I have an intuitive stochastic process as follows, but not sure how to construct it rigorously on some probability space.
Consider the unit interval $[0,1]$, each point $x\in [0,1]$ is associated with an independent Bernoulli($0.5$) random variable. Then each realization can be viewed as a function $f$ from $[0,1]$ to $\{0,1\}$. The level set $\{x: f(x) = 1\}$ is then a random set on $[0,1]$. I want to argue the set is almost surely (Lebesgue)-measurable, and almost-surely has measure $0.5$.
However, the above construction does not seem to be rigorous. Therefore, I want to know if there exists a probability space $\Omega$ that admits the a stochastic process $f(t,\omega) \in \{0,1\}$, such that $1.$ all the finite dimensional distribution is a product of independent Bernoulli. $2.$ For almost every $\omega$, the set $\{x: f(x,\omega) = 1\}$ is measurable, and has probability $0.5$.
It seems the first condition can be guaranteed using the Kolmogorov's extension theorem. But I have not idea how to guarantee the second condition.
Best Answer
It seems this is impossible. Let $\mu$ denote Lebesgue measure on $[0,1]$. If $A(\omega)=\{x: f(x,\omega) = 1\}$ is measurable then $\mu\bigl(A(\omega) \cap [A(\omega)+t]\bigr) \to \mu(A)=1/2$ as $t \to 0$. But $$E\Bigl(\mu\bigl(A(\omega) \cap [A(\omega)+t]\bigr)\Bigr) $$ $$=\int_\Omega \int_0^1 f(x,\omega) f(x-t,\omega) \,dx \,dP $$ $$= \int_0^1 \int_\Omega f(x,\omega) f(x-t,\omega) \,dP \, dx=1/4\,,$$ and this is a contradiction in view of Lebesgue's bounded convergence theorem.