The general question

It is easy to find on the Wikipedia page for Lebesgue measure that Haar measure is a common generalization that preserves the idea of "invariance under some group action". While wondering about the "most natural" way of defining a measure on lines of $\mathbb{R}^2$ (see below for more information), it struck me that it's not always obvious what the "most natural" measure is for certain spaces is like in $\mathbb{R}^n$. I was wondering how this problem is usually tackled, if it even comes up.

Is finding an "appropriate" Haar measure usually easy to find? Is it unique in some sense or usually dependent on the use case? What can we do in the case of the space of lines in $\mathbb{R}^2$ (or general hyperplanes in $\mathbb{R}^n$)?

The specific case of lines in $\mathbb{R}^2$

To give an example, and some motivation, consider lines in $\mathbb{R}^2$. Is there a "unique", "most natural" measure on this space? By this, I mean some invariance under isometries, plus some extra properties that satisfy my intuition of what it should look like. My primary motivation is whether there is actually a "most natural" way to solve Bertrand's paradox.

Specifically something satisfying the properties below would be nice. In what follows, $S$ is a measurable set of lines, $m(S)$ is the measure of $S$, and if we have an operation $f:\mathbb{R}^2\to \mathbb{R}^2$ we can consider $f(S)$ to be the set of $f(L)$ for each $L$ in $S$ (where $f(L)$ for some line $L$ is just the set of $f(p)$ for $p$ in $L$).

  1. If $f$ is a rotation, $m(S)=m(f(S))$
  2. If $f$ is a reflection, $m(S)=m(f(S))$
  3. If $f$ is a translation, $m(S)=m(f(S))$
  4. If $f$ is a dilation with factor $C$ (i.e.- two points $d$ apart get dilated to $C\cdot d$ apart), then $C^2\cdot m(S)=m(f(S))$
  5. If $M$ is a compact subset of $\mathbb{R}^2$, then the set of lines $S$ that intersect $M$ is measurable and $0<m(S)<∞$

I see three cases: Either there are no measures satisfying the above, in which case perhaps there is a weakening of the conditions to give some meaningful content (we might be able to at least satisfy a chosen subset of the above); there is exactly one measure satisfying the above (up to perhaps some relatively trivial modifications) and I am happy; or there are many and perhaps we need some more conditions specifying what a "natural" measure should be.

Of course, this is a relatively open-ended question, so I am satisfied with any related comments. I simply don't want to wade through a couple courses in measure theory just for a chance of learning the answer. :)

  • 5
    $\begingroup$ Grassmanians have a natural "Haar" measure by thinking about them as quotients of $O(n)$. See the "associated measures" section en.m.wikipedia.org/wiki/Grassmannian $\endgroup$ – Alessandro Codenotti May 17 at 15:31
  • 4
    $\begingroup$ @AlessandroCodenotti Since exfret discussing Bertrand's paradox, I think the space of lines must be lines not required to pass through the origin, which is not the Grassmannian (and noncompact). After all, the Grassmannian of lines through the origin for $\mathbb{R}^2$ is just the circle (with its usual rotationally invariant measure). $\endgroup$ – Robert Furber May 17 at 15:46
  • 1
    $\begingroup$ I think the answer is that Haar measure is the natural extension of the concept to (locally compact) groups. You can obtain spaces such as the space of lines as quotients of groups (I.e. quotients by a non-normal subgroup) and you can push forward Haar measure to the quotient in some cases. $\endgroup$ – Anthony Quas May 17 at 21:44
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    $\begingroup$ @RobertFurber The lines in the projective plane (over $\mathbb R$) form a compact space, identified with the Grassmannian of $2$-planes in $\mathbb R^3$. The space of affine lines in $\mathbb R^2$ is the space of projective lines minus a single element (the line at infinity), which shouldn't affect the measure. $\endgroup$ – Andreas Blass May 18 at 3:55
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    $\begingroup$ I think measures on spaces like the space of lines in the plane are studied in integral geometry. For example, there is a theorem (Crofton's formula) saying that the length of any (nice) curve $C$ in the plane equals (up to some normalization factor) the integral over the space of lines of the number of intersections of the line with $C$. $\endgroup$ – Andreas Blass May 18 at 3:58

For obtaining natural measures using invariance under a group action, I am aware of the following theorem from Bourbaki's Integration. It is Chapter VII, Section 2, Proposition 4.

