5. Preferences and Utility

Jun He June 01, 2026

中文English 偏好关系Preference Relation 效用表示Utility Representation 公理Axioms 字典序偏好Lexicographic Preferences 连续性Continuity

5. Preferences and Utility

Note

本章导读本章为需求理论奠定公理基础：先用二元关系 $\succsim$ 定义偏好与各种等高集，给出完备性 $A_1$、传递性 $A_2$ 两条公理使 $\succsim$ 成为偏好关系；再讨论效用表示定理——有限消费集下满足 $A_1,A_2$ 即可被效用函数表示（Thm 5.1），而无限消费集下不一定（字典序偏好反例 Example 5.1），须加连续性公理 $A_3$（Thm 5.2）。最后引入严格单调性 $A_4$、（严格）凸性 $A_5'/A_5$ 以简化证明，并给出 $A_1$–$A_4$ 下连续效用函数存在的完整构造性证明（Thm 5.3）。

5. Preferences and Utility

Note

Overview This chapter lays the axiomatic foundation for demand theory: we define preference via a binary relation $\succsim$ and various contour sets, give completeness $A_1$ and transitivity $A_2$ making $\succsim$ a preference relation; then the utility representation theorem — on a finite consumption set $A_1,A_2$ suffice for a utility representation (Thm 5.1), but on an infinite set they do not (lexicographic-preferences counter-example, Example 5.1), so we add the continuity axiom $A_3$ (Thm 5.2). Finally we introduce strict monotonicity $A_4$ and (strict) convexity $A_5'/A_5$ to ease the proof, and give the full constructive proof that a continuous utility function exists under $A_1$–$A_4$ (Thm 5.3).

5.1 Preference relations and axioms

5.1.1 Notations

$\mathbf{X}=\mathbb{R}_+^n$ 是消费者的消费集；
$\mathbf{x}=\{x_1,x_2,\dots,x_n\}\in\mathbf{X}$ 是一个商品束；
$\succsim$ 是"至少一样好（at-least-as-good-as）"二元关系。对任意 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，$\mathbf{x}^1\succsim\mathbf{x}^2$ 表示商品束 $\mathbf{x}^1$ 至少与 $\mathbf{x}^2$ 一样好；
$\succ$ 是"严格偏好"二元关系：$\mathbf{x}^1\succ\mathbf{x}^2\Leftrightarrow(\mathbf{x}^1\succsim\mathbf{x}^2)\wedge\neg(\mathbf{x}^2\succsim\mathbf{x}^1)$；
$\sim$ 是"无差异"二元关系：$\mathbf{x}^1\sim\mathbf{x}^2\Leftrightarrow(\mathbf{x}^1\succsim\mathbf{x}^2)\wedge(\mathbf{x}^2\succsim\mathbf{x}^1)$。

5.1.2 Preference relation

若 $\succsim$ 满足下面的公理 $A_1$ 与公理 $A_2$，则称 $\succsim$ 为偏好关系。

5.1 Preference relations and axioms

5.1.1 Notations

$\mathbf{X}=\mathbb{R}_+^n$ is the consumer's consumption set;
$\mathbf{x}=\{x_1,x_2,\dots,x_n\}\in\mathbf{X}$ is a bundle of goods;
$\succsim$ is the at-least-as-good-as binary relation. For any $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, $\mathbf{x}^1\succsim\mathbf{x}^2$ means bundle $\mathbf{x}^1$ is at least as good as $\mathbf{x}^2$;
$\succ$ is the strictly-preferred binary relation: $\mathbf{x}^1\succ\mathbf{x}^2\Leftrightarrow(\mathbf{x}^1\succsim\mathbf{x}^2)\wedge\neg(\mathbf{x}^2\succsim\mathbf{x}^1)$;
$\sim$ is the indifference binary relation: $\mathbf{x}^1\sim\mathbf{x}^2\Leftrightarrow(\mathbf{x}^1\succsim\mathbf{x}^2)\wedge(\mathbf{x}^2\succsim\mathbf{x}^1)$.

5.1.2 Preference relation

If $\succsim$ satisfies the following axiom $A_1$ and axiom $A_2$, then we call $\succsim$ a preference relation.

