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Generally speaking, a set $E$ can have boundary of positive measure. For example, the set of rational points in a unit disk. What if $E$ is a star-shaped set? Intuitively the boundary can not be "think", otherwize it's not star-shaped.

I have no idea how to prove or disprove the claim. Any ideas or comments will be appreciated.

By the way, there is an answer here Set $E\subset \mathbb{R}^n$ of positive Lebesgue measure such that the Lebesgue measure of its boundary is zero saying fat Cantor set is a star-shaped set with boundary of positive measure, but actually it's not star-shaped.

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    Hint: connect the center of unit circle with all points which make rational angle with $X$ axis... (btw, you have just deleted and undeleted this question, I'm curious, what was the reason for that?) – Wojowu Nov 22 '16 at 08:04
  • It was because I found a checked answer saying fat cantor set is star-shaped but has boundary of positive measure. Then after deleting my post and thinking for seconds, I realized fat cantor set is not star-shaped, so I posted my question again... – student Nov 22 '16 at 08:07
  • @Wojowu, thank you very much! I got it! – student Nov 22 '16 at 08:10
  • That answer doesn't say the fat Cantor set is star-shaped. Instead, it sayd that after connecting all of its points to $(0,1)$ we get a star-shaped set. – Wojowu Nov 22 '16 at 08:11
  • @Wojowu, I think the "cross" connecting would intersect the removed rectangles inherited from Cantor set, which doesn't satisfy the condition of star-shaped set. – student Nov 22 '16 at 08:15
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    Fat Cantor set is a subset of a line; there are no rectangles there, just intervals. It looks like this, but with the fat Cantor set: https://en.wikipedia.org/wiki/Knaster%E2%80%93Kuratowski_fan – Wojowu Nov 22 '16 at 08:17
  • @Wojowu, I see. I was thinking something else! Thanks for all the useful comment. – student Nov 22 '16 at 08:20

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