Let $X$ be locally compact Hausdorff and let $G$ be a locally compact group with Haar measure $\mu$, and a continuous, proper action on $X$. Let $X/G$ denote the space of orbits with the quotient topology. Given a continuous compactly-supported function $f$ on $X$, let $f^\flat$ denote the function on $X/G$ given by

$$f^\flat(Gx)=\int_G f(gx)d\mu(g)$$

Given a (Radon) measure $\lambda$ on $X/G$, there exists a unique (Radon) measure $\lambda^\sharp$ on $X$ such that for all $f$,

$$\int_X f\ d\lambda^\sharp = \int_{X/G} f^\flat\ d\lambda$$

Moreover, $\lambda^\sharp$ is invariant under the action of $G$.

In nice cases, the space of orbits is discrete, so there is a natural choice of $\lambda$. In fact, if $G$ acts transitively, as in your example, then $X/G$ is a singleton.

  • $\begingroup$ Very good (+1). I like also the framework which is set in the same chapter's subsequent exercise (Bourbaki's Integration Chapter VII § 1 Ex. 14). $\endgroup$ – Duchamp Gérard H. E. May 18 at 7:26

Someone should mention an elementary answer the specific question by OP. An affine hyperplane in $\mathbb{R}^n$ can be uniquely represented as $l_{v,c}:=\{x\in\mathbb{R}^n:v\cdot \mathbb{x}=c\}$, where $v\in\mathbb{R}^n$, $|v|=1$, and $c>0$ (except for planes passing through the origin which can be disregarded as they will have zero measure anyway). Let $dv$ be the uniform measure on the unit sphere and $dc$ the Lebesgue measure on the real line. Then, if $\mathcal{L}$ denotes the map $(v,c)\mapsto l_{v,c}$, the measure on the set of planes defined by $\mu(A)=dv\otimes dc(\mathcal{L^{-1}}(A))$ satisfies all the requested properties*. Indeed, for example, the effect of a translation by a vector $-w$ in these coordinates is the map $(v,c)\mapsto (v,c+v\cdot w)$ or $(v,c)\mapsto (-v,-c-v\cdot w)$, both of which have Jacobian equal to one. The effect of rotation is simply rotating $v$, the effect of scaling is to scale $c$ etc.

The same can be done for affine subspaces of all dimensions - an affine subspace is an element of the Grassmanian shifted by a vector in its orthogonal complement, and you can just take the Haar measure on the Grassmanian times the Lebesgue measure on its orthogonal complement.

*except that the scaling property is different, but it should not be hard to see that there are no non-trivial measures with the requested scaling property.

UPD: To see that such a measure is unique up to scaling, observe that if we have another such measure $\nu$, then $\nu$ is absolutely continuous with respect to $\mu$, thus its pullback to $\{(v,c)\}$ has a density with respect to Lebesgue measure. Clearly, this density is rotationally invariant, thus, it is enough to show that it is invariant under shifts $c\mapsto c+\alpha$. Let $\varepsilon,\delta>0$ be small, let $|v_0|=1, c_0>0$ and consider the set $R_{\varepsilon,\delta} (v_0,c_0)=\{|v-v_0|<\varepsilon,|c-c_0|<\delta\}$. We have $1-\varepsilon^2<v\cdot v_0<1+\varepsilon^2$ for all $v\in R_{\varepsilon,\delta} (v_0,c_0)$. Hence, shifting the corresponding set of lines by $\alpha v_0$ sends $R_{\varepsilon,\delta} (v_0,c_0)$ to a set contained in $R_{\varepsilon,\delta+\varepsilon^2} (v_0,c_0+\alpha)$ and containing $R_{\varepsilon,\delta-\varepsilon^2} (v_0,c_0+\alpha)$. Sending $\varepsilon\to 0$ and using (e. g.) the Lebesgue differentiation theorem completes the proof (of course, one can also do it without densities).

  • $\begingroup$ I’m quite tired at the moment, so excuse me if I’m wrong, but isn’t this in fact not translation invariant? It’s hard to believe that something that by nature assigns the same “size” to the “smaller” number of lines close to the unit disk than, say, the “larger” number far away would satisfy translation invariance. $\endgroup$ – exfret May 19 at 5:09
  • $\begingroup$ Nevermind, I am in fact tired. I often hear as a first response to send lines to their nearest point to the origin, and thought this was the same approach. Though yours works fine! Very nice! $\endgroup$ – exfret May 19 at 5:12
  • $\begingroup$ As for the scaling property, my intuition was that we could translate, say, a square four times to get the result of a dilation by two. If the four smaller squares are far enough apart, then we should get a quadratic increase in measure. $\endgroup$ – exfret May 19 at 5:21
  • $\begingroup$ It is easy to see the flaw with this now, though, since the squares are forced to be close together, and the lines that pass through them will actually overlap quite a lot because of this! $\endgroup$ – exfret May 19 at 5:22
  • $\begingroup$ One last thing: is this measure unique? $\endgroup$ – exfret May 19 at 5:24

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