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Since my question was not answered on MSE, I would like to ask it here.

Let $\mu$ be a finite Borel measure on the plane. Does there exist a characterization of the property that almost all (wrt rotations) projections of $\mu$ to lines on the plane are absolutely continuous with respect to the Lebesgue measure? I think this could be equivalent to the fact that $\mu e=0$ for any set $e$ of zero linear Hausdorff measure? (This property is obviously necessary, but I don'k know if it is also sufficient.)

References to related results will also be appreciated.

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    $\begingroup$ It is not sufficient: The 1/4 Cantor square with the $H^1$ measure on it has almost every projection singular. We can even thicken it a bit and still have this property. $\endgroup$ – fedja Sep 17 '13 at 12:28
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The answer to the question is that it almost exists a characterization, but (as far as I know) there is a gap.

There are a number of projection theorems in geometric measure theory, which have the following flavor: if a measure or set in $\mathbb{R}^2$ satisfies some property, then almost all orthogonal projections satisfy this other property.

Let me start with the slightly more transparent problem for sets. Marstrand's Projection Theorem of 1954 says that if $A$ has Hausdorff dimension strictly larger than $1$, then almost all of its orthogonal projections indeed have positive length (one-dimensional Lebesgue measure). If $A$ has Hausdorff dimension strictly less than $1$, then so do all of its projections, which in particular have zero length.

What about the critical case in which $A$ has Hausdorff dimension $1$? Here the situation is more subtle. There are sets of Hausdorff dimension $1$ for which all orthogonal projections have positive length (a circle) and sets of Hausdorff dimension $1$ for which almost all projections have zero length (the "four courner" Cantor set constructed by replacing the unit square by 4 squares of side-length 1/4 placed in the corners, and continuing inductively). Among the sets of finite (or $\sigma$-finite) linear Hausdorff measure $\mathcal{H}^1$, however, Besicovitch characterized sets for which almost all projections have positive length: they are sets which have positive $\mathcal{H}^1$ measure themselves, and are not purely unrectifiable. A Borel set is called purely unrectifiable if it meets every Lipschitz curve in zero length. Arbitrary Borel sets can be decomposed into a rectifiable (countable union of Lipschitz curves) and a purely unrectifiable part; what Besicovitch really proved is that for purely unrectifiable sets, almost all projections have zero length.

The situation where $A$ has Hausdorff dimension $1$ but non-$\sigma$-finite linear Hausdorff measure to my knowledge is not completely understood.

The situation for measures is analogous, once one figures out the correct notions of dimension and pure unrectifiability to consider. For dimension, one needs the lower Hausdorff dimension of a measure, defined as the infimum of the Hausdorff dimensions of sets of positive measure. For purely unrectifiability, the correct definition is that a Borel measure in the plane is purely unrectifiable if it is absolutely continuous with respect to $\mathcal{H}^1$, and gives zero mass to all Lipzchitz curves. Then again, if a Radon measure $\mu$ has lower Hausdorff dimension strictly larger than $1$, then almost all projections are absolutely continuous (and this is the weakest bound purely in terms of dimension that guarantees this). And among measures absolutely continuous with respect to $\mathcal{H}^1$, those who have almost all projections absolutely continuous are precisely the ones which are rectifiable.

Again, I don't think the situation is well understood for measures which have lower Hausdorff dimension $1$ but are not absolutely continuous with respect to $\mathcal{H}^1$.

As Fedja pointed out, the natural measure on the four-corner Cantor set is absolutely continuous with respect to $\mathcal{H}^1$, but, being purely unrectifiable, almost all of its projections are singular.

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  • $\begingroup$ In particular, as far as I understood, there is no expicit criterion. Many thanks! $\endgroup$ – limanac Sep 17 '13 at 19:07
  • $\begingroup$ That's right, as far as I know there isn't a simple explicit criterion. There is one condition (large dimension) that arises when the measure is "large" and a quite different one (unrectifiability) that arises in the critical dimension; it is not clear (to me at least) how the two of them "meet in the middle". $\endgroup$ – Pablo Shmerkin Sep 18 '13 at 18:08
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A (relatively) elementary reference for Pablo's purely unrectifiable material: K. Falconer, The Geometry of Fractal Sets.

Some have complained that he does only projection from dimension 2 to 1, and waves his hands for higher dimensions (as he had to to make it this elementary). But since you are only asking about this case, that should not bother you.

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  • $\begingroup$ Okay, although I took the two-dimensional case as an example. Thanks for the reference! $\endgroup$ – limanac Sep 17 '13 at 19:08

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