<p>I want to generalize:  <span class="math-container">$\sum_nu_n(x_n)$</span>.</p>
<p>Here <span class="math-container">$x_1,x_2,..,x_n,...$</span> are commodities. There are infinite number of commodities: <span class="math-container">$n\in\mathbb N$</span> or <span class="math-container">$n\in \mathbb R_+$</span></p>
<p>The following <strong>not</strong> a candidate: <span class="math-container">$\int_{n\in\mathbb R}u_n(x_n)d\mu$</span>. It is not a candidate because we need <strong>additional</strong> assumptions for <span class="math-container">$\mu$</span> being a measure over sigma-field. By adding additional assumption that is too strong, we cannot say it is a good generalization.</p>
<p>What is too strong: assuming sigma-algebra or countably additive measure. Sigma algebra makes no sense when <span class="math-container">$n$</span> finite.</p>
<p>What assumption is acceptable: assuming non-sigma field, or any other assumption commonly assumed within finite settings.</p>
<p>One thing have in my mind is: <span class="math-container">$x\succsim y \iff \sum_{n\in\mathbb N}(u_n(x_n)-u_n(y_n))\geq0$</span>. In this case we have <span class="math-container">$x\succ y$</span> when <span class="math-container">$x_n=2n$</span> and <span class="math-container">$y_n=n$</span></p>
<p>Relation to literature:</p>
<ol>
<li><p><a href="https://link.springer.com/article/10.1007/s11238-013-9391-2" rel="nofollow noreferrer">https://link.springer.com/article/10.1007/s11238-013-9391-2</a> This paper focuses on the infinite dimensional representation <span class="math-container">$\sum_n(u(x_n)-u(y_n))\geq 0$</span> which is similar to the representation above.</p>
</li>
<li><p><a href="https://www.sciencedirect.com/science/article/abs/pii/0022249674900315" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/abs/pii/0022249674900315</a>
This paper is only distantly related because my question requires that all directions are separable.</p>
</li>
<li><p><a href="https://www.jstor.org/stable/1909948" rel="nofollow noreferrer">https://www.jstor.org/stable/1909948</a> Fishburn's ECMA</p>
</li>
<li><p>link.springer.com/chapter/10.1007/978-3-642-48773-6_3
Debreu 1997</p>
</li>
</ol>

CC BY-SA 4.0
2023-08-17T10:52:14.373Z
Could you explain what additional assumptions you consider 'too strong'? So it would be easier to understand your question. Because, when you pass  from finite to infinity necessarily there must be new assumptions, the passage from finite to infinity isn't 'painless'. Your question is interesting but needs details.
BakerStreet
CC BY-SA 4.0
2023-08-17T11:38:38.007Z
@BakerStreet Sure please bear with me for a moment
High GPA
CC BY-SA 4.0
2023-08-17T11:39:39.160Z
Of course, when you have time.
BakerStreet
CC BY-SA 4.0
2023-08-17T12:10:46.703Z
@BakerStreet Does this look better?
High GPA
CC BY-SA 4.0
2023-08-17T12:20:43.443Z
Yes, I think so. But, how can you define an integral and a measure without a sigma-algebra?
BakerStreet
CC BY-SA 4.0
2023-08-17T12:24:08.380Z
@BakerStreet I added two papers. The first one might be very helpful. I just find it a few minutes ago.
High GPA
CC BY-SA 4.0
2023-08-17T12:40:49.860Z
Moreover, I’m not sure that a sigma-algebra makes no sense for finite sets. Actually, finite is a particular case of countably infinite, where you add to the *n* finite sets an infinite number of empty sets. For example, the finite union of *n* sets $A_i$ can be written as $\cup _{i=0}^{n} A_i$= $(\cup _{i=0}^{n}A_i )\cup \emptyset \cup \emptyset \cup \emptyset ….$
BakerStreet
CC BY-SA 4.0
2023-08-17T12:59:11.597Z
Therefore, sigma (countable)- additivity implies finite additivity.
BakerStreet
CC BY-SA 4.0
2023-08-17T13:49:02.580Z
You talk about additional assumptions but not about the assumptions you are willing to make to start with. I'm not sure I understand what the question asks for.
Michael Greinecker
CC BY-SA 4.0
2023-08-18T11:56:06.777Z
@BakerStreet Sigma algebra is usually neither observable nor testable, though.
High GPA
CC BY-SA 4.0
2023-08-18T12:23:23.060Z
Of course, as many mathematical objects. It is just a definition, in set theory. I was just saying that a sigma-algebra is not necessarily something that pertains to countable infinite collections of sets, but also to finite collections: *a collection of subsets of $\mathbb{R}$ is called a sigma-algebra provided it contains $\mathbb{R}$ and is closed with respect to the formation of complements and countable unions*. But, as I said above, a finite collection can be considered a countable collection if  infinite empty sets are added to the union.
BakerStreet
CC BY-SA 4.0
2023-08-18T12:24:14.210Z
So,  your objection that a sigma-algebra makes no sense for finite collections isn’t true, and it can be considered a generalization, in the sense that it encompasses also the finite case.
BakerStreet
CC BY-SA 4.0
2023-08-18T14:40:02.923Z
Maybe you can begin with this article, of 1997, by Debreu about additive preferences, that refers also to the infinite commodities case. 


