<p>Sigma additivity requires that if you have a countable collection <span class="math-container">$(B_i)_{i \ge 0}$</span> of mutually disjoint sets (<span class="math-container">$B_i \cap B_j = \emptyset$</span> for <span class="math-container">$i \ne j$</span>) then:
<span class="math-container">$$
\sum_{i \ge 0} \mu(B_i) = \mu(\cup_{i \ge 0} B_i).
$$</span></p>
<p>The sets <span class="math-container">$A_i$</span> are nested, hence not mutually disjoint. For the proof, the idea is to construct from these a collection of mutually disjoint sets whose union equals the union of the <span class="math-container">$A_i$</span>.</p>
<p>(<span class="math-container">$\to$</span>) Let <span class="math-container">$\mu$</span> be <span class="math-container">$\sigma$</span>-additive. Let <span class="math-container">$(A_i)_{i \in \mathbb{N}}$</span> be a countable collection of nested sets such that <span class="math-container">$A_i \uparrow A$</span>.</p>
<p>Iteratively define <span class="math-container">$B_0 = A_0$</span> and:
<span class="math-container">$$
B_n = A_n \setminus A_{n-1}.
$$</span>
Note that <span class="math-container">$(B_i)_{i \in \mathbb{N}}$</span> are mutually disjoint and that for all <span class="math-container">$n$</span>, <span class="math-container">$\cup_{i = 1}^n B_i = A_n$</span>. Also <span class="math-container">$\cup_{i \ge 0} B_i = \cup_{i \ge 0} A_i = A$</span>.</p>
<p>From <span class="math-container">$\sigma$</span>-additivity, we have:
<span class="math-container">$$
\begin{align*}
\mu(A) &amp;= \mu(\cup_{i \ge 0} A_i),\\
&amp; = \mu(\cup_{i \ge 0} B_i),\\
&amp; = \sum_{i = 0}^\infty \mu(B_i),\\
&amp; = \lim_{n \to \infty} \sum_{i = 0}^n \mu(B_i),\\
&amp; = \lim_{n \to \infty} \mu(A_n).
\end{align*}
$$</span></p>
<p><span class="math-container">$(\leftarrow)$</span>. Let <span class="math-container">$(B_i)_{i \ge 0}$</span> be a countable collection of mutually disjoint measurable sets. Define for all <span class="math-container">$n \ge 0$</span>, <span class="math-container">$A_n = \cup_{0 \le i \le n} B_i$</span>. Note that <span class="math-container">$A \equiv \cup_{i \ge 0} A_i = \cup_{i \ge 0} B_i$</span> and that <span class="math-container">$A_n \uparrow A$</span>.</p>
<p>So:
<span class="math-container">$$
\begin{align*}
\sum_{i \ge 0} \mu(B_i) &amp; = \lim_{n \to \infty} \sum_{i = 0}^n \mu(B_i),\\
&amp;= \lim_{n \to \infty} \mu(A_n), \tag{$\mu$ is additive}\\
&amp;= \mu(A),\\
&amp;= \mu(\cup_{i \ge 0} B_i).
\end{align*}
$$</span>
So, <span class="math-container">$\mu$</span> is <span class="math-container">$\sigma$</span>-additive.</p>

