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It is well-known that a rational, continuous and monotone preference relation $\succeq$ defined on $\mathbb{R}^L$ admits a utility representation.

I would like to understand why monotonicity is required. In other words, what is an example of a rational and continuous but non-monotone preference relation that does not have a utility representation?

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    $\begingroup$ I don't think monotonicity is required for utility representation, at least in Debreu’s representation theorem. $\endgroup$ – Herr K. Jan 14 at 20:05
  • $\begingroup$ Thank you, you are right. I had never seen this result without the monotonicity axiom, which is probably used for simplifying the proof. I'll accept this answer if you post it. $\endgroup$ – Oliv Jan 14 at 20:30
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A necessary condition for a preference relation to be represented by a utility function is that the preference relation is rational (where a "utility function" is a real-valued function that assigns a higher or equal numerical value to bundle A than to bundle B, when A is weakly preferred to B).

Preferences are rational when they are complete (I can express my preferences for all conceivable bundles), and transitive (If I weakly prefer A to B and B to C, then I weakly prefer A to C).

Given rationality, a sufficient condition is that the rational preference relation is continuous, i.e. if it is preserved under limits.

Then, the related utility function is also continuous.

Monotonicity is a property where we go from quantities to preferences, after assuming desirability ("goods" rather than "bads"). Local nonsatiation is actually the weaker assumption that is required for most of the theory.

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    $\begingroup$ (+1) Two related results may be worth noting. First, if $\succsim$ is defined on a finite or countable set, then rationality of $\succsim$ alone is both necessary and sufficient for utility representation. Second, if $\succsim$ is defined on an uncountable set, then rationality and continuity together are necessary and sufficient for (continuous) utility representation. $\endgroup$ – Herr K. Jan 15 at 1:46

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