Projection of Borel set from $R^2$ to $R^1$ 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
 A: 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.
A: 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.)
A: 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.)
A: 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.
A: 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
