Sufficient Condition for Existence of a Uniform Distribution on a Set

Question

What are some simple sufficient conditions for the existence of a uniform distribution on a set?

Specifically, what are some conditions on the set that aren't much more complicated than what you might see in undergraduate analysis (compact, connected, etc) that might guarantee the existence of a uniform distribution?

Answer 1

I am not sure if this is what you are looking for, but I'll give it a try. Assume that you have a Borel set $B \subseteq \Bbb R$ such that $B$ has finite and positive Lebesgue measure, i.e. $\lambda(B)\in (0, \infty)$. Then the normed and restricted measure $$ \mu: \mathcal B \to [0,1]: A \mapsto \frac 1{\lambda(B)}\lambda(A\cap B) $$ is a uniform distribution on $B$.

On the other hand, you can define a discrete uniform distribution on every finite set $\{x_1,\dots,x_n\} \subset \Bbb R$.

Moreover, there is no uniform distribution on unbounded sets.

Answer 2

Think of these three possible solutions. I provide the claims and the intuition, but not the proofs.

Condition 1. (On $\mathbb{R}^{n}$ with Riemann-integrals, without measure theory)

If $S\subseteq\mathbb{R}^{n}$ is such that the indicator function $\boldsymbol{1}_{S}$ is Riemann-integrable over $\mathbb{R}^{n}$ and the integral $c\stackrel{\text{def}}{=}\int_{\mathbb{R}^{n}}\boldsymbol{1}_{S}(x)\,dx$ yields $c\in(0,\infty)$, then $S$ admits the uniform distribution given by the density $x\mapsto1/c$.

Condition 2. (On $\mathbb{R}^{n}$, with topology)

Let $S\subseteq\mathbb{R}^{n}$ be a countable union of closed boxes, where each box is the product of $n$ closed intervals. If there are $U,K\subseteq\mathbb{R}^{n}$ such that $U$ is open and non-empty, $K$ is compact and $U\subseteq S\subseteq K$, then $S$ admits a uniform distribution.

Condition 3. (On a metric space, using measure theory)

Let $(X,\text{dist})$ be a metric space and $S\subseteq X$ a measurable set. If there is $d\geq0$ such that $H^{d}(S)=c\in\left(0,\infty\right)$, where $H^{d}(S)$ is the $d$-dimensional Hausdorff measure of $S$, then $S$ admits a uniform distribution.

Any of these conditions is sufficient, but they do not capture all examples.

1. Condition 1 fails to capture $[0,1]\cap(\mathbb{R}\setminus\mathbb{Q})$, any finite set in $\mathbb{R}$ or a segment in $\mathbb{R}^{2}$. The beauty of this condition is that it can be introduced without measure theory, in which case the notion of probability of an event $A\subseteq X$ should be understood as the Riemann-integral $\int_{X}\boldsymbol{1}_{A}(x)\,dx$, and would be defined only when the integral exists.
2. Condition 2 fails to capture the finite set in $\mathbb{R}$, the segment in $\mathbb{R}^{2}$, etc., the set $\bigcup_{i=1}^{\infty}\left[i,i+1/i^{2}\right]$ in $\mathbb{R}$ and in general those that are unbounded but small in volume. But it does capture $[0,1]\cap(\mathbb{R}\setminus\mathbb{Q})$. The intuition behind the proof is that being a countable union of closed boxes implies measurability, containing $U$ implies non-nullity ($\text{vol}(S)>0$), and being contained in $K$ implies finite-measure ($\text{vol}(S)<\infty$).
3. Condition 3 captures everything that comes to my mind, but it uses rather advanced tools.

This question is excellent because although it is simple to state, solving it will require you to think of many other fundamental questions of measure theory and topology.