Lebesgue non measurability in the plane Suppose A is contained in the unit square of R^2, and the projection of A on any line outside the unit square is not Lebesgue measurable in R. Does that imply that A is not Lebesgue measurable in the plane?
 A: No, it doesn't. Take a non-measurable set $S$ on a line segment, and make this line segment to be a side of $A$. Then $S$ as a subset of the plane is measurable (it is of measure zero), while its projection on the line segment (i.e. itself) is not.
EDIT: As Pietro Mayer says one needs a union of a few isometric copies of my $S$ to yield an example whose projection on every line is non-measurable.
A: This is an answer to Gerald Edgar's question (a) (see comment just after main question). To produce a set such that all but two projections are non-measurable, take a subset X of [0,1] that's not measurable in any interval and put a copy of X into the segment from (0,0) to (1,0) and a copy of the complement of X into the segment from (1,0) to (1,1). If we project to the Y axis we get two points, so it's measurable. If we project to the X axis, we get the interval [0,1], so it's measurable. If we project in any other direction, then the images of the two segments are intervals of positive length that do not share their endpoints, so by the local non-measurability of X the resulting projection is non-measurable. My guess is that similar tricks can be used to do other things but I haven't thought about this.
Edit: I also have a very silly answer to Gerald Edgar's question (b), which is that the existence of such an example is consistent with ZF. It is known that there can be non-measurable sets of cardinality less than the continuum (of course, this means that CH fails). Let $K$ be such a set and let $X=K\times\mathbb{R}$. Now let $Y$ be the union of uncountably many rotations of $K$, but not continuum many. Finally, let $Z$ be the complement of $Y$ in the plane. If one of the rotates of $K$ is through an angle of $\theta$, then the projection of $Z$ on to the line $L_\theta$ that makes an angle of $\theta$ with the x-axis misses all points in $K$ (or rather the obvious copy of $K$ along that line). It also contains each point not in $K$, since if P is such a point, then we have removed fewer than continuum many points from the line perpendicular to $L_\theta$ that goes through P. Therefore, this projection is just a copy of the complement of $K$ and so non-measurable. If $L$ is any line that is not perpendicular to one of the $L_\theta$ for which $\theta$ is one of the angles we chose, then again we have removed fewer than continuum many points from $L$. Therefore, if we project onto $L_\phi$ for some $\phi$ that is not one of our chosen angles, then we obtain the whole of $L_\phi$, which is of course measurable. So we have uncountably many measurable projections and uncountably many non-measurable ones. 
I don't know what happens if we ask for continuum many measurable projections and continuum many non-measurable projections ... 
A: The subset S is not as required.
Its projection on the X-axis is not measurable, but it certainly does satisfy it for every line segment outsides the unit squrae. For instance, its progection on a line parallel to the y-axis is a single point, which is measurable in R.
A: Partial answere (anyway I guess no). 
Take $S\subset [0,1]$ non mesaurabile and let $A:=S\times ${0}$\cup${0}$\times S, \ B:=\mathbb{R}\times ${0}$\cup${0}$\times \mathbb{R}$, then $A\subset B\subset \mathbb{R}^2$  and $A$ and $B$ have measure 0. If the projection is  bijective is a  omeomorphism from  $B$ to the line, then the image of this proiection is no measurable:
we have that  $A\subset B$ isnt measurable (on $B$ consider the linear lebesgua measure, if we suppose $A$ measurable the also $A\setminus (0\times \mathbb{R})=S\times 0$ is measurable, absurd) 
further a projection is   continuous, then  meaurable i.e. the inverse image of a measurble subset is still measurable. If a  projection is a  omeomorphism  then establishes an identification (or bijection) between the measurable sets. 
THe some   follow  if the line is vertical or horizontal (the image of prjection is essentally $S$).
Let now $f: B\to l $ a projection , no  injective and $l$ no vertical or horizontal.
then $f(S\times 0)$ and $f(0\times S)$ are a homothetic (dilatate or traslate) copies of $S$. 
Then the problem reduces to finding a no-measurable subset $ S $ such that the union of two of its homothetic images is  no-measurable.   
