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Let us consider the problem 3.D. in the book

$\textbf{The Elements of Integration and Lebesgue Measure}$

of Robert G. Bartle

Let $X=\mathbb{N}$ and $\mathcal{A}$ be the $\sigma-$algebra of all subsets of $\mathbb{N}$. If $(a_n)$ is a sequence of nonnegative real numbers and if we define $\mu$ by $$ \mu(\emptyset)=0; \quad \mu(E)=\sum_{n\in E}a_n, \quad E\ne\emptyset, $$ then $\mu$ is a measure on $\mathcal{A}$. Conversely, every measure on $\mathcal{A}$ is obtained in this way for some sequence $(a_n)$ in $\overline{R}^+$.

I am stuck in proving the countably additive property of $\mu$.

My attempt. I intend to use the theory of double index series to prove countably additive property.

Thank you for all solutions.

impartialmale
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  • can you see finite additivity? –  Nov 02 '14 at 10:57
  • Nope, I cannot see the finite additivity. – impartialmale Nov 02 '14 at 11:01
  • Maybe it's too obvious. If we have $A, B \subseteq \mathbb N$ with $A \cap B = \emptyset$, then $\mu(A \cup B) = \sum_{n \in A \cup B} a_n = \sum_{n \in A} a_n + \sum _{n \in B} a_n = \mu(A) + \mu(B)$, since A and B are disjoint. That's finite additivity. Countable additivity is completely analogous. – jflipp Nov 02 '14 at 11:11
  • Thank you for your comments. But I do not think it is too obvious for the case finite additivity and countable additivity. We have to use some results related to theory of real series to prove two properties. – impartialmale Nov 02 '14 at 11:15
  • That's true. But luckily, we have $a_n \geq 0$, so any $\sum a_j$ can only diverge to $+\infty$. That's the lesson to learn here. – jflipp Nov 02 '14 at 16:14

2 Answers2

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The proof of countable additivity: \begin{align*} \mu\left( \bigcup_{i=1}^\infty E_i\right)&=\sum_{n\in\bigcup _{i=1}^\infty E_i}a_n \end{align*} Notice that $(E_i)$ is a disjoint sequence, then n belongs to one and only one $E_i$, which means: $$\sum_{n\in\bigcup _{i=1}^\infty E_i}a_n = \sum_{i=1}^{\infty}\sum_{n\in E_i}a_n = \sum_{i=1}^{\infty} \mu(E_i)$$

The proof of every measure on A is obtained in this way for some sequence $(a_n)$ in $\overline{R}^+$:

For every $E\subset N$, $E=\bigcup_{i=1}^{\infty}{\{n_i\}}$, where $n_i\in N$ or $\{n_i\}=\emptyset$. From the countable additivity of $\mu$ we know that: $$\mu(E) = \sum_{n \in E}\mu(\{n\})$$ We set $\mu(\{n\}):=a_n$, which establishs the proposition.

horatio1587
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We have to check the following

  • $\mu(\emptyset)=0$, by definition

  • $\mu(E)=\sum_{n\in E}a_n\geq 0$, since $a_n\geq 0$, for all $n$.

  • Let $\{E_k\}$ be disjoint, then\begin{align*} \mu\left( \bigcup_{k=1}^\infty E_k\right)&=\sum_{n\in\bigcup E_k}a_n=\sum_{k=1}^\infty \sum_{n\in E_k}a_n=\sum_{k=1}^\infty\mu (E_k). \end{align*}

Thus, $\mu$ is a measure on X. Conversely, for $(a_n)$ in $\overline{\mathbb{R}}^+$, we can obtain any measure on X using the enunciated definition of $\mu$.

sam wolfe
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