# Closure of the support

Let $X$ be [0,1], function $\phi:X\times X\to\mathbb{R}^+$ is Lipschitz continuous on $X\times X$. Define $$G = supp(\phi) = \overline{\{(x,y)|\phi(x,y)>0\}}.$$ and its section $$s(x) = \{y\in X|(x,y)\in G\}.$$

Is it true that $s(x) = \overline{\{y\in X|\phi(x,y)>0\}}$?

Edited: Still is it true that $\overline{\{y\in X|\phi(x,y)>0\}}\subset s(x)$? If yes, what can we say about $s(x)\setminus \overline{\{y\in X|\phi(x,y)>0\}}$?

Edited2: Are we sure now that $\partial A'$ is of measure zero, where $$\bigcup\limits_{x\in A}s(x)$$ for an arbitrary set $A$?

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## 2 Answers

No. Hint: find $\phi$ such that $s(\frac12) =\{\frac12\}$ and $\{y\in X\,|\,\phi(\frac12,y)>0\}$ is empty.

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I've edited. Seems obvious for me, but want to be sure. –  Ilya Jun 8 '11 at 16:27
I've edited one more time ) –  Ilya Jun 8 '11 at 16:48

No. Let $\phi(x,y)=x$ and consider $s(0)$.

The support of $\phi$ will be all of $X\times X$, but the section of $x=0$ is in the support because of non-zero values of $x$.

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Thank you for the explanation. –  Ilya Jun 8 '11 at 16:51