I learnt about the concept of the Lebesgue measure some time ago, and one question was always confusing to me (and was never answered).
What is the Lebesgue measure of zero-dimensional space?
What I know is that the Lebesgue measure is defined as the standard way of measuring a subset of some Euclidean space $R^n$, where $n$ is the dimension of the space.
This means that:
- The standard way to measure the subset (a “part” that is well-defined in the space) of a line ($R^1$) is length; makes sense
- The standard way to measure the subset of a plane ($R^2$) is area; makes sense
- The standard way to measure the subset of a solid space ($R^3$) is volume; makes sense
- And this continues for dimensions $d > 3$ in some $R^d$.
But when It comes to $R^0$, there are two questions that often come unanswered and can really confuse you:
- Q1 If Lebesgue measure is defined on $n$-dimensional space as a standard way to assign measure to a proper subset of $R^n$, how would we define the Lebesgue measure of $0$-dimensional space? This amounts to measuring the subset of a point, which kind of doesn’t make sense in my mind.
- Q2 I think we define the dimension of some $S^n$, where $S$ can be some space, as the number of coordinates (also: coordinate axis) needed to specify the location of some point in/on $S_n$. This makes sense for $1$-dimensional space, as you just need one number on the number line to find a location of a point on the number line, and for $2$-dimensional space, you need two: the $x$ and $y$-axis to specify a point on a plane. Their logic continues for higher dimensions. But how does it make sense that we need “no coordinates” to specify a point on a point?
Or is all of this above stuff nonsense and a point is just zero length? If that’s a case, why would we assign a new dimension to something that can be measured in a Lebesgue measure defined in another dimension?
I couldn’t find any useful information anywhere else for these questions, hence
Any help or clarification is appreciated!
I would suspect that most authors who define the Lebesgue measure for $\mathbb R^n$ are not thinking about the case $n=0$. But if you do define it for $n=0$ (which I think is sensible), then the zero-dimensional Lebesgue measure of $\mathbb R^0$ is $1$ according to standard mathematical conventions about vacuous/trivial cases.
One standard way to define the Lebesgue outer measure in $\mathbb R^n$ is as follows. First, we define a closed rectangle $R$ in $\mathbb R^n$ to be a subset of $\mathbb R^n$ of the form $[a_1,b_1]\times \dots\times [a_n,b_n]$, with $a_i\le b_i$ for each $i$. The "volume" $\DeclareMathOperator{\vol}{vol}\vol(R)$ of $R$ is defined to be $$ \vol(R)=\prod_{i=1}^{n}(b_i-a_i) \, . $$ If $E\subseteq\mathbb R^n$, then the outer measure $\lambda^*(E)$ of $E$ is defined to be $$ \lambda^*(E)=\inf\left\{\sum_{R\in\mathcal C}\vol(R):\text{$\mathcal C$ is a countable collection of closed rectangles which covers $E$}\right\} \, . $$ (For me, finite sets are countable. This is actually essential to the definition when $n=0$.)
Finally, if $E$ satisfies the Carathéodory criterion, that $\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^c)$ for all $A\subseteq\mathbb R^n$, then we say that $E$ is Lebesgue measurable, and the Lebesgue measure $\lambda(E)$ of $E$ is defined as $\lambda^*(E)$.
So what about the case $n=0$? This is just a matter of following the definitions carefully. First, a closed rectangle in $\mathbb R^0$ is a Cartesian product of zero sets, which is $\{()\}$, the set containing the empty tuple. The "volume" of this rectangle is the empty product $1$. There are two subsets of $\mathbb R^0$, namely $\varnothing$ and $\mathbb R^0=\{()\}$. The empty set is covered by the empty collection of closed rectangles, and $\sum_{R\in\varnothing}\vol(R)$ is the empty sum $0$. Thus, $\lambda^*(\varnothing)=0$. It is also routine to check that $\lambda^*(\{()\})=1$ (this time, the computation is less vacuous). Both $\varnothing$ and $\{()\}$ satisfy the Carathéodory criterion (there are four cases to consider in total), and so we find that $\lambda(\varnothing)=0$ and $\lambda(\{()\})=1$. Thus, as Didier mentions in the comments, in zero dimensions, the Lebesgue measure coincides with the counting measure.
While this might all sound very silly, I can think of one example where defining the zero-dimensional Lebesgue measure is marginally useful. If $A$ is an $n\times n$ matrix over $\mathbb R$, then the determinant of $A$ ought to equal the $n$-dimensional Lebesgue measure of the image of the unit cube under $A$. The determinant of the $0\times 0$ matrix over $\mathbb R$ is $1$, which does indeed equal the Lebesgue measure of the image of $\{()\}$.