# Does the Lebesgue measure on $\mathbf{R}^{n}$ require that the space be equipped with the Euclidean metric?

In my mind, $\mathbf{R}^{n}$ can look very different from Euclidean space as a metric space, depending on the metric we choose to impose on it. Suppose we have a metric that gives the length of the unit interval in $\mathbf{R}^{n}$ to be $\pi$. Then in this space the Lebesgue measure of the unit interval disagrees with the length given by this different metric. Thus it seems (to me at least) that when we define the Lebesgue measure on $\mathbf{R}^{n}$, which involves giving the unit interval measure $1$, we secretly suppose that we're talking about $\mathbf{R}^{n}$ as the Euclidean space.

But when I look at the full definition of Lebesgue measure, the metric is never mentioned. Only the set-theoretic structure of subsets of $\mathbf{R}^{n}$ is talked about.

In other words, I would only find it reasonable that the set $[0,1]^{p}$ be called the unit $p$-cube if the metric on $\mathbf{R}^{n}$ is Euclidean. But the Lebesgue measure is agnostic about the metric...

## Origin of question

I am trying to carefully consider which concepts in multivariate calculus depend on the Euclidean structure, and which do not. For example, differential calculus only involves the affine and topological structure of $\mathbf{R}^{n}$, and can be recast in more general spaces with these properties (Banach space). Vector calculus is largely dependent on the special dimension $n=3$, orientation, and metric.

But what about integration? For example, do the ordinary multiple integrals involve the metric in some way? This question led me to the more general Lebesgue measure, hence this question.

I think that what you're looking for is the following.

If $V$ is a finite dimensional vector space of dimension $n$ on the real numbers, is there a canonical way to define the Lebesgue measure?

The answer is no. In fact, you can define the Lebesgue measure only after you have choosen a base $\{e_1,...,e_n\}$ of $V$ that gives you an isomorphism with $\mathbb{R}^n$. However, different choices of the base give different measures on $V$, that differs by $|\det(A)|$ where $A$ is the change of base matrix.

However, what if we add a norm $\|\|$ on $V$ and we require that the base we choose to define the Lebesgue measure on $V$ has the property that $\| e_1 \|=1,...,\| e_n \|=1$?

The answer is no again. In fact, there exists norm on $V$ and there exists two bases $\{e_1,...,e_n\}$ and $\{f_1,...,f_n\}$ that satisfy such a property and they are such that the change of base matrix $A$ is such that $|\det(A)|\neq 1$

However, what if the norm comes from an inner product $\langle, \rangle$ and we require that the base we choose to define Lebesgue measure on $V$ is orthonormal?

The answer now is yes. In fact, given two orthonormal bases of $(V,\langle, \rangle)$, the change of base matrix has the property that $|\det(A)|=1$ and so, picking two orthonormal bases of $(V,\langle, \rangle)$, you get the same Lebesgue measure on $V$. In conclusion, the arbitrariness in defining the Lebesgue measure on a finite dimensional $V$ is due only to the choice of an inner product on $V$.

Yes and no.

You can define a measure on $\mathbb{R}^n$, without reference to the metric, that will equal Lebesgue measure up to a constant. One way to do this is by considering $\mathbb{R}^n$ as a locally compact topological group (under vector addition) and then use the associated Haar measure. Lebesgue measure is compatible with the topology (i.e. is Radon) and is translation-invariant, so by uniqueness of Haar measure (up to a constant), in this way we get Lebesgue measure up to a constant. This makes no use of the metric whatsoever, just the topology and group structure.

In the euclidean case, we choose to normalize the measure so that $[0,1]^n$ has measure 1. This seems geometrically intuitive, since an interval $[a,b]$ can be considered (trivially) as a curve in $\mathbb{R}^1$ and its length (with respect to the euclidean metric!) is $b-a$. However, note that even the metric is normalized in a certain manner. You can multiply the euclidean metric by any positive scalar and get a metric that still satisfies pretty much all the properties we like about euclidean space. The length of a curve in this metric will be a multiple, by the same scalar, of the euclidean length and consequently you might say that the appropriate measure to give $[0,1]$, when working with this metric, is the aforementioned scalar.

Lebesgue measure on ${\mathbb R}^n$ is defined as product measure resulting from Lebesgue measure on ${\mathbb R}$. The latter is strongly tied to the features present in ${\mathbb R}$: It gives $[0,1]$ the measure $1$, is translation invariant, and behaves as expected under scaling.

It follows that Lebesgue measure in ${\mathbb R}^n$ gives the unit cube $[0,1]^n$ the volume $1$, is translation invariant, and behaves as expected under scaling.

The unit cube is tied to the way ${\mathbb R}^n$ is defined, and can be regarded as the most natural basic set for volume measurement. This cube and its translates, etc., are not affected by any metric you may choose later.

Note that there are different notions of "metric" in this context: (i) a distance function $d:\>{\mathbb R}^n\times {\mathbb R}^n\to{\mathbb R}_{\geq0}$, and (ii) a Riemannian metric defined on some manifold $X$ via a a metric tensor $g$ on the tangent bundle TX.

(i) If you define on the plane ${\mathbb R}^2$ a metric by letting $d(x,y)$ be a somehow established road distance between $x$ and $y$ then the Lebesgue area measurement on ${\mathbb R}^2$ is in now way affected by any roads , sideroads, and capillaries present in the plane.

(ii) The toy example of such a Riemann metric is a constant Riemannian metric on $X:={\mathbb R}^n$ itself. It is given by a positive definite symmetric matrix $[g]=[g_{ik}]$ that has to be provided by you. This metric defines a scalar product $$\langle x,y\rangle:=\sum_{i,k} x_i\, g_{ik}\, y_k$$ on $X$, so that you can do Euclidean geometry on $X$ according to this metric. Now this Euclidean metric on $X$ automatically defines a $d$-dimensional volume for arbitrary $d$-dimensional parallelograms $P\subset X$ via the so-called Gram determinant of the spanning vectors. In particular, the $n$-dimensional $g$-volume of $n$-dimensional parallelotopes $P\subset X$ is a constant multiple of the $n$-dimensional Lebesgue measure $\lambda(P)$; in fact one has $${\rm vol}_g(P)=\sqrt{\det[g]}\>\lambda(P)\ .$$

• What do you mean by "The unit cube is tied to the way $\mathbf{R}^{n}$ is defined"? – Maxis Jaisi Jul 23 '18 at 8:27
• Without specifying the metric before constructing the measure, how do you know you're not talking about some other space? For example, how do you know that you're not defining the Lebesgue measure on the punctured sphere instead of the plane, since both are indistinguishable (same $\mathbf{R}^{2}$ as sets)? – Maxis Jaisi Jul 26 '18 at 12:28
• Last bit, before I accept your answer: you say "if you define on the plane $\mathbf{R}^{2}$ a metric...". Is "plane" just a synonym for the set of pairs of real numbers, or more than that, i.e. it has a topological structure generated by open disks? – Maxis Jaisi Jul 27 '18 at 12:04