Existence of utility representation of a rational but discontinuous preference

Question

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.

Answer 1

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.

Answer 2

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).

Answer 3

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 $x^* \in X^*$.

Answer 4

@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 $\hat{U}(h') = U(h') < \hat{U}(h)$.