This is related to Do discontinuous preferences imply no continuous utility function?

I think the title of the above-linked question is phrased in such a way that obscures a subtly different but more interesting question which the OP also hinted at in the body. I'd like to ask that explicitly here.

Does there exist a rational but discontinuous preference relation that is representable by a (potentially discontinuous) utility function?

In other words, if $\succsim$ satisfies completeness and transitivity but violates continuity, can we still find a utility function to represent it?

From known results, the answer does not seem obvious.

  • We know that continuous utility representation exists if and only if preference is complete, transitive, and continuous. But this doesn't tell us what happens when preference is not continuous.
  • We know that utility representation does not exist for some discontinuous preference (e.g. the lexicographic preference). But can this conclusion be generalized?

Finally, I want to note that the requirement for $\succsim$ to violate continuity means we're ruling out finite (and countable?) domains.


I think a basic problem is that any utility function defines a preference, and discontinuous utility functions can be used to define discontinuous preferences. Hence there are many discontinuous preferences that can be represented by utility functions. An example:

Let $U(x,y)$ be a continuous utility function that maps from $\mathbb{R}^2$ to $(0,1)$. This latter may seem arbitrary, but the strictly monotononicaly increasing function $\frac{x}{x+1}$ maps from $\mathbb{R}_{++}$ to $(0,1)$, so it should be fine. Also define a closed set $H \subset \mathbb{R}^2$. Let $$ \hat{U}(x,y) = \left\{ \begin{array}{ll} U(x,y) & \mbox{if} (x,y) \notin H \\ \\ U(x,y)+1 & \mbox{if} (x,y) \in H. \end{array} \right. $$ Obviously $\hat{U}(x,y)$ is not continuous and neither are the preferences defined by it. (The boundary of $H$ is prefered to anything outside $H$.) But the way these preferences were generated seems fairly general. So a large class of discontinuous preferences exists for which there is a utility representation.

Future question: How 'large' is this class, what measure can be used?

EDIT: As @NicolasPinto points out in his answer it is also necessary to specify that $H$ is such that $$ \exists x \in H, \exists y \notin H: y \succ x, $$ so $H$ is not the upper contour set of some point, otherwise $U(x,y)$ and $\hat{U}(x,y)$ would in fact represent the same continuous preferences.


Consider the (rational) preference on $[0,1]$ defined by $u(0)=0$, $u(x)=1$ for $x>0$. It is not continuous, but it can obviously be represented (by $u$).

In fact, many preferences that fail to be continuous are representable. This is due to a variety of well known representation results that do not require continuity. For instance, the following general result accomodates the example above:

Theorem. Let $X$ be a topological space with a countable base. Then, every preference on $X$ that has closed "not-better-than" sets is representable by a utility function.

Of course, if $X$ is a separable metric space, then the topological condition holds automatically. This result is essentially Theorem 1 in the following paper:

Rader, T. (1963) "The Existence of a Utility Function to Represent Preferences" Review of Economic Studies, 30, 229-232.

(the only difference is that Rader proves it for "not-worse-than" sets and I would have to "reverse" the preference in the example above).

  • $\begingroup$ Welcome to Economics SE. Please format your future answers in MathJax. Also, is not the closed-ness of the "not-better-than" sets a typical definition of the continuity of preferences? $\endgroup$ – Theoretical Economist Feb 25 '18 at 17:53
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    $\begingroup$ I will, thanks for the edit. The definition of continuity (I think) you have in mind (i.e. the one used by G. Debreu in his "Theory of value") also requires that the "not-worse-than" sets are closed. In other words, $\succsim$ is said to be continuous when all the upper and lower sections are closed. This is what fails in the example above, since the set of alternatives "not-worse-than" $x=1$ is $(0,1]$, not a closed set in $[0,1]$. $\endgroup$ – john Feb 25 '18 at 18:59
  • $\begingroup$ I would upvote this if you would actually name the general result and not just state it. $\endgroup$ – Giskard Feb 25 '18 at 19:42
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    $\begingroup$ You are right, but I didn't remember a source. I added one now. The theorem I am quoting now is more general. I also had omitted that a metric space has to be separable for this work. $\endgroup$ – john Feb 26 '18 at 12:28

denesp's answer above asks:

How large is the class of preferences that are not continuous but have a utility representation?

This is answered by proposition 1.12 in Kreps, Microeconomic Foundations I. I believe the result is originally due to Debreu. Here is the (verbatim) statement:

Suppose $\succsim$ is a complete and transitive preference relation on a set $X$. The relation $\succsim$ can be represented by a utility function if and only if some countable subset $X^*$ of $X$ has the property that if $x \succ y$ for $x$ and $y$ from $X$, then $x \succsim x^* \succ y$ for some $x^* \in X^*$.

  • $\begingroup$ Was hoping someone hit this part. It's a size issue, exactly. $\endgroup$ – Pete Caradonna Apr 11 '18 at 23:05

@denesp answer for this question( that was accepted) is not correct. It is not true that a non continuous utility function imply a non continuous preference relation.

Looking at the example given in the answer: let $U(x,y)$ be a utility representation of rational and continuous preferences over $\mathbb{R}^2$ , $x^* \in \mathbb{R}^2 $and $H$ = $ \{ x \in \mathbb{R}^2: x \succsim x^*\} $ then $H$ is closed by continuity of $\succsim$. Define $\hat{U}(x,y)$ as before:

$$ \hat{U}(x,y) = \left\{ \begin{array}{ll} U(x,y) & \mbox{if} (x,y) \notin H \\ \\ U(x,y)+1 & \mbox{if} (x,y) \in H. \end{array} \right. $$

Then $\hat{U}(x,y)$ represents the same preferences of $U(x,y)$ that are, indeed, continuous. The main point is you can represent continuous preferences with discontinuous functions.

Getting back to the question: you could, of course, use the same method to find a really discontinuous preference relation. Just choose $H$ to be a set consisting of only one point $h$ in $\mathbb{R^2}$ whose strictly upper contour set is not empty. That is a finite set and so is closed. The resulting utility function modified as before would result in discontinuous preferences. That would work because we can create a sequence $h_n$ in the lower contour set that is converging to $h$ and another sequence in the strictly upper contour set that is only the repetition of the point $h'$ such that $U(h_n) < U(h') \,\,\forall \, n \in \mathbb{N}$ but $\hat{U}(h') = U(h') < \hat{U}(h)$.

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    $\begingroup$ Indeed! To be precise: I did not claim that "a non continuous utility function imply a non continuous preference relation", I gave a very special non continuous utility function. But you are right that the proof was not exact, I also needed to specify $H$ is such that $$ \exists x \in H, \exists y \notin H: y \succ x. $$ $\endgroup$ – Giskard Mar 11 '18 at 8:22

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