<p>The first direction is fine, the second has some small problems and some scope for improvement.</p>
<p>There is the notational issue that <span class="math-container">$\cup_{n=1}^\infty \in \mathcal{F}$</span> should be <span class="math-container">$\cup_{n=1}^\infty A_n \in \mathcal{F}$</span>.</p>
<p><span class="math-container">$\tilde{A_2} = A_2 \cap (\cup_{i \neq 2} A_i^c)$</span> means that <span class="math-container">$\tilde{A_2}$</span> is the set of all points that are in <span class="math-container">$A_2$</span> and are not in some <span class="math-container">$A_i$</span> with <span class="math-container">$i\neq 2$</span>. This does in general not work, let <span class="math-container">$A_1=\{0\}$</span>, and <span class="math-container">$A_i=\{1\}$</span> for <span class="math-container">$i\neq 1$</span>. Then <span class="math-container">$\tilde{A_i}=\emptyset$</span> for <span class="math-container">$i\neq 1$</span>. What works is <span class="math-container">$\tilde{A_i} = A_i \cap (\cup_{j&lt;i} A_j^c)$</span>. With this change, the proof works.</p>
<p>However, the contradiction argument is superfluous. You are then directly showing that a field that is closed under disjoint countable unions is closed under arbitrary countable unions.</p>

CC BY-SA 4.0
2024-02-13T21:10:11.827Z
yeah, that makes sense. Thanks a lot for your feedback!!
Calogero Sortino
CC BY-SA 4.0
2024-02-13T12:56:26.290Z
2024-02-13T12:56:26.290Z
Michael Greinecker
<p>I'm trying to prove the following statement:
<a href="https://i.sstatic.net/4v32m.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4v32m.png" alt="enter image description here" /></a></p>
<p>I did the following proof but i donno if it makes sense:</p>
<p><strong>Proof of</strong> <span class="math-container">$(\Rightarrow)$</span> <strong>direction:</strong></p>
<p>Assume <span class="math-container">$\mathcal{F}$</span> is a <span class="math-container">$\sigma$</span>-algebra. Assume also <span class="math-container">$\{A_n\} \subset \mathcal{F}$</span> and <span class="math-container">$A_i \cap A_j = \emptyset$</span> for all <span class="math-container">$i \neq j$</span>. Hence, from the definition of a <span class="math-container">$\sigma$</span>-algebra, <span class="math-container">$\cup_{n=1}^\infty \in \mathcal{F}$</span>.</p>
<p><strong>Proof of</strong> <span class="math-container">$(\Leftarrow)$</span> <strong>direction:</strong></p>
<p>Assume <span class="math-container">$\{A_n\} \subset \mathcal{F}$</span> and <span class="math-container">$A_i \cap A_j = \emptyset$</span> for all <span class="math-container">$i \neq j \Rightarrow \cup_{n=1}^\infty \in \mathcal{F}$</span>. We need to show that <span class="math-container">$\mathcal{F}$</span> is a sigma algebra. Assume by contradiction that <span class="math-container">$\mathcal{F}$</span> is not a <span class="math-container">$\sigma$</span>-algebra. Since <span class="math-container">$\mathcal{F}$</span> is an algebra, assuming that <span class="math-container">$\mathcal{F}$</span> is not a sigma algebra means assuming that <span class="math-container">$\{A_n\} \subset \mathcal{F}$</span> and <span class="math-container">$\cup_{n=1}^\infty \notin \mathcal{F}$</span>. By definition of <span class="math-container">$\{A_n\} \subset \mathcal{F}$</span>, we have that <span class="math-container">$\forall A_i \in \{A_n\} \Rightarrow A_i \in \mathcal{F}$</span>. Moreover, since <span class="math-container">$\mathcal{F}$</span> is an algebra, <span class="math-container">$A_i^c \in \mathcal{F}$</span>, <span class="math-container">$\cup_{n=1}^n A_i \in \mathcal{F}$</span>, and <span class="math-container">$\cup_{n=1}^n A_i^c \in \mathcal{F}$</span>. Now define <span class="math-container">$\tilde{A_1} = A_1$</span>, <span class="math-container">$\tilde{A_2} = A_2 \cap (\cup_{i \neq 2} A_i^c)$</span> and so on. But then <span class="math-container">$\tilde{A_i} \cap \tilde{A_j} = \emptyset$</span> for all <span class="math-container">$i \neq j$</span>. Hence, by hypothesis, <span class="math-container">$\cup_{n=1}^\infty \in \mathcal{F}$</span> so we reach a contradiction.</p>
<p>Could you please review the proof and provide me feedback?</p>
<p>Thanks a lot!</p>

CC BY-SA 4.0
2024-02-14T11:52:06.287Z
In addition to the suggestions of Michael Greineker (avoid the  proof by contradiction) you should explicitely point out that  $\cup_{n=1}^\infty A_i= \cup_{n=1}^\infty \tilde A_i$.

