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I'm trying to understand the answer to this question:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a (general) function. Is there an $N\subset \mathbb{R}^2$ with $\lambda^2(N)=0$, such that $\{(x,f(x)):x\in \mathbb{R}\}$ $\subset N$ ?

First line of the answer, found there: "No function can have a graph with positive measure or even positive inner measure, since every function graph has uncountably many disjoint vertical translations, which cover the plane. "

I don't understand why the fact that "every function graph has uncountably many disjoint vertical translations, which cover the plane" (I agree with that) implies the desired result.

Thank you for your help.

Eric Wofsey
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Alphonse
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  • It seems the OP made a mistake: the accepted answer does not prove that the graph has measure zero. It proves that the inner measure of the graph is zero. Can anyone confirm and make that clear? Thanks – Colas Aug 07 '20 at 11:27

2 Answers2

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If a function graph $G$ of positive inner measure existed, then choosing $K\subset G$ measurable of positive measure and considering the translates of $K$, you would get uncountably many disjoint measurable sets of positive measure. This is impossible in any $\sigma$-finite measure space.

Indeed, let $X=\bigcup X_n$ be a $\sigma$-finite measure space with $\mu(X_n)<\infty$ and let $\mathcal{C}$ be an uncountable collection of disjoint measurable subsets of $X$ of positive measure. Then for each $A\in\mathcal{C}$, there exist $m,n\in\mathbb{N}$ such that $\mu(A\cap X_n)>1/m$. Thus there must exist some pair $(m,n)\in\mathbb{N}^2$ such that $\mu(A\cap X_n)>1/m$ for uncountably many different elements $A\in \mathcal{C}$. But since the sets $A\cap X_n$ are all disjoint, this would imply $\mu(X_n)=\infty$ (say by choosing a countably infinite collection of such $A$ and using countable additivity).

Eric Wofsey
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  • Thank you very much! The compact $K$ exists because Lebesgue measure is inner regular (this is just for me to remember). But why are the translates of $K$ disjoint? I agree for the translates of the graph, for $K$ I'm not sure. – Alphonse Aug 21 '16 at 21:17
  • Actually, compactness of $K$ is totally irrelevant to the argument--all you need is measurability. I'm going to change that since it's potentially distracting. – Eric Wofsey Aug 21 '16 at 21:19
  • If I understand well, here $N$ is the graph of the function, supposed to have a positive inner measure (for a contradiction). We find a measurable subset of $N$ (of positive measure). Is that right? Then obviously I understand better why the translates of $K$ are disjoint. – Alphonse Aug 21 '16 at 21:21
  • Oops, yes, I got myself confused and thought you were referring to the graph as $N$. I'll fix that. – Eric Wofsey Aug 21 '16 at 21:22
  • It seems the OP made a mistake: your answer does not prove that the graph has measure zero. It proves that the inner measure of the graph is zero. Can you confirm and make that clear in your answer? Thanks – Colas Aug 07 '20 at 11:27
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    @Colas: That's correct (so it proves the graph has measure zero if it is measurable). I'll edit the question to make this clearer. – Eric Wofsey Aug 07 '20 at 14:32
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Suppose the graph had positive measure. Then we can cover the plane with uncountably many disjoint translates of this positive-measure set. Then this question completes the proof.

Patrick Stevens
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