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

5 Answers 5


This is false; take a look at https://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. Another good reference is A second course on real functions, by van Rooij and Schikhof.

Here are some additional details (that I added to another thread). Recall that the analytic sets are the empty set, and the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$. This notion makes sense in arbitrary topological spaces, and has been particularly studied in Polish spaces (separable completely metrizable spaces).

  • The analytic sets are closed under countable unions and intersections, and contain the Borel sets. In all uncountable Polish spaces, there are analytic sets whose complement is not analytic. Suslin proved that the Borel sets of uncountable Polish spaces are precisely those analytic sets whose complement is analytic.

  • In Baire space $\mathcal N=\mathbb N^{\mathbb N}$ (the countable product of discrete $\mathbb N$), analytic sets are the projections of closed subsets of ${\mathcal N}^2$. This is false with $\mathbb R$ in place of $\mathcal N$. Note that in that case, closed sets are countable union of compact sets, so their projections are $F_\sigma$. Baire space plays a special role in the theory, since a nonempty set in a Polish space is analytic iff it is the continuous image of $\mathcal N$.

The actual results in ${\mathbb R}$ are as follows:

  • A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

  • A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

  • There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

  • A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

  • 3
    $\begingroup$ In fact the general analytic subset of $\mathbb{R}$ is the projection of some $G_\delta$ set in $\mathbb{R} \times \mathbb{R}$. $\endgroup$ Aug 1, 2010 at 22:30
  • $\begingroup$ It should be noted that the empty set is an image of a Borel subset of R, i.e, itself, by any Borel function. So you need not specify it separately. $\endgroup$
    – Alon Gurny
    Mar 24 at 6:39

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 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.)

  • 2
    $\begingroup$ 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. $\endgroup$ Aug 2, 2010 at 22:02

This was a famous mistake made by Lebesgue (see also Gerald Edgar's answer to this MO question).

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.


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.

In my translation it reads:

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.

That's it. Evidently he just didn't think about the projection of an intersection.

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    $\begingroup$ Great specimen for "Common false beliefs" collection! $\endgroup$ Aug 3, 2010 at 0:09

This question has been answered Jul 24 '13 at 20:14(see the construction of the set $B_0$ after Fact 2) on cite https://math.stackexchange.com/questions/78628/is-projection-of-a-measurable-subset-in-product-sigma-algebra-onto-a-componen?answertab=active#tab-top

  • $\begingroup$ The question had a satisfactory answer when the question was raised on the former site. Now this question is moved on other site and the satisfied answer I remained on a former site is on the previous place without knowing for what there it is. $\endgroup$
    – George
    Jan 5, 2014 at 12:59
  • 1
    $\begingroup$ Sorry, but I’m lost. What was moved where and remained elsewhere and what knows what is where? $\endgroup$ Jan 5, 2014 at 17:18

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