More on Lebesgue non-measurability - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:53:02Z http://mathoverflow.net/feeds/question/71983 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71983/more-on-lebesgue-non-measurability More on Lebesgue non-measurability gowers 2011-08-03T11:02:45Z 2011-08-06T03:52:47Z <p>I have just read <a href="http://mathoverflow.net/questions/71976/lebesgue-non-measurability-in-the-plane" rel="nofollow">the following question about measurable and non-measurable sets</a>. 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 $\mathbb{R}^2.$ </p> <p>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.</p> <p>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 $\pi(X\Delta Y).$</p> http://mathoverflow.net/questions/71983/more-on-lebesgue-non-measurability/71985#71985 Answer by Emil Jeřábek for More on Lebesgue non-measurability Emil Jeřábek 2011-08-03T11:12:50Z 2011-08-03T11:12:50Z <p>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.</p> http://mathoverflow.net/questions/71983/more-on-lebesgue-non-measurability/71993#71993 Answer by Péter Komjáth for More on Lebesgue non-measurability Péter Komjáth 2011-08-03T13:06:32Z 2011-08-03T13:06:32Z <p>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 <code>$\{(L_\alpha,P_\alpha):\alpha&lt;2^\omega\}$</code>. 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 <code>$\{x_\beta:\beta&lt;\alpha\}$</code> 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 ($\beta&lt;\alpha$). Then we can choose $x_\alpha\in K$ such that it is not in any of the lines $L_\beta$ ($\beta\leq\alpha$). Finally, $B={x_\alpha:\alpha&lt;2^\omega}$ is as required. </p> http://mathoverflow.net/questions/71983/more-on-lebesgue-non-measurability/72222#72222 Answer by Michael Hardy for More on Lebesgue non-measurability Michael Hardy 2011-08-06T03:52:47Z 2011-08-06T03:52:47Z <p>"is there a way of somehow ruling out any use of this principle and thereby obtaining a more challenging question?"</p> <p>How about this: Does there exist a Borel subset of $\mathbb{R}^2$ all of whose projections are non-measurable?</p>