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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 $\bigcup_{n=1}^{\infty}A_n \in \mathcal{C}$, $\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$.

Claims to be proven:

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

  1. $\bar{\mu}$ is independent of the representation of $A$ as $A = \bigcup_{i=1}^{k}B_i$.
  2. $\bar{\mu}$ is finitely additive on $\mathcal{A}$.

Attempts:

  1. Suppose $A \in \mathcal{A}$ is such that $A = \bigcup_{i=1}^{k}B_i = \bigcup_{j=1}^{\ell}C_j$, where $B_i, C_j \in \mathcal{C}$ are pairwise disjoint. Then obviously $\bar{\mu}(A) = \sum_{i=1}^{k}\mu(B_i)$ and $\bar{\mu}(A) = \sum_{j=1}^{\ell}\mu(C_j)$. Thus the claim is proven.
  2. Let $A, B \in \mathcal{A}$ be disjoint. Since $\mathcal{A}$ is an algebra, $A \cup B \in \mathcal{A}$. Write $A = \bigcup_{i=1}^{k}A_i$ and $B = \bigcup_{j=1}^{\ell}B_j$. I'm struggling with how I should write $A \cup B$ as a union of pairwise disjoint sets.
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    $\begingroup$ Ad (1): No. You have to prove that the sums are equal. $\endgroup$ – Friedrich Philipp Mar 4 '16 at 16:54
  • $\begingroup$ @FriedrichPhilipp Sorry, can you clarify? I don't see this beyond the trivial fact that $\bar{\mu}(A) = \bar{\mu}(A)$. We have to show that the sums themselves are equal? How? $\endgroup$ – Clarinetist Mar 4 '16 at 16:56
  • $\begingroup$ It is not clear from the scratch that this definition is proper. You could have C's and B's such that both unions equal A but $\sum\mu(B_i)\neq\sum\mu(C_i)$. Therefore, it has to be proved that the sums actually give the same value. $\endgroup$ – Friedrich Philipp Mar 4 '16 at 16:59
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    $\begingroup$ You can't use $\bar{\mu}$ in your proof, as its goal is to prove that it is well defined. What you have to prove is that if $A=\bigcup_{i=1}^kB_i=\bigcup_{i=1}^lC_i$, then $\sum_{i=1}^k\mu(B_i)=\sum_{i=1}^l\mu(C_i)$. To do this, use the properties of $\mu$, which is defined on $\mathcal{C}$. $\endgroup$ – Augustin Mar 4 '16 at 17:00
  • $\begingroup$ @Augustin Okay, I follow that. Thanks. Do you have any suggestions on 2? $\endgroup$ – Clarinetist Mar 4 '16 at 17:03
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Let it be that $A=\bigcup_{i=1}^{k}B_{i}=\bigcup_{j=1}^{m}C_{j}$ where $B_{1},\dots,B_{k}$ are pairwise disjoint elements of $\mathcal{C}$ and also where $C_{1},\dots,C_{m}$ are pairwise disjoint elements of $\mathcal{C}$ .

$\mu$ is additive on $\mathcal{C}$ and as a semialgebra $\mathcal{C}$ is closed under finite intersections. So the sets $B_i\cap C_j$ are mutually disjoint elements of $\mathcal C$ and we find:

$$\sum_{i=1}^{k}\mu B_{i}=\sum_{i=1}^{k}\sum_{j=1}^{m}\mu\left(B_{i}\cap C_{j}\right)=\sum_{j=1}^{m}\sum_{i=1}^{k}\mu\left(B_{i}\cap C_{j}\right)=\sum_{j=1}^{m}\mu C_{j}$$

Proved is now that $\bar{\mu}$ is independent of the representation of $A$.

Let it be that $A=\bigcup_{r=1}^{n}A_{r}$ where $A_{1},\dots,A_{n}\in\mathcal{A}$ are mutually disjoint.

As elements of the algebra generated by semialgebra $\mathcal{C}$ every $A_{r}$ can be written as $A_{r}=\bigcup_{l=1}^{k_{r}}B_{r,l}$ where the $B_{r,1},\dots,B_{r,k_{r}}$ are mutually disjoint elements of $\mathcal{C}$.

Then: $$A=\bigcup_{r=1}^{n}A_{r}=\bigcup_{r=1}^{n}\bigcup_{l=1}^{k_{r}}B_{r,l}$$ and: $$\bar{\mu}A=\sum_{r=1}^{n}\sum_{l=1}^{k_{r}}\mu B_{r,l}=\sum_{r=1}^{n}\bar{\mu}A_{r}$$ Proved is now that $\bar{\mu}$ is finitely additive on $\mathcal{A}$.

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