Forming a subset of $\mathbb{R}$ by coin tossing

We form a subset $X$ of $\mathbb{R}$ as follows: for every $x\in\mathbb{R}$ we toss a coin. if heads then we put it in $X$. Assume two tosses are independent of each other. What is the probability that $X$ is a (Lebesgue) measurable subset of $\mathbb{R}$?

I just started learning probability theory so I am not sure how to formulate this properly, I mean I get uncountably many random variables ... . Will really appreciate any help with this. Feel free to use any advanced theorems, I will look them up.

EDIT: Thanks to users angryavian & StephenMontgomery-Smith for pointing out that it makes sense only to talk about probability of $X$ being measurable. I have edited the same.

Thanks

Solution 1:

I assert that the answer is undefined.

Let's describe $(\Omega,\Sigma,\Pr)$, the probability space of $2^{\aleph_0}$ coins being tossed, one for each real number. The "Borel" sets in $\Sigma$ are the $\sigma$-algebra generated by sets of the form $\Pi_{r \in \mathbb R} F_r$ where finitely many $F_r$ are not $\emptyset$ or $\{H,T\}$, and the probability of this set of $0$ if any $F_r = \emptyset$, and otherwise $\frac12$ to the power of the number of $r$ such that $F_r$ has exactly one element in it. Then $\Sigma$ is the completion of the "Borel" sets, by adding in subsets of "Borel" sets of measure zero.

Let $$ B = \{ w : X(w) \text{ is measurable} \} .$$ I assert that $B$ is not measurable.

We say that $A \subset \Omega$ depends upon $Y \subset \mathbb R$ if for every $(c_r) \in \Omega$, we have that $(c_r) \in A$ is completely determined by the values of $c_r$ for $r \in Y$. We can see that all Borel subsets of $\Omega$ depend upon a countable subset of $\mathbb R$. So we can see that $B$ isn't a "Borel" set.

Lemma: If $B \subset A$ and $A$ is "Borel," then $A = \Omega$.

Suppose $(c_r) \in \Omega \setminus A$. Suppose $A$ depends upon $Y \subset \mathbb R$ where $Y$ is countable. Let $$ Z = \{r \in Y : c_r = H \} .$$ Then unwinding the definitions, we see that there exists a countable $Y$ and $Z \subset Y$ such that for all $X \subset \mathbb R$, that $X \cap Y = Z$ implies $X$ is not measurable. But clearly this is not true. $\square$

Corollary: The outer measure of $B$ is 1.

Exactly the same argument can be applied to $\Omega \setminus B$, so:

Lemma: The inner measure of $B$ is 0.

So $B$ is non-measurable in the worst possible way. And $\Pr(B)$ is undefined.

Note 1: You could ask far more mundane questions like, "What is the probability that $X$ has measure $0$?" Answer: undefined.

Note 2: I put Borel in quotes. True Borel sets are the sigma algebra created by open sets in the product topology, which is not quite what I have. For example, the subset of sequences $(c_r)$ such that at $c_r = T$ for at least one $r \in \mathbb R$ is open in the product topology, but not "Borel" using the definition I have given. I think that my "Borel sets" are equivalent to Baire sets (the $\sigma$-algebra generated by close $G_\delta$ sets, also defined as the minimal $\sigma$-algebra such that continuous functions from $\Omega \to \mathbb R$ are continuous.)

Solution 2:

I'm going to interpret the question a little bit differently, and I try to will address the original question of whether or not the set of $x$ for which you get a heads is almost surely Lebesgue measurable.

I will show that, in a certain convoluted sense, yes, you can force the set to not only be Lebesgue measurable, but it can even be a countable set or a co-countable set. From an intuitive perspective, this obviously sounds like complete nonsense, even stupidity. Half the points have to be heads after all! But the paradox here essentially amounts to the fact that probability theory and measure theory aren't really equipped to "properly" handle uncountably many independent events like this, something which is inherent in the countability conditions in the definition of sigma algebra. I guess the lesson here is that "in probability theory, we can only ever hope to observe countably many things at once."

