Given $p\in[0,1]$ form a set $\Omega$ and a function $f$ such that $\{f = 1\} = p$ and $\{f = 0\} = 1-p$

Question

Given $p\in[0,1]$ is it always possible to form a set $\Omega$ and a function $f$ such that the expressions below are true? $$ \begin{align} \frac{|\{\omega\in\Omega\,:\, f(\omega) = 1\}|}{|\Omega|} &= p \\ \frac{|\{\omega\in\Omega\,:\, f(\omega) = 0\}|}{|\Omega|} &= 1-p \end{align} $$ Ideally, I would like $\Omega$ to always be finite, but I recon this could be difficult when $p$ is irrational, althogh the fact that one can choose $f$ might actually make this possible.

Context

I am trying to write down a detailed definition of a Bernoulli random variable and of its distribution using measure theory. I am working "backwards" from the distribution, to figure out all the measurable spaces and functions involved.

• Bernoulli Distribution:
  • Discrete Probability Measure: Here's the definition of a discrete measure. Let $(\mathsf{E}, \Sigma_\mathsf{E})$ be any measurable space, let $\mathsf{D}\subset\mathsf{E}$ be a countable subset of $\mathsf{E}$. For any $x\in\mathsf{D}$ let $m(x)$ be a positive number, such that $\sum_{x\in\mathsf{D}} m(x) = 1$. Then $$ \eta(\mathsf{A}) = \sum_{x\in\mathsf{D}} m(x) \delta_x(\mathsf{A}) \qquad \mathsf{A}\in\Sigma_\mathsf{E} $$ is a valid probability measure, and we call all distributions of this form discrete probability measures.
  • Bernoulli Distribution: This makes me believe that the Bernoulli distribution, for $p\in[0, 1]$ is defined as $\text{Ber}_p:\Sigma_\mathsf{E}\to\{p, 1-p\}$ $$ \text{Ber}_p(\mathsf{A}) = \sum_{x\in\{0, 1\}} p^x (1 - p)^{1-x}\delta_x(\mathsf{A}) $$ for any measurable set $\mathsf{A}$. The challenge is understanding what sigma algebra these sets come from, i.e. what is $(\mathsf{E}, \Sigma_\mathsf{E})$ in this case? By definition $\{0, 1\}\subset\mathsf{E}$, but there are infinitely many such sets, so one has to choose a sensible one.
  • Counting Measure: To try and figure out $(\mathsf{E}, \Sigma_\mathsf{E})$ I tried to think about a dominating measure on the same space. The classical dominating measure is the counting measure $$ d\text{count}(\mathsf{A}) = \begin{cases} \text{number of elements in } \mathsf{A} & \text{ if } \mathsf{A} \text{ is finite.} \\ \infty & \text{ if } \mathsf{A} \text{ if infinite.}. \end{cases} $$ This doesn't tell us what $(\mathsf{E}, \Sigma_\mathsf{E})$ is though.
• Bernoulli Random Variable: To try and figure this out, I tried thinking about the Bernoulli random variable $X$ that has $\text{Ber}_p$ as its distribution. Of course, the random variable must take values in $\{0, 1\}$, but the domain of this random variable is unclear to me. However, by definition, if $\text{Ber}_p$ is the distribution of a random variable, there must exist another probability space $(\Omega, \Sigma_\Omega, \lambda)$ such that $$ \text{Ber}_p = \lambda\circ X^{-1} $$ These two things combine tell me that $(\mathsf{E}, \Sigma_\mathsf{E}) = (\{0, 1\}, 2^{\{0, 1\}})$, where $2^{\{0, 1\}}$ is the power set of $\{0, 1\}$, since we know the values that $X$ takes. However, I now need to figure out $(\Omega, \Sigma_\Omega, \lambda)$. My guess is that $\lambda$ is the counting measure on some countable measurable set.

Then, I tried to guess that given $p\in[0, 1]$ one would construct $\Omega$ and $X$ so that the expressions in the original question hold.

Answer 1

[Note: this answer was given to an earlier, much simpler version of the question in which it was not immediate the questioner was familiar with measure theory. I will leave it phrased as is for the benefit of less advanced readers.]

This depends on how you want to define $|\cdot|$.

If you define $|\cdot|$ to give the number of elements in a set (what we call the "cardinality"), then your first equation forces $p$ to be a rational number, since the left hand side is a ratio of two whole numbers.

Moreover, this is the only restriction, as any rational number will work, for if $p=\frac{a}{b}$, then let $\Omega=\{1,2,\cdots,b-1,b\}$ and let

$$f(\omega) = \begin{cases} 1 & \omega\leq a\\ 0 & \omega>a\text{.} \end{cases} $$

However, if you really want to get the irrational numbers too, you can use the same idea but with a different notion of "size" of a set. If $\Omega\subseteq \mathbb R$, we can define the Lebesgue measure of a set. Lebesgue measure in general is a somewhat advanced topic but for an interval, the Lebesgue measure is simply its length.

So then you can define $\Omega=[0,1]$, and given any $p\in [0,1]$ you can define $$f(\omega) = \begin{cases} 1 & \omega\leq p\\ 0 & \omega>p\text{.} \end{cases} $$

and you get your desired result.