Because the point is that an arbitrary union of a countable collection of sets in an algebra is also the union of a collection of (two by two) disjoint sets of the algebra: I think this is the idea underlying  your proof.
BakerStreet
<mathematical-economics><proof><real-analysis><proof-reference>
Is this proof correct (measure theory)?
Calogero Sortino
# Is this proof correct (measure theory)?

(Asked by: username_0 on 2024-02-12)

I'm trying to prove the following statement:
[![enter image description here](https://i.sstatic.net/4v32m.png)](https://i.sstatic.net/4v32m.png)

I did the following proof but i donno if it makes sense:

**Proof of** $(\Rightarrow)$ **direction:**

Assume $\mathcal{F}$ is a $\sigma$-algebra. Assume also $\{A\_n\} \subset \mathcal{F}$ and $A\_i \cap A\_j = \emptyset$ for all $i \neq j$. Hence, from the definition of a $\sigma$-algebra, $\cup\_{n=1}^\infty \in \mathcal{F}$.

**Proof of** $(\Leftarrow)$ **direction:**

Assume $\{A\_n\} \subset \mathcal{F}$ and $A\_i \cap A\_j = \emptyset$ for all $i \neq j \Rightarrow \cup\_{n=1}^\infty \in \mathcal{F}$. We need to show that $\mathcal{F}$ is a sigma algebra. Assume by contradiction that $\mathcal{F}$ is not a $\sigma$-algebra. Since $\mathcal{F}$ is an algebra, assuming that $\mathcal{F}$ is not a sigma algebra means assuming that $\{A\_n\} \subset \mathcal{F}$ and $\cup\_{n=1}^\infty \notin \mathcal{F}$. By definition of $\{A\_n\} \subset \mathcal{F}$, we have that $\forall A\_i \in \{A\_n\} \Rightarrow A\_i \in \mathcal{F}$. Moreover, since $\mathcal{F}$ is an algebra, $A\_i^c \in \mathcal{F}$, $\cup\_{n=1}^n A\_i \in \mathcal{F}$, and $\cup\_{n=1}^n A\_i^c \in \mathcal{F}$. Now define $\tilde{A\_1} = A\_1$, $\tilde{A\_2} = A\_2 \cap (\cup\_{i \neq 2} A\_i^c)$ and so on. But then $\tilde{A\_i} \cap \tilde{A\_j} = \emptyset$ for all $i \neq j$. Hence, by hypothesis, $\cup\_{n=1}^\infty \in \mathcal{F}$ so we reach a contradiction.

Could you please review the proof and provide me feedback?

Thanks a lot!

**Tags:** mathematical-economics, proof, real-analysis, proof-reference
> **Comments:**
> **username_1**: In addition to the suggestions of Michael Greineker (avoid the  proof by contradiction) you should explicitely point out that  $\cup_{n=1}^\infty A_i= \cup_{n=1}^\infty \tilde A_i$.  Because the point is that an arbitrary union of a countable collection of sets in an algebra is also the union of a collection of (two by two) disjoint sets of the algebra: I think this is the idea underlying  your proof.  

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## 1 answers

### Answer by username_2 on 2024-02-13. Score: 3

The first direction is fine, the second has some small problems and some scope for improvement.

There is the notational issue that $\cup\_{n=1}^\infty \in \mathcal{F}$ should be $\cup\_{n=1}^\infty A\_n \in \mathcal{F}$.

$\tilde{A\_2} = A\_2 \cap (\cup\_{i \neq 2} A\_i^c)$ means that $\tilde{A\_2}$ is the set of all points that are in $A\_2$ and are not in some $A\_i$ with $i\neq 2$. This does in general not work, let $A\_1=\{0\}$, and $A\_i=\{1\}$ for $i\neq 1$. Then $\tilde{A\_i}=\emptyset$ for $i\neq 1$. What works is $\tilde{A\_i} = A\_i \cap (\cup\_{j. With this change, the proof works.

However, the contradiction argument is superfluous. You are then directly showing that a field that is closed under disjoint countable unions is closed under arbitrary countable unions.

> **Comments:**
> **username_0**: yeah, that makes sense. Thanks a lot for your feedback!!  

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