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
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 


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



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. 


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.) 


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. 191192 of "Sur les fonctions representables analytiquement" in J. de math. pures et appl. (1905). Lebesgue calls Borel sets "Bmeasurable 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:
That's it. Evidently he just didn't think about the projection of an intersection. 


This question has been answered Jul 24 '13 at 20:14(see the construction of the set $B_0$ after Fact 2) on cite http://math.stackexchange.com/questions/78628/isprojectionofameasurablesubsetinproductsigmaalgebraontoacomponen?answertab=active#tabtop 

