## Projection of Borel set from $R^2$ to $R^1$

Hello

This should be easy to prove but i have no idea how to do it:

If $X \subseteq \mathbb{R}^2$ is borel then $f(X)$ is borel where $f(x,y) = x$

Thanks Tobias

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It's so easy that it's wrong, one would say. :) – Mark Schwarzmann Aug 2 2010 at 21:49
True if you replace "Borel" by "$F_\sigma$", since compact sets will project to compact sets. (But only because $\mathbb R$ is $\sigma$-compact.) – Goldstern May 23 2011 at 15:32

This is false; take a look at http://en.wikipedia.org/wiki/Analytic_set for a quick introduction. For details, look at Kechris's book on Classical Descriptive Set Theory. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of descriptive set theory.

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In fact the general analytic subset of $\mathbb{R}$ is the projection of some $G_\delta$ set in $\mathbb{R} \times \mathbb{R}$. – Gerald Edgar Aug 1 2010 at 22:30

Suslin showed that a plane Borel set exists whose projection is not a Borel set. See the references to the original article by Suslin and related works here.

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I once tracked down the exact mistake Lebesgue made in his published "proof" that the projection of a Borel set in the plane is a Borel set in the line. It came down to his claim that if $\{A_n\}$ is a decreasing sequence of subsets in the plane with intersection $A$, the the projected sets in the line intersect to the projection of $A$. Of course this is nonsense. Lebesgue knew projection didn't commute with countable intersections, but apparently thought that by requiring the sets to be decreasing this would work. (My answer is a slight expansion of Edgar's comment alluded to above.)

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Thanks for your explanation of that. I think everyone in the area knows about this "mistake" but I've never heard any sort of explanation of the cause before. – Carl Mummert Aug 2 2010 at 22:02

Since this error of Lebesgue has come up on MO a few times, it may be of interest to see his mistaken argument in its entirety. It is on pp. 191--192 of "Sur les fonctions representables analytiquement" in J. de math. pures et appl. (1905). Lebesgue calls Borel sets "B-measurable sets", and he builds them from closed intervals (or their cartesian products, in higher dimensions) by what he calls operations I and II, which are countable union and intersection.

I wish to prove that, if $E$ is B-measurable, then so is its projection. This is evident when $E$ is an interval, because then $e$ [the projection] is one too. But every B-measurable set comes from intervals by applications of operations I and II, which are preserved by projection, so the proposition is established.