# Representing these lexicographic preferences by non-continuous utility function?

I have the following preferences and I am wondering whether I could represent them by a utility function:

$R_1 \succ R_2$ <-> $R_1\leq Q$ and $R_2 > Q$,

$R_1 \succ R_2$ <-> $R_1 \leq Q$ and $R_2 \leq Q$ and $R_1>R_2$,

$R_1 \sim R_2$ <-> $R_1 > Q$ and $R_2 > Q$

R is positive and real-valued, Q is kind of the "threshold" of R for the first criterion (below Q everything prefered to above Q). These are lexicographic preferences, right?

Can't I represent these preferences by the utility function:

$U(R)=a+bR$ for $R\leq Q$ and positive a,b

$U(R)=0$ otherwise ?

Because now

$U(R_1)>U(R_2)$ <-> $R_1 \succ R_2$

$U(R_1)=U(R_2)$ <-> $R_1 \sim R_2$

So, is it finally possible to represent lexicographic preferences by anon-continuous utility function?

Thanks a lot for help!

Felix

• Could you be more specific on the dimensions of R1? If it is single dimensional, I have no clue how you are trying to construct lexicographic preferences here. – HRSE Feb 11 '16 at 3:38

As @HRSE pointed out, your terminology is a bit misleading, and these preferences are not exactly what we refer to when we use the word 'lexicographic'. You might be tempted to consider every $R>0$ as a two-dimensional vector $R=(\mathbb{1}_{R \leq Q},R)$ and to apply the lexicographic order on these vectors, but I would say it is a bit unnatural since the two dimensions of the vector are not independent (you cannot vary the second freely without affecting the first).

That said, the usual observation that lexicographic preferences can be represented by a utility function (but not by a continuous utility function) applies to your case as well. The common feature of both situations is that preferences satisfy standard rationality requirements (weak order) but are discontinuous. You can refer to this question for a formal discussion of these issues:

Lexicographic preference relation cannot be represented by a utility function

• Oh ok, thx a lot for that. So, this utility functio works. I think, I ll read up a bit more on lexicographic preferences. ;-) – intasys Feb 21 '16 at 12:24