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 $\mathcal{A}$.


  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.
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $\mathcal{A}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.