Does $\int_{[0,1]} f(x,y)\,dx = 0$ imply $\int_{[0,1]^2} f = 0$?

Question

I always believed that $\int_{[0,1]} f(x,y)\,dx = 0$ (for all $y$) implies $\int_{[0,1]^2} f = 0$ by Fubini-Tonelli. However, in this upvoted answer here it says the opposite.

So in particular, you shouldn't believe that $\int_{[0,1]^2} f = 0$ follows from $\int_{[0,1]} f(x,y)\,dx = 0$ for every $y$ and $\int_{[0,1]} f(x,y)\,dy = 0$ for every $x$ (take $f$ to be the characteristic function of this $E$).

Can someone clarify?

I think the function in that answer is not measurable on $[0,1]^{2}$, so the integral over $[0,1]^{2}$ is not defined. — geetha290krm, Jan 30 '24 at 09:35
But he said $f$ is the characteristic function of $E$. So if $f$ wasn't measureable then $E$ wouldn't be measurable a measurable set right? So how can $E$ have positive measure then? — LordOfNumbers, Jan 30 '24 at 09:39
From that article: "Edit: It is worth noting that the pathologic behaviour is due to the resulting sets not being measurable. For measurable sets such things cannot happen due to Fubini's Theorem." — coffeemath, Jan 30 '24 at 09:39
So this means $E$ is not a measurable set and only has positive Lebesgue outer measure? — LordOfNumbers, Jan 30 '24 at 09:44
'We cannot conclude that $\int_{[0,1]^2} f =0$' doesn't mean that the integral has a value different from $0$. It could also mean that the integral is not defined. — geetha290krm, Jan 30 '24 at 09:46
As geetha290krm recalled, the post you linked to (False beliefs about Lebesgue measure on $\mathbb{R}$) already answered your question, which I am therefore voting to close as a duplicate of it. — Anne Bauval, Jan 30 '24 at 10:08

Does $\int_{[0,1]} f(x,y)\,dx = 0$ imply $\int_{[0,1]^2} f = 0$?

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