CC BY-SA 4.0
2024-02-27T09:00:13.197Z
2024-02-27T09:00:13.197Z
tdm
<p>I'm here to kindly ask u a check about the proof of the following statement: <br />
Let <span class="math-container">$\mathcal{A}$</span> be an algebra and <span class="math-container">$\mu : A \rightarrow [0,\infty]$</span> and additive function. Then <span class="math-container">$\mu$</span> is <span class="math-container">$\sigma$</span> additive iff for any sequence <span class="math-container">$\{A_n\} \subset \mathcal{A} $</span> such that <span class="math-container">$A_n \uparrow A \in \mathcal{A}$</span> we have <span class="math-container">$\mu(A_n) \uparrow \mu(A)$</span>.</p>
<p>Proof: <br />
<strong><span class="math-container">$(\Rightarrow)$</span> direction</strong>:<br />
Assumre <span class="math-container">$\mu$</span> is <span class="math-container">$\sigma$</span> additive s.t. <span class="math-container">$\mu(\cup_{n = 1}^{\infty}A_n)$</span> = <span class="math-container">$\sum_{n = 1}^{\infty} \mu(A_n)$</span>. Take an arbritary sequence <span class="math-container">$\{A_n\} \subset \mathcal{A}$</span> s.t. <span class="math-container">$A_n \uparrow A \in \mathcal{A}$</span>.Hence, <span class="math-container">$\cup_{n = 1}^{\infty}A_n = A$</span> so that  <span class="math-container">$\mu(\cup_{n = 1}^{\infty}A_n) = \mu(A)$</span>. Rewrite <span class="math-container">$\sum_{n = 1}^{\infty} \mu(A_n)$</span> as <span class="math-container">$lim_{N \rightarrow \infty}\sum_{n = 1}^{N} \mu(A_n)$</span> =   <span class="math-container">$lim_{N \rightarrow \infty}\mu(\cup_{n = 1}^{N} A_n) \text{(by additivity)} = lim_{N \rightarrow \infty} \mu(A_n) = \mu(A) \text{(by convergence)}$</span>.</p>
<p><strong><span class="math-container">$(\Leftarrow)$</span> direction</strong>:<br />
Consider a sequence <span class="math-container">$ {A_n} \subset \mathcal{F} $</span> s.t. <span class="math-container">$A_n \uparrow A$</span>, <span class="math-container">$A = \cup_{n = 1} ^ {\infty} A_n \in \mathcal{A}, \mu(A) = \mu(\cup_{n = 1}^{\infty} A_n) = lim_{N \rightarrow \infty} \mu(\cup_{n = 1}^{N} A_n)$</span>. We need to show that <span class="math-container">$\mu(\cup_{n = 1} ^ {\infty} A_n) = \sum_{n = 1}^{\infty} \mu(A_n) $</span>.</p>
<p>Since <span class="math-container">$A_n \uparrow A$</span>, then <span class="math-container">$\cup A_n = A$</span>. Hence, we need to show that <span class="math-container">$\mu(A) = \sum_{n = 1}^{\infty} \mu(A_n)  $</span> by show that it is both <span class="math-container">$\geq, \leq$</span>.<br />
But, we know that <span class="math-container">$\mu(A) = \mu(\cup_{n = 1} ^ {\infty} A_n) \leq \sum_{n = 1}^{\infty} \mu(A_n)$</span>. Moreover, take <span class="math-container">$B_n = A_n \backslash  \cup_{j&lt;n} A_j$</span> such that <span class="math-container">$B_i \cap B_j = \emptyset \forall i \neq j$</span>. But then, since <span class="math-container">$\{B_n\}$</span> are pairwise disjoint, we have that <span class="math-container">$\mu(\cup_{n = 1} ^ {\infty} B_n) = \sum_{n = 1} ^ {\infty} \mu(B_n)$</span>. Now <span class="math-container">$B_n \subset A_n$</span> so <span class="math-container">$\sum_{n = 1} ^ {\infty} \mu(B_n) \leq  \sum_{n = 1} ^ {\infty} \mu(A_n)$</span>. Moreover, <span class="math-container">$\mu(\cup_{n = 1} ^ {\infty} B_n) =\mu(\cup_{n = 1} ^ {\infty} A_n)$</span>. Hence, <span class="math-container">$\mu(A) = \mu(\cup_{n = 1} ^ {\infty} A_n) = \mu(\cup_{n = 1} ^ {\infty} B_n)  = \sum_{n = 1} ^ {\infty} \mu(B_n) \leq \sum_{n = 1} ^ {\infty} \mu(A_n) $</span>.</p>

CC BY-SA 4.0
2024-02-27T04:48:40.197Z
MichaelGreinecker if these are on topic here it might be a good idea to create a measure-theory tag.
Giskard
<mathematical-economics><proof><real-analysis>
Proving an additive function is sigma additive (Measure Theory)
Calogero Sortino
# Proving an additive function is sigma additive (Measure Theory)

(Asked by: username_0 on 2024-02-26)

I'm here to kindly ask u a check about the proof of the following statement:   
Let $\mathcal{A}$ be an algebra and $\mu : A \rightarrow [0,\infty]$ and additive function. Then $\mu$ is $\sigma$ additive iff for any sequence $\{A\_n\} \subset \mathcal{A} $ such that $A\_n \uparrow A \in \mathcal{A}$ we have $\mu(A\_n) \uparrow \mu(A)$.

Proof:   
**$(\Rightarrow)$ direction**:  
Assumre $\mu$ is $\sigma$ additive s.t. $\mu(\cup\_{n = 1}^{\infty}A\_n)$ = $\sum\_{n = 1}^{\infty} \mu(A\_n)$. Take an arbritary sequence $\{A\_n\} \subset \mathcal{A}$ s.t. $A\_n \uparrow A \in \mathcal{A}$.Hence, $\cup\_{n = 1}^{\infty}A\_n = A$ so that $\mu(\cup\_{n = 1}^{\infty}A\_n) = \mu(A)$. Rewrite $\sum\_{n = 1}^{\infty} \mu(A\_n)$ as $lim\_{N \rightarrow \infty}\sum\_{n = 1}^{N} \mu(A\_n)$ = $lim\_{N \rightarrow \infty}\mu(\cup\_{n = 1}^{N} A\_n) \text{(by additivity)} = lim\_{N \rightarrow \infty} \mu(A\_n) = \mu(A) \text{(by convergence)}$.

