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Let $A \subset \mathbb{R}^n$ and $x \in \mathbb{R}^n$( arbitrary). Is $B \setminus(A+x) = B \cap (A+x)^c = B \cap A^c + x$?($A+x=\{a+x: a \in A \}$ and $A\setminus B = A \cap B^c$)

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up vote 2 down vote accepted

I do not think this is true. Let $A=B=[0,1]$, $x=0.5$. Then $A+x=[0.5,1.5]$, and $B\setminus(A+x)=[0,0.5)$. But we know $B\cap A^{c}=\emptyset$, therefore $B\cap A^{c}+x=\emptyset$. So the two sets are not equal.

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user32240: with assumptions above does $(A+x)^c = A^c+x^c$ or $(A+x)^c = A^c+x$ hold? What about does $B \cap (A+x) = B \cap A+x$ hold? – alvoutila Jan 26 '13 at 11:21
@alvoutila: Let $x=0$, then $A^{c}\not=A^{c}+R/\{0\}$ in general. The second one looks fine to me. The third is not right - Let $B=[3,4],A=[1,2]$, then $B\cap A+2=\emptyset$, but $B\cap (A+2)=B$. – Bombyx mori Jan 26 '13 at 11:32
User32240: What about if $A \subset \mathbb{R}^n $, $x \in \mathbb{R}^n$ and A is measurable; how to show that $x+A$ is measurable? some hint. – alvoutila Jan 26 '13 at 19:08
@alvoutila: I suggest you take a look at translation invariance of Lesbegue measure. – Bombyx mori Jan 27 '13 at 0:43
User32240: Yes, but one have to use also information that it is enough to prove that $m*(A) \geq m*(A \cap E) + m*(A \setminus E)$ – alvoutila Jan 27 '13 at 11:05

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