then $\int_A f(x,\alpha) , dx$ exists for almost every $\alpha \in B$ and $\int_B f(\alpha,y) , dy$ exists for almost every $\alpha \in A$. – RRL Mar 08 '22 at 21:57
  • @RRL, in this context, almost every means except for finitely many ? – Matcha Latte Mar 08 '22 at 22:10
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    It means except for $\alpha$ in a set of Lebesgue measure zero which could possibly be infinite. – RRL Mar 08 '22 at 22:12
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    So if $\int_{A \times B} f$ exists as a Riemann integral you could never have $\int_A f(x,\alpha) ,dx$ not exist for $\alpha$ in some interval or any set of positive measure. – RRL Mar 08 '22 at 22:17
  • @RRL, thank you! I'm going to think more about it in the morning and than maybe follow-up, if that's okay. – Matcha Latte Mar 08 '22 at 22:19

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Make the simplest possible function that is Riemann integrable on the square and yet fails to be Riemann integrable on both $x=0$ and $y=0$. Take, for example, the function $$f(x,y) = \begin{cases} 0, & x=0, y\in\Bbb Q \text{ or } y=0, x\in\Bbb Q \\ 1, & x=0, y\notin\Bbb Q \text{ or } y=0, x\notin\Bbb Q \\ 5, & \text{otherwise} \end{cases}.$$

Ted Shifrin
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