# Must the Minkowski sum of a Borel set and a *closed* ball be Borel?

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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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.
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