In Advanced Microeconomic theory by Jehle and Reny there is a proof of the theorem which states the existence of utility function.

In order to prove the existence of a utility function $u(\mathbf{x})$ which represents binary relation $\succeq$ if it is complete, transitive, continuous, and strictly monotonic, it is suggested to consider a mapping

$u: \mathbb{R_+^n} \to \mathbb{R}$ such that $u(\mathbf{x})e\sim \mathbf{x}$ is satisfied, where $\mathbf{x}$ is a bundle, $u(\mathbf{x})$ is some number and $\mathbf{e}$ is a bundle which contains one of every good.

So first we need to show that there always exists such number $u(\mathbf{x})$. To do this consider two sets:

$A \equiv \{t \geq 0 \mid t\mathbf{e} \succeq \mathbf{x}\}$

$B \equiv \{t \geq 0 \mid t\mathbf{e} \preceq \mathbf{x}\}$

if $t^* \in A \cap B$, then $t^*\mathbf{e} \sim \mathbf{x}$, so we need to show that $A \cap B$ is nonempty.

The continuity of $\succeq$ implies that both A and B are closed in $\mathbb{R_+}$. By strict monotonicity, $t \in A$ implies $t' \in A,$ $\forall$ $ t'\geq t$. So $A=[\underline{t}, \infty)$. Similarly $B=[0, \overline{t}]$

For any $t \geq 0$, completeness of $\succeq$ implies that either $t\mathbf{e} \succeq \mathbf{x}$ or $t\mathbf{e} \preceq \mathbf{x}$, i.e., $t \in A \cup B$

$\mathbb{R_+} = A \cup B = [0, \overline{t}] \cup [\underline{t}, \infty)$.

So in order to $A \cap B$ be nonempty it should be that $\underline{t} \leq \overline{t}$.

But is it always like this? I can't see why the last inequality should hold always.

  • 4
    $\begingroup$ I think you pretty much finished it. If $\underline{t} > \overline{t}$, then $A \cup B$ does not contain any numbers in the (nonempty) interval $(\overline{t}, \underline{t})$. But this contradicts the fact you already showed that $A \cup B = \mathbb{R}_+$. So we must have $\underline{t} \leq \overline{t}$. $\endgroup$
    – usul
    Oct 26, 2016 at 3:59
  • $\begingroup$ Just one further reminder, this is far from the only axiomization of the utility function. You are complete. $\endgroup$ Apr 13, 2017 at 21:16

1 Answer 1


Το take this one out of the Unanswered queue:

The completeness property of the preference relation, implies that for all non-negative $t$ we will be able to form and declare the preference relation. Hence

$$A \cup B = \mathbb{R_+}$$

By monotonicity we have $B=[0, \overline{t}],\;\; A=[\underline{t}, \infty)$.

Ad absurdum, assume that $\overline{t} < \underline{t}$. Then there exists an open interval $(\overline{t},\underline{t})$ that does not belong to the union of $A$ and $B$. But then we arrive at $A \cup B \neq \mathbb{R_+}$ which contradicts the implications of the completeness property.

So we conclude that $\overline{t} \geq \underline{t}$, which implies that the intersection $A \cap B$ is non-empty (even if it may contain a single element, when $\overline{t} = \underline{t}$).


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