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I have just read the following question about measurable and non-measurable sets. Does there exist a measurable subset of $\mathbb{R}^2$ all of whose projections are non-measurable? As with many such questions, there is an easy solution based on the general principle that measure-zero sets in the plane can be very nasty: you can just take your favourite non-measurable set in $\mathbb{R}$ and think of it as a subset of $\mathbb{R}^2.$

Just for fun, here is a meta-question: is there a way of somehow ruling out any use of this principle and thereby obtaining a more challenging question? One idea that fails miserably is to insist that the subset of $\mathbb{R}^2$ has positive measure. That fails because all you have to do is take the union of a nasty measure-zero set with a token nice set of positive measure that doesn't cause any of the projections to become measurable. And that is easy.

Here is a different idea, which comes with a warning that I've only just thought of it so the question has a very good chance of not being interesting. Let X be a measurable subset of the plane. Does there necessarily exist a measure-zero subset Y of X and a projection $\pi$ such that $\pi(X\setminus Y)$ is measurable? In case asking for Y to be a subset of X is too much of a restriction, an alternative question would be for Y to be an arbitrary set of measure zero and consider $\pi(X\Delta Y).$

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also, one may strengthen the requirement of the linked question asking for a set all of whose projections are "locally non-measurable" sets, meaning that they meet any non-empty open set of the line in a non-measurable set. –  Pietro Majer Aug 3 '11 at 11:11
    
Perhaps, is it more natural to ask "almost all projections" instead of "all projections"? –  Qfwfq Aug 3 '11 at 12:38
    
If you take your favorite non-measurable subset of $\mathbb{R}$ and think of it as a subset of $\mathbb{R}^2$, then one of its projections is measurable, and has measure 0. At least if you mean a subset of a line in $\mathbb{R}^2$. –  Michael Hardy Aug 6 '11 at 12:11
    
.....but then two copies of it on non-parallel lines takes care of that. –  Michael Hardy Aug 6 '11 at 12:32
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3 Answers

The answer to the third paragraph is positive. Any measurable $X$ contains an $F_\sigma$ set $Z$ of the same measure. Take $Y=X\smallsetminus Z$. Since continuous images of compacts are compact, any projection of a $F_\sigma$ subset of the plane is again $F_\sigma$, hence measurable.

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Sigh -- I should have thought harder about the question before posting it. But thanks for the answer. –  gowers Aug 3 '11 at 14:36
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I have a construction which is probably overkill, but may be flexible enough to answer the "real" question (whatever that is). Let $A$ be a measure-zero planar set which meets all lines in continuum many points. (It is easy to find, e.g., take the union of countably many products of the interval and the Cantor-set) We select a subset $B\subseteq A$ such that all projections of $B$ are Bernstein sets (a set on a line is Bernstein if neither it nor its complement contains a perfect set). Enumerate all pairs consisting of a line $L$ and a perfect set $P$ on $L$ as $\{(L_\alpha,P_\alpha):\alpha<2^\omega\}$. At step $\alpha$ we choose a point $x_\alpha$ and a line $M_\alpha$ with the intention of promising $x_\alpha\in B$ and $B\cap M_\alpha=\emptyset$. Assume we arrived at step $\alpha$. We have some less than continuum many points $\{x_\beta:\beta<\alpha\}$ which are promised into $B$, and some less than continuum many lines, which are promised to be disjoint from $B$. We are given a line $L_\alpha$ and a perfect set $P_\alpha$ on it. Choose a point $z$ in $P_\alpha$ which is not the projection of some $x_\beta$ (possible, as $P_\alpha$ has cardinality continuum, and less than that many points have been chosen). Erect a line on $z$, perpendicular to $L_\alpha$, that will be $M_\alpha$. Next, choose an element $y\neq z$, $y\in P_\alpha$ such that the line $K$ throu $y$, perpendicular to $L_\alpha$ is not one of the $M_\beta$'s ($\beta<\alpha$). Then we can choose $x_\alpha\in K$ such that it is not in any of the lines $L_\beta$ ($\beta\leq\alpha$). Finally, $B=\{x_\alpha:\alpha<2^\omega\}$ is as required.

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"is there a way of somehow ruling out any use of this principle and thereby obtaining a more challenging question?"

How about this: Does there exist a Borel subset of $\mathbb{R}^2$ all of whose projections are non-measurable?

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Projections of Borel sets are Lebesgue measurable. –  Andreas Blass Aug 6 '11 at 4:36
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