Uniform probability over real numbers?

Question

I'm a bit conflicted by two answers I read.

This first one, regarding the implications of different infinite cardinalities, explains that

We often talk about a "uniform" probability over [0,1] where every singleton has probability zero. If [0,1] is countable, we can't do that.

This to me implies we can define a uniform probability distribution over the interval.

However, this question hints at the fact that we can't define a uniform probability distribution over the positive reals. Since one can create a bijection from $[0,1]$ to $\mathbb{R}^+$, how are they different? And if so, why can we define a uniform distribution on the interval but not the positive reals? I have yet to take a course with measure theory.

Answer 1

Bijections just care about the cardinality of the two sets. Once you have a bijection it doesn't tell you much about how measure transforms unless it is differentiable.

As an example, let us make a bijection between $[0,1]$ and $[0,2]$ as $$f(x)= \begin {cases} x&0\le x \le 0.9\\11(x-0.9)&x \gt 0.9 \end {cases}$$ If we pull a random number $x$ uniformly in $[0,1]$ the chance that $f(x)$ is in a small interval less than $0.9$ is the length of the interval. The chance that $f(x)$ is in a small interval greater than $0.9$ is $\frac 1{11}$ times the length of the interval. When both intervals are finite there is a linear relation between that keeps the measure of intervals proportional to their length. If one interval is infinite that is not possible. You can define a measure on $\Bbb R^+$ but it will not be uniform.

Answer 2

If $B\subseteq\mathbb R$ is measurable with $0<\lambda(B)<\infty$ (where $\lambda$ denotes the Lebesgue measure) then we can induce a well defined uniform probability by setting: $$\mathsf P(A)=\frac{\lambda(A)}{\lambda(B)}$$for measurable subsets of $A$.

Unfortunately this does not work if $\lambda(B)=+\infty$.