<p>I believe we get it from the axioms you have provided with a few technical conditions on the spaces we are working with.</p>
<p>Let <span class="math-container">$X$</span> be a separable, metrizable space (like your <span class="math-container">$\mathbb{R}$</span>) and let <span class="math-container">$\Delta(X)$</span> be the set of countably additive probability distributions, endowed with the weak convergence topology.</p>
<p>The complete preorder <span class="math-container">$\succeq$</span> on <span class="math-container">$\Delta(X)$</span> satisfies Continuity and Independence if and only if it has an expected utility representation:</p>
<p><span class="math-container">$$p \succeq q \quad \text{ if and only if } \quad \int_X u(x) dp \geq \int_X u(x)dq$$</span></p>
<p>where <span class="math-container">$u : X \rightarrow \mathbb{R}$</span> is continuous and bounded.</p>

CC BY-SA 4.0
2024-07-12T16:38:14.000Z
Hi, the usual continuity conditions for EU are mixture continuity and Archimedean. Could you please explain how is your continuity stronger than the mixture continuity?
dodo
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2024-07-12T17:07:29.873Z
Its stronger in the sense that we require the preference to be continuous with respect to the weak convergence topology. i.e. $(p_n) \rightarrow p$ if for every bounded and continuous $f$, $\int f(x) dp_n \rightarrow \int f(x) dp$.
Walrasian Auctioneer
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2024-07-12T12:23:53.283Z
2024-07-12T12:23:53.283Z
Walrasian Auctioneer
<p>Consider EU on monetary outcomes. Say we have a utility function <span class="math-container">$u:\mathbb R\to\mathbb R$</span></p>
<p>The common axioms of EU are continuity, independence and weak order.</p>
<p>These axioms do not imply that the utility function is continuous.</p>
<p>My question: how to reach a continuous utility?</p>
<p>Edit: the usual continuity conditions for EU are mixture continuity and Archimedean.</p>

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2024-07-12T09:03:07.060Z
Expected Utility is linear is probabilities and therefore, it is continuous.
Amit
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2024-07-12T11:25:25.343Z
This is a fairly technical issue. There is a nice discussion in the "Notes on Theory of Choice" by Kreps.
Michael Greinecker
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2024-07-12T16:33:43.603Z
@Amit We are talking about the continuity of small u function.
dodo
CC BY-SA 4.0
2024-07-12T22:02:31.560Z
@dodo Thanks. Yes, I misinterpreted.
Amit
<microeconomics><theory><expected-utility>
how to reach Continuous Expected Utility (EU)?
dodo
# how to reach Continuous Expected Utility (EU)?

(Asked by: username_0 on 2024-07-12)

Consider EU on monetary outcomes. Say we have a utility function $u:\mathbb R\to\mathbb R$

The common axioms of EU are continuity, independence and weak order.

These axioms do not imply that the utility function is continuous.

My question: how to reach a continuous utility?

Edit: the usual continuity conditions for EU are mixture continuity and Archimedean.

**Tags:** microeconomics, theory, expected-utility
> **Comments:**
> **username_1**: Expected Utility is linear is probabilities and therefore, it is continuous.  
> **username_2**: This is a fairly technical issue. There is a nice discussion in the "Notes on Theory of Choice" by Kreps.  
> **username_0**: @Amit We are talking about the continuity of small u function.  
> **username_1**: @dodo Thanks. Yes, I misinterpreted.  

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## 1 answers

### Answer by username_3 on 2024-07-12. Score: 1 | Accepted answer

I believe we get it from the axioms you have provided with a few technical conditions on the spaces we are working with.

Let $X$ be a separable, metrizable space (like your $\mathbb{R}$) and let $\Delta(X)$ be the set of countably additive probability distributions, endowed with the weak convergence topology.

The complete preorder $\succeq$ on $\Delta(X)$ satisfies Continuity and Independence if and only if it has an expected utility representation:

$$p \succeq q \quad \text{ if and only if } \quad \int\_X u(x) dp \geq \int\_X u(x)dq$$

where $u : X \rightarrow \mathbb{R}$ is continuous and bounded.

> **Comments:**
> **username_0**: Hi, the usual continuity conditions for EU are mixture continuity and Archimedean. Could you please explain how is your continuity stronger than the mixture continuity?  
> **username_3**: Its stronger in the sense that we require the preference to be continuous with respect to the weak convergence topology. i.e. $(p_n) \rightarrow p$ if for every bounded and continuous $f$, $\int f(x) dp_n \rightarrow \int f(x) dp$.  

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