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Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}$ be such that for every $a,b \in \mathbb{R}$ the fuctions $F(a,\cdot), F(\cdot,b) \in L^1(\mathbb{R})$.

Consider the set $A:= \left\{a:\lim\limits_{r \rightarrow 0}\frac{1}{\mu{(B(a,r))}}\int_{B(a,r)}F(a,y) d y = F(a,a) \right\}$, where $\mu$ is the Legesgue measure on $\mathbb{R}$.

Is $\mu(\mathbb{R}/A)=0$?

P.S. By Lebesgue differentiation theorem for every $a\in \mathbb{R}$ the set of all Legesgue points of the function $F(a,\cdot)$ denoted by $A_a$ is measurable and satisfies $\mu (\mathbb{R} / A_a)=0$ .

The set $A$ defined above is also given by $A= \cup_{a\in \mathbb{R}} (A_a \cap \{a \})$. But I am unable to conclude the value of $\mu(\mathbb{R}/A)$

Any help would be greatly appreciated. Thanks in advance.

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  • $\begingroup$ How is your $A_a$ defined? $\endgroup$
    – Kenny Wong
    Nov 20 at 19:20
  • $\begingroup$ Thanks. Updated the question $\endgroup$
    – Veronica
    Nov 21 at 12:34

1 Answer 1

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Try $F(x,y) = 0$ for $x\ne y$, $1$ for $x=y$.

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  • $\begingroup$ I did not follow you. Is this a counter example? $\endgroup$
    – Veronica
    Nov 21 at 12:35
  • $\begingroup$ @Veronica I think you get that $A=\emptyset$ in this example, which is why its a counter example to your statement. $\endgroup$
    – Keen-ameteur
    Nov 21 at 13:26

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