<p>If the preference is continuous, reflexive, transitive and complete, then there exists a continuous utility representation <span class="math-container">$g$</span>, and since <span class="math-container">$U$</span> also represents the same preference, so there must be a strictly increasing function <span class="math-container">$f$</span> such that <span class="math-container">$U= f\circ g$</span>.</p>

CC BY-SA 4.0
2022-04-07T16:30:58.597Z
Can you please elaborate the second part where you say that such an $f$ must exist? It may be true, but I can't convince myself that it really is the case. A proof would be appreciated!
not tdm's twin
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2022-04-07T17:03:47.303Z
Take it as an exercise. Try to construct such a $f$, it is easy.
Amit
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2022-04-08T00:18:31.093Z
The existence of $g$ comes from Debreu's theorem. Then define $f$ such that $f(g(x)) = U(x)$. Since $g$ and $U$ denote the same preferences, it follows that $f(g(x)) > f(g(y)) \iff U(x) \geq U(y) \iff g(x) \iff g(y)$ which further implies that $f$ is monotone. Does this work?
not tdm's twin
CC BY-SA 4.0
2022-04-08T02:09:31.900Z
$f$ defined as $f(g(x)) = U(x)$ works provided you want $f:\text{Range}(g)\rightarrow\mathbb{R}$. Here $\text{Range}(g)$ will be an interval because $g$ is continuous. If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, then you just need to complete the construction of $f$ outside the range of $g$.
Amit
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2022-04-08T07:30:44.087Z
@Pocket Cat If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, see Michael's answer.
Amit
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2022-04-07T15:03:20.270Z
2022-04-07T15:03:20.270Z
Amit
<p>I took the question as asking for a representation with <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span>. If we only require <span class="math-container">$f$</span> to be defined on a subset of <span class="math-container">$\mathbb{R}$</span>, Amit's answer solves the problem.</p>
<p>Here is a proof for the case that the domain is connected (and second countable): Under these assumptions, there exists a continuous utility representation <span class="math-container">$V:X\to\mathbb{R}$</span>. Since continuous functions map connected sets to connected sets, the image <span class="math-container">$V(X)$</span> is a, possibly unbounded, interval. I do the special case that <span class="math-container">$V(X)=[a,b)$</span> for some real numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, the other cases can be dealt with by analogous methods. Now define the surjective continuous function <span class="math-container">$g:X\to [a,\infty)$</span> by <span class="math-container">$$g(x)=\frac{V(x)-a}{b-V(x)}.$$</span>
Note that <span class="math-container">$g$</span> represents the preference ordering. Let <span class="math-container">$r^*=U(x)$</span> for any (and hence all) <span class="math-container">$x\in g^{-1}(a).$</span></p>
<p>Now define <span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span> so that for <span class="math-container">$r\geq a$</span> we have <span class="math-container">$f(r)=U(x)$</span> for any (and hence all) <span class="math-container">$x\in g^{-1}(r)$</span> and for <span class="math-container">$r&lt;a$</span> we have <span class="math-container">$f(r)=r^*-|r-a|$</span>. It is easy to verify that <span class="math-container">$f$</span> is strictly increasing and <span class="math-container">$U=f\circ g$</span>.</p>

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2022-04-10T12:36:20.393Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/135445/discussion-on-answer-by-michael-greinecker-discontinuous-function-u-with-conti).
1muflon1
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2022-04-07T17:16:38.617Z
2022-04-07T17:16:38.617Z
Michael Greinecker
<blockquote>
<p><strong>Conjecture:</strong> Every discontinuous utility function <span class="math-container">$U$</span> representing continuous preferences can be written as <span class="math-container">$U = f \circ g$</span> for some continuous <span class="math-container">$g$</span> and discontinuous strictly monotone <span class="math-container">$f$</span>.</p>
</blockquote>
<p>The goal is to prove or disprove this conjecture.</p>
<p>We know that continuous utility implies continuous preferences, and running it through a strict monotone gives us the same preferences but with a discontinuous utility function. The conjecture is inspired from this.</p>