**$(\Leftarrow)$ direction**:  
Consider a sequence $ {A\_n} \subset \mathcal{F} $ s.t. $A\_n \uparrow A$, $A = \cup\_{n = 1} ^ {\infty} A\_n \in \mathcal{A}, \mu(A) = \mu(\cup\_{n = 1}^{\infty} A\_n) = lim\_{N \rightarrow \infty} \mu(\cup\_{n = 1}^{N} A\_n)$. We need to show that $\mu(\cup\_{n = 1} ^ {\infty} A\_n) = \sum\_{n = 1}^{\infty} \mu(A\_n) $.

Since $A\_n \uparrow A$, then $\cup A\_n = A$. Hence, we need to show that $\mu(A) = \sum\_{n = 1}^{\infty} \mu(A\_n) $ by show that it is both $\geq, \leq$.  
But, we know that $\mu(A) = \mu(\cup\_{n = 1} ^ {\infty} A\_n) \leq \sum\_{n = 1}^{\infty} \mu(A\_n)$. Moreover, take $B\_n = A\_n \backslash \cup\_{j such that $B\_i \cap B\_j = \emptyset \forall i \neq j$. But then, since $\{B\_n\}$ are pairwise disjoint, we have that $\mu(\cup\_{n = 1} ^ {\infty} B\_n) = \sum\_{n = 1} ^ {\infty} \mu(B\_n)$. Now $B\_n \subset A\_n$ so $\sum\_{n = 1} ^ {\infty} \mu(B\_n) \leq \sum\_{n = 1} ^ {\infty} \mu(A\_n)$. Moreover, $\mu(\cup\_{n = 1} ^ {\infty} B\_n) =\mu(\cup\_{n = 1} ^ {\infty} A\_n)$. Hence, $\mu(A) = \mu(\cup\_{n = 1} ^ {\infty} A\_n) = \mu(\cup\_{n = 1} ^ {\infty} B\_n) = \sum\_{n = 1} ^ {\infty} \mu(B\_n) \leq \sum\_{n = 1} ^ {\infty} \mu(A\_n) $.

**Tags:** mathematical-economics, proof, real-analysis

> **Comments:**
> **username_1**: MichaelGreinecker if these are on topic here it might be a good idea to create a measure-theory tag.  

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## 1 answers

### Answer by username_2 on 2024-02-27. Score: 2

Sigma additivity requires that if you have a countable collection $(B\_i)\_{i \ge 0}$ of mutually disjoint sets ($B\_i \cap B\_j = \emptyset$ for $i \ne j$) then:
$$
\sum\_{i \ge 0} \mu(B\_i) = \mu(\cup\_{i \ge 0} B\_i).
$$

The sets $A\_i$ are nested, hence not mutually disjoint. For the proof, the idea is to construct from these a collection of mutually disjoint sets whose union equals the union of the $A\_i$.

($\to$) Let $\mu$ be $\sigma$-additive. Let $(A\_i)\_{i \in \mathbb{N}}$ be a countable collection of nested sets such that $A\_i \uparrow A$.

Iteratively define $B\_0 = A\_0$ and:
$$
B\_n = A\_n \setminus A\_{n-1}.
$$
Note that $(B\_i)\_{i \in \mathbb{N}}$ are mutually disjoint and that for all $n$, $\cup\_{i = 1}^n B\_i = A\_n$. Also $\cup\_{i \ge 0} B\_i = \cup\_{i \ge 0} A\_i = A$.

From $\sigma$-additivity, we have:
$$
\begin{align\*}
\mu(A) &= \mu(\cup\_{i \ge 0} A\_i),\\
& = \mu(\cup\_{i \ge 0} B\_i),\\
& = \sum\_{i = 0}^\infty \mu(B\_i),\\
& = \lim\_{n \to \infty} \sum\_{i = 0}^n \mu(B\_i),\\
& = \lim\_{n \to \infty} \mu(A\_n).
\end{align\*}
$$

$(\leftarrow)$. Let $(B\_i)\_{i \ge 0}$ be a countable collection of mutually disjoint measurable sets. Define for all $n \ge 0$, $A\_n = \cup\_{0 \le i \le n} B\_i$. Note that $A \equiv \cup\_{i \ge 0} A\_i = \cup\_{i \ge 0} B\_i$ and that $A\_n \uparrow A$.

So:
$$
\begin{align\*}
\sum\_{i \ge 0} \mu(B\_i) & = \lim\_{n \to \infty} \sum\_{i = 0}^n \mu(B\_i),\\
&= \lim\_{n \to \infty} \mu(A\_n), \tag{$\mu$ is additive}\\
&= \mu(A),\\
&= \mu(\cup\_{i \ge 0} B\_i).
\end{align\*}
$$
So, $\mu$ is $\sigma$-additive.


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