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My question is almost the same as this question.

Let $f(\cdot, \cdot): R^2 \rightarrow R$. Suppose that:

(a) Fix any $y \in R, f(\cdot, y)$ is continuous almost everywhere.

(b) Fix any $x \in R, f(x, \cdot)$ is continuous almost everywhere.

Is $f(\cdot, \cdot): R^2 \rightarrow R$ is continuous almost everywhere?

I didn't really understand the reasoning in the comments of the linked question. Any help is most appreciated.

Canine360
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  • I think I have a counter example. Define $f(x,y):=0$ if $x$ or $y$ are irrational and otherwise, if $x=a/b$ and $y=c/d$ (in irreducible fractions), define $f(x,y):=1/(|b-d| + 1)$. Then $f(x,.)$ and $f(.,y)$ are still continuous at every irrational number, but $f(x,y)$ is discontinuous at every point. If $x$ and $y$ are rational then it's clear, and otherwise there are arbitrarily close rational points $x'$, $y'$ with the same denominator in their irreducible fraction. – fweth Jul 03 '23 at 16:45
  • The result, or a counterexample, may be mentioned in one or both of the references I cite in my answer to Functions continuous in each variable (one is a published paper, the other is a 2005 sci.math post of mine), but I don't have time now to look into (or give any thought to) this. – Dave L. Renfro Jul 03 '23 at 17:28

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