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?

  • $\begingroup$ This may be a really silly question, but is $S$ countable? $\endgroup$ Jun 27, 2017 at 13:42
  • $\begingroup$ @SystematicDisintegration For $n=1$ we get something in the flavor $S=[0,1]$, so I believe the answer is no. $\endgroup$
    – omerbp
    Jun 27, 2017 at 16:09
  • $\begingroup$ How, then, have you sorted the elements of $S$ into an n-tuple? Isn't this an enumeration? $\endgroup$ Jun 27, 2017 at 16:51
  • $\begingroup$ @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 $i<j$, $s'_{(i)}\leq s'_{(j)}$. $\endgroup$
    – omerbp
    Jun 27, 2017 at 17:31
  • 1
    $\begingroup$ 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). $\endgroup$ Jun 28, 2017 at 15:29


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