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2022-04-07T14:12:22.643Z
I recommend that you delete your counter-example and the edit from the body of the Q.
Giskard
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2022-04-07T14:14:40.930Z
Are the continuous preferences themselves monotone?
Giskard
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2022-04-07T14:16:31.900Z
@Giskard The preferences need not be monotone. (I have edited the post as well.)
not tdm's twin
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2022-04-07T14:36:47.690Z
Any assumptions on the domain of $U$?
Michael Greinecker
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2022-04-07T14:57:57.617Z
@MichaelGreinecker The domain $X \subseteq \mathbb{R}^n$. But I would like to see these two cases, if possible: (1) $X \subseteq \mathbb{R}^n$ and (2) $X = \mathbb{R}^n$.
not tdm's twin
<microeconomics><econometrics><preferences><choice-theory>
Discontinuous function $U$ with continuous preferences can be written as a composition of discontinuous & monotone function and a continuous function
not tdm's twin
# Discontinuous function $U$ with continuous preferences can be written as a composition of discontinuous & monotone function and a continuous function

(Asked by: username_0 on 2022-04-07)

> **Conjecture:** Every discontinuous utility function $U$ representing continuous preferences can be written as $U = f \circ g$ for some continuous $g$ and discontinuous strictly monotone $f$.

The goal is to prove or disprove this conjecture.

We know that continuous utility implies continuous preferences, and running it through a strict monotone gives us the same preferences but with a discontinuous utility function. The conjecture is inspired from this.

**Tags:** microeconomics, econometrics, preferences, choice-theory

> **Comments:**
> **username_1**: I recommend that you delete your counter-example and the edit from the body of the Q.  
> **username_1**: Are the continuous preferences themselves monotone?  
> **username_0**: @username_1 The preferences need not be monotone. (I have edited the post as well.)  
> **username_2**: Any assumptions on the domain of $U$?  
> **username_0**: @username_2 The domain $X \subseteq \mathbb{R}^n$. But I would like to see these two cases, if possible: (1) $X \subseteq \mathbb{R}^n$ and (2) $X = \mathbb{R}^n$.  

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## 2 answers

### Answer by username_2 on 2022-04-07. Score: 3 | Accepted answer

I took the question as asking for a representation with $f:\mathbb{R}\to\mathbb{R}$. If we only require $f$ to be defined on a subset of $\mathbb{R}$, username_3's answer solves the problem.

Here is a proof for the case that the domain is connected (and second countable): Under these assumptions, there exists a continuous utility representation $V:X\to\mathbb{R}$. Since continuous functions map connected sets to connected sets, the image $V(X)$ is a, possibly unbounded, interval. I do the special case that $V(X)=[a,b)$ for some real numbers $a$ and $b$, the other cases can be dealt with by analogous methods. Now define the surjective continuous function $g:X\to [a,\infty)$ by $$g(x)=\frac{V(x)-a}{b-V(x)}.$$
Note that $g$ represents the preference ordering. Let $r^\*=U(x)$ for any (and hence all) $x\in g^{-1}(a).$

Now define $f:\mathbb{R}\to\mathbb{R}$ so that for $r\geq a$ we have $f(r)=U(x)$ for any (and hence all) $x\in g^{-1}(r)$ and for $r<a$ we have $f(r)=r^\*-|r-a|$. It is easy to verify that $f$ is strictly increasing and $U=f\circ g$.

> **Comments:**
> **username_4**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/135445/discussion-on-answer-by-username_2-username_2-discontinuous-function-u-with-conti).  

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### Answer by username_3 on 2022-04-07. Score: 4

If the preference is continuous, reflexive, transitive and complete, then there exists a continuous utility representation $g$, and since $U$ also represents the same preference, so there must be a strictly increasing function $f$ such that $U= f\circ g$.

> **Comments:**
> **username_0**: Can you please elaborate the second part where you say that such an $f$ must exist? It may be true, but I can't convince myself that it really is the case. A proof would be appreciated!  
> **username_3**: Take it as an exercise. Try to construct such a $f$, it is easy.  
> **username_0**: The existence of $g$ comes from Debreu's theorem. Then define $f$ such that $f(g(x)) = U(x)$. Since $g$ and $U$ denote the same preferences, it follows that $f(g(x)) > f(g(y)) \iff U(x) \geq U(y) \iff g(x) \iff g(y)$ which further implies that $f$ is monotone. Does this work?  
> **username_3**: $f$ defined as $f(g(x)) = U(x)$ works provided you want $f:\text{Range}(g)\rightarrow\mathbb{R}$. Here $\text{Range}(g)$ will be an interval because $g$ is continuous. If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, then you just need to complete the construction of $f$ outside the range of $g$.  
> **username_3**: @Pocket Cat If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, see username_2's answer.  

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