https://link.springer.com/chapter/10.1007/978-3-642-48773-6_3
BakerStreet
CC BY-SA 4.0
2023-08-27T13:25:52.477Z
@BakerStreet Let me try!
High GPA
<microeconomics><mathematical-economics><preferences><theory><decision-theory>
Generalization of Debreu's additive utility function $\sum_nu_n(x_n)$ with infinite number of commodities
High GPA
# Generalization of Debreu's additive utility function $\sum_nu_n(x_n)$ with infinite number of commodities

(Asked by: username_0 on 2023-08-17)

I want to generalize: $\sum\_nu\_n(x\_n)$.

Here $x\_1,x\_2,..,x\_n,...$ are commodities. There are infinite number of commodities: $n\in\mathbb N$ or $n\in \mathbb R\_+$

The following **not** a candidate: $\int\_{n\in\mathbb R}u\_n(x\_n)d\mu$. It is not a candidate because we need **additional** assumptions for $\mu$ being a measure over sigma-field. By adding additional assumption that is too strong, we cannot say it is a good generalization.

What is too strong: assuming sigma-algebra or countably additive measure. Sigma algebra makes no sense when $n$ finite.

What assumption is acceptable: assuming non-sigma field, or any other assumption commonly assumed within finite settings.

One thing have in my mind is: $x\succsim y \iff \sum\_{n\in\mathbb N}(u\_n(x\_n)-u\_n(y\_n))\geq0$. In this case we have $x\succ y$ when $x\_n=2n$ and $y\_n=n$

Relation to literature:

1. <https://link.springer.com/article/10.1007/s11238-013-9391-2> This paper focuses on the infinite dimensional representation $\sum\_n(u(x\_n)-u(y\_n))\geq 0$ which is similar to the representation above.
2. <https://www.sciencedirect.com/science/article/abs/pii/0022249674900315>
   This paper is only distantly related because my question requires that all directions are separable.
3. <https://www.jstor.org/stable/1909948> Fishburn's ECMA
4. link.springer.com/chapter/10.1007/978-3-642-48773-6\_3
   Debreu 1997

**Tags:** microeconomics, mathematical-economics, preferences, theory, decision-theory
> **Comments:**
> **username_1**: Could you explain what additional assumptions you consider 'too strong'? So it would be easier to understand your question. Because, when you pass  from finite to infinity necessarily there must be new assumptions, the passage from finite to infinity isn't 'painless'. Your question is interesting but needs details.  
> **username_0**: @BakerStreet Sure please bear with me for a moment  
> **username_1**: Of course, when you have time.  
> **username_0**: @BakerStreet Does this look better?  
> **username_1**: Yes, I think so. But, how can you define an integral and a measure without a sigma-algebra?  
> **username_0**: @BakerStreet I added two papers. The first one might be very helpful. I just find it a few minutes ago.  
> **username_1**: Moreover, I’m not sure that a sigma-algebra makes no sense for finite sets. Actually, finite is a particular case of countably infinite, where you add to the *n* finite sets an infinite number of empty sets. For example, the finite union of *n* sets $A_i$ can be written as $\cup _{i=0}^{n} A_i$= $(\cup _{i=0}^{n}A_i )\cup \emptyset \cup \emptyset \cup \emptyset ….$  
> **username_1**: Therefore, sigma (countable)- additivity implies finite additivity.  
> **username_2**: You talk about additional assumptions but not about the assumptions you are willing to make to start with. I'm not sure I understand what the question asks for.  
> **username_0**: @BakerStreet Sigma algebra is usually neither observable nor testable, though.  
> **username_1**: Of course, as many mathematical objects. It is just a definition, in set theory. I was just saying that a sigma-algebra is not necessarily something that pertains to countable infinite collections of sets, but also to finite collections: *a collection of subsets of $\mathbb{R}$ is called a sigma-algebra provided it contains $\mathbb{R}$ and is closed with respect to the formation of complements and countable unions*. But, as I said above, a finite collection can be considered a countable collection if  infinite empty sets are added to the union.  
> **username_1**: So,  your objection that a sigma-algebra makes no sense for finite collections isn’t true, and it can be considered a generalization, in the sense that it encompasses also the finite case.  
> **username_1**: Maybe you can begin with this article, of 1997, by Debreu about additive preferences, that refers also to the infinite commodities case.    https://link.springer.com/chapter/10.1007/978-3-642-48773-6_3  
> **username_0**: @BakerStreet Let me try!  

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## 0 answers

