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Let A be a Borel set in R^n. Must then A + B(0,1) be Borel? Here B(0,1) is the closed ball centered at 0 of radius 1.

I know that Erdos and Stone gave an example of a compact set (it is Cantor) and a G_\delta set, whose Minkowski's sum is not Borel. But can we have an example with one of them being a closed ball?

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up vote 8 down vote accepted

Idea from http://arxiv.org/abs/1211.0430 (Example 2.4).

Take a Borel set $A' \subset [0,1]^2$ with the property that its projection to the first coordinate is not Borel. Now put this set on a cylinder in $\mathbb{R}^3$ and call it $A$. The set $$(A + B(0,1)) \cap (\mathbb{R}\times\{0\}^2)$$ (the Minkowski sum intersected with the axis of the cylinder) is the same as the non-Borel projection. Hence $A + B(0,1)$ is not Borel.

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On the duplicate thread on math.SE math.stackexchange.com/q/298494 user 5PM asked: "The answer by Tapio Rajala applies to $n \geq 3$ only. What if $n=2$? (The case n=1 has an easy affirmative answer)." –  Martin Feb 10 '13 at 0:15
    
I don't have an answer (at least yet) for $n=2$. What one can immediately say is that the intersection of the set with any line is Borel (it is the union of the translates of intersections of the set $A$ with the two lines with distance 1 from the line, and a countable number of intervals). –  Tapio Rajala Feb 10 '13 at 6:45
    
Thanks, Tapio! Actually, I also realized that the example from [Luiro, Parviainen, Saksman] paper works for my question! In fact, I posted the question while trying to find a counterexample to their Lemma 3.1 when open balls over which we take the supremum are replaced by closed balls. Incidentally, one can prove this lemma elementarily (in their proof they refer to a more complicated selection theorem). –  Liz Feb 10 '13 at 7:00
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