Let's do the construction which "makes the set of heads a countable set." The result for co-countable sets is symmetric. I think the proof works, but let me know if there's a problem.

For any subset $B \subset \Bbb R$, let $\mathcal C_B$ be the collection of all countable subsets of $B$. For $x \in B$ define $\pi_x : \mathcal C_B \to \{0,1\}$ by $\pi_x(A):= 1_A(x)$. Let $\mathcal F_B$ be the $\sigma$-algebra on $\mathcal C_B$ generated by the maps $\{\pi_x:x\in\Bbb R\}$. I will interpret the question as asking whether or not there exists a probability measure $\mu$ on $(\mathcal C_{\Bbb R}, \mathcal F_{\Bbb R})$ such that the maps $\{\pi_x\}_{x \in \Bbb R}$ are mutually independent Bernoulli(1/2) variables under $\mu$.

As Stephen Montgomery-Smith already pointed out above, any event in $\mathcal F_{\Bbb R}$ can only depend on a countable set of $x$. More precisely, for any event $E \in \mathcal F_{\Bbb R}$, there exists a countable set $D \subset \Bbb R$ and an event $F \in \mathcal F_D$ such that $E = \pi_D^{-1}(F)$, where $\pi_D:\mathcal C_{\Bbb R} \to \mathcal C_D$ is given by $A \mapsto A\cap D$. The proof can be found in a comment below the other answer.

For each countable subset $D \subset \Bbb R$ we define $\mu_D$ to be the measure on $(\mathcal C_D,\mathcal F_D)$ which is a product of iid Bernoulli measures on the points of $D$. There is no ambiguity here since $\mathcal C_D = 2^D \simeq \{0,1\}^D$, since $D$ is countable.

We then define a measure $\mu$ on $(\mathcal C_{\Bbb R},\mathcal F_{\Bbb R})$ by the formula $$\mu(E) := \mu_D(F),\;\;\;\; \text{whenever }E = \pi_D^{-1}(F) \text{ for some countable } D \subset \Bbb R \text{ and }F \in \mathcal F_D.$$

In other words, we are defining $\mu$ by the condition that $(\pi_D)_*\mu = \mu_D$ for all countable $D\subset \Bbb R$ (algebraically, $\mu$ is the projective limit of all the $\mu_D$'s as "$D$ approaches $\Bbb R$ along countable sets"). It's easy to see that $\mu$ is well-defined (i.e., if $\pi_D^{-1}(E) = \pi_{D'}^{-1}(E')$ then $\mu_D(E) = \mu_{D'}(E')$) and that $\mu$ satisfies the definition of a probability measure.

Thus $\mu$ defines a random countable subset $C$ of $\Bbb R$ with the seemingly absurd property that the probability of any fixed real number lying in $C$ is $1/2$. More generally, the probability under $\mu$ that any (deterministic) finite subset $F$ of $\Bbb R$ is contained in $C$ equals $2^{-|F|}$. (Fun fact: by the $\pi$-$\lambda$ theorem applied to such events, it's easy to see that $\mu$ is the unique such measure on $(\mathcal C_{\Bbb R},\mathcal F_{\Bbb R})$ and that $\mu$ is also translationally invariant).

So what exactly is happening here? As the other answer shows, if you construct the probability measure on the larger space $2^{\Bbb R}$ then the event "the set of heads is countable" is non measurable in a very bad way. You can never hope to extend the measure to a larger sigma algebra in such a way that this event becomes measurable. Therefore we can just impose the condition that the set of heads is countable by constructing the measure on this smaller collection of subsets of $\Bbb R$, and no mathematical contradiction will arise from doing so.

But there is still something more to be said. If this works, then what's to stop us from constructing the measure on more interesting families of sets? Like open sets or closed sets or even finite sets? Well, the problem is that a contradiction will arise in all of those cases. The definition of $\mu$ is where that happens. If you try to make the set finite for instance, then $\mu$ will not be well defined because the measures $\mu_D, \mu_{D'}$ will fail to be compatible when $D$ is finite and $D'$ is countably infinite (so it makes no sense to talk about projective limits at all). Similar problems occur in the open/closed cases.