Law of Large Numbers for [0,1]

Question

Since arbitrary products of probability spaces exist, we can take a set of Bernoulli probability spaces (I'm not sure if that term's in use, but by "Bernoulli probability space" I mean a probability space $(\{0,1\},\mathcal{P}(\{0,1\}),\mu)$), each with success rate $\frac{1}{2}$ (straightforwardly, I mean $\mu(\{1\})$ by success rate), indexed by $[0,1]$ and take their product. We get outcomes as function $f:[0,1]\to\{0,1\}$, but there is a canonical bijection $X$ from $\{0,1\}^{[0,1]}$ to $\mathcal{P}([0,1])$, so we can view this as randomly selecting a subset of $[0,1]$. Intuitively, one would expect a LLN-type result: almost surely, half of $[0,1]$ is selected. This would be formalised by putting a natural measure $\nu$, presumably Lebesgue, on $[0,1]$. The result would then be that, almost surely, $\nu(X)=\frac{1}{2}$. Looking at it in this light, this seems very surprising for at least two reasons: first, measurable sets seem rare, and second, the product space has seemingly nothing to do with where it's being mapped. We could try it with any measurable space of appropriate cardinality, and it would not work in general as we could simply ban sets with probability $\frac{1}{2}$. So does it hold under any circumstance, and if not, is there some other reasonable analogue of the LLN for $[0,1]$?

Answer 1

You can show that if the random variables $Y_t$, $0\le t\le 1$, are i.i.d (and Bernoulli($1/2$), defined on a product space as you suggest), then $(\omega,y)\mapsto Y_t(\omega)$ can't be jointly measurable. This makes $X(\omega) :=\{t:Y_t(\omega)=1\}$ problematic as a "random set". There are some fairly sophisticated ways to try to tackle this issue; see for example "Joint measurability and the one-way Fubini property for a continuum of independent random variables" by P.J. Hammond and Y. Sun [Proceeding of the AMS, vol. 134 (2005) pp. 737–747] and the references therein; especially the paper by Judd.

Answer 2

Your definition of Bernoulli random sets encounters difficulties, described in the answer by @JohnDawkins. However, it can easily be modified, so the question will actually make sense.

Consider the following construction:

Suppose $\Omega$ is a set and $F$ is a $\sigma$-algebra over $\Omega$, that satisfies a following “separability” condition:

$$\forall a, b \in \Omega \exists f \in F (a \in f \& b \notin f)$$ Consider a $\sigma$-algebra $\Sigma(F)$ over $F$, generated by the collection of subsets $\{ \{f \in F|a \in f\}| a \in \Omega\}$. It is not hard do see, that all nonempty elements of $\Sigma(F)$ will be of the form:

$$\bigcup_{n=1}^\infty \{f \in F|A_i \subset f, B_i \subset \Omega \setminus f\}$$

Where $A_1, A_2, … , B_1, B_2, … $ are at most countable subsets of $\Omega$, $A_i \cap B_i = \emptyset$.

On this $\sigma$-algebra we can easily define such a probability measure, that $\forall a_1, … , a_n \in A$ the values of $\mathbb{I}_{a_i \in f}$ are distributed i.i.d. $\sim Bern(p)$. It is done by assigning to any event $\{f \in F|A_i \subset f, B_i \subset \Omega \setminus f\}$ probability $p^{|A_i|}(1-p)^{|B_i|}$. Let’s call such distribution $Bern_p(F)$.

Then the following will hold:

Suppose $(\Omega, F, \mu)$ is a discrete probability space ($|\Omega| \leq \aleph_0$, $F = P(\Omega)$) and $S \sim Bern_p(F)$ is its random measurable subset. Then $E[\mu(S)] = p$.

Proof:

Suppose, $\Omega = \{a_1, a_2, … \}$ (if $|\Omega| < \aleph_0$ consider $a_{n+1}, a_{n+2}, … $ to be dummy elements with zero measure). Then the measurable function $\mu(S)$ is approximated by the simple functions $\mu(S \cap \{a_1, …, a_n\})$. Thus

$$E[\mu(S)] = \lim_{n \to \infty} E[\mu(S \cap \{a_1, …, a_n\}] = \lim_{n \to \infty} p(\sum_{I=1}^n \mu(\{a_i\})) = p(\sum_{I=1}^\infty \mu(\{a_i\})) = p$$

Q.E.D.

That is the fact of the form you need (but only for discrete spaces).

Can it be generalised to non-discrete probability spaces (like $[0;1]$ from your question)? Unfortunately, the answer is still "No". In that case talking about distribution of $\mu(S)$ will be pointless, because $\{f \in F| \mu(f) \leq t\}$ will generally be non-measurable due to the following fact:

Suppose $(\Omega, F, \mu)$ is a non-atomic probability space ($|\Omega| > \aleph_0$, $\forall a \in \Omega$ $\mu(\{a\}) = 0$). Then every non-empty element of $\Sigma(F)$ contains a set of measure $0$ and a set of measure $1$.

Proof:

Suppose, $T \in \Sigma(F) \setminus \{\emptyset\}$. Then, $T = \bigcup_{n=1}^\infty \{f \in F|A_i \subset f, B_i \subset \Omega \setminus f\}$ for some countable subsets $A_1, A_2, … , B_1, B_2, … $, where $A_i \cap B_i = \emptyset$. Notice that

$$A_1, \Omega \setminus B_1 \in T$$

$$\mu(A_1) = 0$$

$$\mu( \Omega \setminus B_1) = 1$$

Q.E.D.