In practice, how is the Lebesgue measure usually generalized?

Question

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. :)

Answer 1

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.

Answer 2

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).