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Helpful comments I received here lead me to re-asking the question, hopefully better this time.

Let $S=\left\{ (s^1,s^2,\dots s^n) \in [0,1]^n : s^1 \leq s^2 \cdots \leq s^n \right\}$, and let $x$ be a probability distribution over $(S,\mathcal B(S))$, where $\mathcal B(S)$ is the standard Borel $\sigma$-algebra over $S$.

Define $\mu: 2^{[0,1]}\rightarrow [0,n]$ such that for every Lebesgue measurable set $A\subseteq [0,1]$

$$ \mu(A) = \int_{S}\sum_{i=1}^n \mathbb 1_A(s^i)dx(s),$$

where $\mathbb 1_A$ is the characteristic function of the set $A$, and $s=(s^1,\dots, s^n)$ is the name of the variable we integrate over. My questions are:

(1) How can one show that $\mu$ is a measure? (see my solution below)

(2) Am I using the right notations in the definition of $\mu$? is there a better way to formulate the problem?

Referring to (1), due to Fubini's theorem, we can swap the order of summation. Since $x$ is a probability distribution, using the right argument one can claim that the projection of $x$ onto every entry is a measure. Hence, $\mu$ is a sum of measures, thus a measure by itself.

However, I am not sure how to show this formally.

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    $\begingroup$ Could you clarify what the $s^{i}$ are in the definition of your measure on a set $A$? $\endgroup$
    – Dan
    Jul 4 '17 at 7:19
  • $\begingroup$ @Dan Thanks, I added a description. $\endgroup$
    – omerbp
    Jul 4 '17 at 7:26
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We don't need Fubini's b/c we have a finite sum and instead can use linearity to get a sum of integrals. As far as showing a function taking $A \mapsto \int_{S} 1_{A} (s^{i}) \ dx$ is a measure, I'm not quite sure what this notation would mean. Perhaps the biggest of our worries is that $x$ is a multivariate probability distribution, and we're integrating with respect to only one $s^{i}$. Let's suppose that $$x = \overbrace{x' \times \cdots \times x'}^{n \text{ times}}$$ where $x'$ is a probability distribution on $([0,1],\mathcal{B}([0,1]))$. The way I'm interpreting $\int_{S} 1_{A} (s^{i}) \ dx$ is instead $\int_{S} 1_{A} (s^{i}) \ dx'$ and the way I'm interpreting this is $\int_{S} 1_{A} (s^{i}) \ dx' = \int_{S_{i}} 1_{A} \ dx' = x'(A \cap S_{i}),$ where by $S_{i}$ I mean the $i$-component of $S$ (subset of $[0,1]$). So, notationally, there are still some things to be worked out, I think. Am I reading this correctly? We have $S_{i}$ and $A$ are measurable, so $A \cap S_{i}$ is measurable. It follows pretty quickly (from the fact that $x'$ is a measure) that a function taking $A \mapsto \int_{S'} 1_{A} \ dx'$ is a measure. If this definition of $\mu$ is what you wanted, then you'd have $\mu(A) = \sum_{i=1}^{n} \int_{S_{i}} 1_{A} \ dx'$.

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  • $\begingroup$ Thanks for your response. I agree with the first part, however notice the $x$ is not necessarily a product distribution. Therefore, the problem is how to refer to the projection of $x$ on each of its components. $\endgroup$
    – omerbp
    Jul 4 '17 at 8:59
  • $\begingroup$ If you want to restrict $x$ in such a way, I think the Disintegration Theorem might be useful. $\endgroup$
    – Dan
    Jul 4 '17 at 10:18
  • $\begingroup$ Also, I didn't really notice that you specify that $A$ is a Lebesgue measurable set. Why do you make it specifically Lebesgue measurable? Then what do you mean by "standard $\sigma$-algebra"? $\endgroup$
    – Dan
    Jul 4 '17 at 10:39
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    $\begingroup$ I meant Borel $sigma$-algebra, thanks. The Disintegration theorem is indeed helpful, I'll try to use it. $\endgroup$
    – omerbp
    Jul 4 '17 at 10:44

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