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I'm new to formal writing, so please be patient :).

Let $S=\left\{ s \subset [0,1] : |s|=n \right\}$, and let $x$ be a probability distribution over the elements of $S$. Namely, $x$ is a distribution over all sets of size $n$ (for a constant $n$) of elements from $[0,1]$.

Define $\mu: 2^{[0,1]}\rightarrow [0,n]$ such that for every Lebesgue measurable set $A\subset [0,1]$ $$\mu(A) = \int_{S}|s\cap A|dx(s)=\mathbb E_{s\sim x}(|s\cap A|) $$

(1) How can one show that $([0,1],\mathcal B([0,1]),\mu)$ is a measure space?

(2) Am I using the right notations? is there a better introduction of this problem?

Referring to (1), I thought about the following:

  • sort every element $s$ to obtain a tuple,$(s_1,\dots,s_n)\in [0,1]^n$.
  • look at the projection of $s$ on each of the components. If the projection on every component $i$, $P_i$, is itself a measure, $\mu$ can be expressed as sum of indicators, $$\mu(A) =\sum_{i=1}^n\int_{S}\mathbb 1_{s_i \in A}dP_i(s_i), $$ and since countable sum of measures is itself a measure (see, e.g., here) we are done.

I'm not sure though that this is right. Another option is to state straightforward that $$\mu(A) =\int_{S}\sum_{i=1}^n\mathbb 1_{s_i \in A}dx(s), $$ and to flip the order of summation, which is possible due to non-negativity using Fubini's theorem.

Any ideas?

AvidLearner
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  • This may be a really silly question, but is $S$ countable? – SystematicDisintegration Jun 27 '17 at 13:42
  • @SystematicDisintegration For $n=1$ we get something in the flavor $S=[0,1]$, so I believe the answer is no. – AvidLearner Jun 27 '17 at 16:09
  • How, then, have you sorted the elements of $S$ into an n-tuple? Isn't this an enumeration? – SystematicDisintegration Jun 27 '17 at 16:51
  • @SystematicDisintegration Note that I suggested to sort an element $s$ and not the set $S$. For given $s={s_1,\dots,s_n}\in S$, define $s'=(s_{(1)},\dots,s_{(n)})$ such that for all $i<j$, $s'{(i)}\leq s'{(j)}$. – AvidLearner Jun 27 '17 at 17:31
  • Ah, that clarifies things. The OP seemed to suggest otherwise to me. – SystematicDisintegration Jun 27 '17 at 17:40
  • Can you be more specific as to what you mean by "a probability distribution over $S$"? In particular you have to specify a $\sigma$-algebra on $S$. – Nate Eldredge Jun 28 '17 at 15:18
  • @NateEldredge This is one of the places where I get lost. Will the sigma algebra which is the powerset of $S$ work? One thing I can also do is to define a distribution over $[0,1]^n , \mathcal{B}([0,1]^n)$, and state that it support is over tuples with distinct elements and symmetric with respect to all dimensions. Correct? I wish to prove the statement to every possible distribution over $S$. – AvidLearner Jun 28 '17 at 15:27
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    Almost certainly not, unless you want a distribution which is very simple (e.g. discrete). For instance, with $n=1$ you are asking for a probability distribution on all subsets of $[0,1]$, and those typically don't exist with useful properties (remember Lebesgue non-measurable sets for instance). – Nate Eldredge Jun 28 '17 at 15:29
  • @NateEldredge I edited the post. I wish to analyze "measurable probability distributions". Does it make sense? – AvidLearner Jun 28 '17 at 15:38
  • No, it doesn't; I have no idea what that combination of words means. A probability distribution on a set is by definition a $\sigma$-algebra on the set and a countably additive probability measure on that $\sigma$-algebra. – Nate Eldredge Jun 28 '17 at 15:39

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