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Note that the definition of a measure here is different than in most contexts.

Let $\mathcal{C}$ be a semialgebra. A function $\mu: \mathcal{C} \to [0, \infty]$ is a measure if:

  1. $\mu(\emptyset) = 0$
  2. For any sequence of pairwise disjoint sets $\{A_n\}_{n=1}^{\infty}\subset\mathcal{C}$ with $\bigcup_{n=1}^{\infty}A_n \in \mathcal{C}$, $\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$.

I don't understand the following proposition:

Let $\mu$ be a measure on a semialgebra $\mathcal{C}$ and let $\mathcal{A}$ be the smallest algebra generated by $\mathcal{C}$. For each $A \in \mathcal{A}$, set $$\bar{\mu}(A) = \sum_{i=1}^{k}\mu(B_i)$$ if the set $A$ has the representation $A = \bigcup_{i=1}^{k}B_i$ for some pairwise disjoint $B_1, \dots, B_k \in \mathcal{C}$, $k < \infty$. Then $\bar{\mu}$ is independent of the representation of $A$ as $A = \bigcup_{i=1}^{k}B_i$.

Can someone explain "$\bar{\mu}$ is independent of the representation of $A$ as $A = \bigcup_{i=1}^{k}B_i$" to me?

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  • 3
    $\begingroup$ It means that, for a fixed set $A \in \mathcal A$, it doesn't matter which sets $B_i$ you choose, as long as $A = \bigcup_{i=1}^k B_i$, the value you get for $\overline \mu(A)$ will always be the same. $\endgroup$ – Dustan Levenstein Mar 3 '16 at 16:06
  • $\begingroup$ There can be more ways to write $A$ is a finite union of pairwise disjoint elements of $\mathcal C$. So it is not directly clear that $\overline{\mu}$ is well defined. In this context saying that there is independence is the same as saying that $\overline{\mu}$ is well defined. $\endgroup$ – drhab Mar 3 '16 at 16:10
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$A\in\mathcal{A}$ means that there exists $B_1,\dots,B_k\in\mathcal{C}$ such that $$A=\bigcup_{i=1}^kB_i$$ This is a representation of $A$. But it's not necessarily unique. You could also have an other representation $$A=\bigcup_{i=1}^lC_i$$ What is said here is that the result that you get for $\bar{\mu}(A)$ does not depend on the representation that you choose for $A$ but only on $A$, that is: $$\sum_{i=1}^k\mu(A_i)=\sum_{i=1}^l\mu(C_i)$$ This is necessary in order for $\bar{\mu}$ to be well defined.

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  • $\begingroup$ What is the difference between $\mu$ and $\bar\mu$ ? Is $\bar\mu$ the completed measure ? $\endgroup$ – nicomezi Mar 3 '16 at 16:10
  • $\begingroup$ $\mu$ is only defined over $\mathcal{C}$, wheras $\bar(\mu)$ is an extension of $\bar{\mu}$ to the algebra generated by $\mathcal{C}$. Of course if $A\in\mathcal{C}$, $\bar{\mu}(A)=\mu(A)$. $\endgroup$ – Augustin Mar 3 '16 at 16:15

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