Important

公理（偏好关系）/ Axioms (preference relation) $A_1$（完备性，Completeness）：对任意一对商品束 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，要么 $\mathbf{x}^1\succsim\mathbf{x}^2$，要么 $\mathbf{x}^2\succsim\mathbf{x}^1$（此处"或"是相容的，两者可同时成立）。
$A_2$（传递性，Transitivity）：对任意三个商品束 $\mathbf{x}^1,\mathbf{x}^2,\mathbf{x}^3\in\mathbf{X}$，若 $\mathbf{x}^1\succsim\mathbf{x}^2$ 且 $\mathbf{x}^2\succsim\mathbf{x}^3$，则 $\mathbf{x}^1\succsim\mathbf{x}^3$。$A_1$ (Completeness): for any pair of bundles $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, either $\mathbf{x}^1\succsim\mathbf{x}^2$ or $\mathbf{x}^2\succsim\mathbf{x}^1$ (here "or" is inclusive, both can be true).
$A_2$ (Transitivity): for any three bundles $\mathbf{x}^1,\mathbf{x}^2,\mathbf{x}^3\in\mathbf{X}$, if $\mathbf{x}^1\succsim\mathbf{x}^2$ and $\mathbf{x}^2\succsim\mathbf{x}^3$, then $\mathbf{x}^1\succsim\mathbf{x}^3$.

Important

Definition 5.1（等高集 / contour sets）对满足 $A_1,A_2$ 的二元关系 $\succsim$，定义至少一样好集（上等高集，upper-contour set）$\succsim(\mathbf{x}^0)\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\succsim\mathbf{x}^0\}$，并类似定义不优于集、被偏好集、劣于集、无差异集（见下式）。For a binary relation $\succsim$ satisfying $A_1,A_2$, define the at-least-as-good-as set (upper-contour set) $\succsim(\mathbf{x}^0)\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\succsim\mathbf{x}^0\}$, and similarly the no-better-than, preferred-to, worse-than, and indifference sets (below).

$$ \begin{aligned} \succsim(\mathbf{x}^0)&\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\succsim\mathbf{x}^0\}&&\text{(at-least-as-good-as / upper-contour set)}\\ \precsim(\mathbf{x}^0)&\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\precsim\mathbf{x}^0\}&&\text{(no-better-than set)}\\ \succ(\mathbf{x}^0)&\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\succ\mathbf{x}^0\}&&\text{(preferred-to set)}\\ \prec(\mathbf{x}^0)&\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\prec\mathbf{x}^0\}&&\text{(worse-than set)}\\ \sim(\mathbf{x}^0)&\equiv\{\mathbf{x}\in\mathbf{X}:\mathbf{x}\sim\mathbf{x}^0\}&&\text{(indifference set)} \end{aligned} $$

5.2 Utility representation theorem

5.2.1 Finite consumption set

5.2 Utility representation theorem

5.2.1 Finite consumption set

Important

Definition 5.2（Utility representation）称效用函数 $u:\mathbf{X}\to\mathbb{R}$ 表示 $\succsim$，当且仅当对任意 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，$u(\mathbf{x}^1)\ge u(\mathbf{x}^2)\Leftrightarrow\mathbf{x}^1\succsim\mathbf{x}^2$。We say a utility function $u:\mathbf{X}\to\mathbb{R}$ represents $\succsim$ if and only if for any $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, $u(\mathbf{x}^1)\ge u(\mathbf{x}^2)\Leftrightarrow\mathbf{x}^1\succsim\mathbf{x}^2$.

Important

Theorem 5.1 设 $\succsim$ 满足公理 $A_1$ 与 $A_2$，且 $\mathbf{X}$ 有限。则存在效用函数 $u:\mathbf{X}\to\mathbb{R}$ 表示 $\succsim$。Suppose $\succsim$ satisfies axiom $A_1$ and axiom $A_2$, and that $\mathbf{X}$ is finite. Then there exists a utility function $u:\mathbf{X}\to\mathbb{R}$ representing $\succsim$.

Note

证明 / Proof (Theorem 5.1)

Thm 5.1 是显然的。由于 $\mathbf{X}$ 有限，故商品束的两两配对有限。$\succsim$ 满足 $A_1,A_2$，思路是总能给所有有限商品束排序，再按排名赋数即得效用函数。形式上，设 $\mathbf{X}$ 中有 $n$ 个商品束，$n$ 为有限正整数，先写

Theorem 5.1 is obvious. Since $\mathbf{X}$ is finite, there are finitely many distinct pairs of bundles. Since $\succsim$ satisfies $A_1,A_2$, the idea is that we can always rank all finite bundles, then assign numbers per ranking to get the utility function. Formally, suppose $n$ bundles in $\mathbf{X}$, $n$ a finite positive integer, write

$$\mathbf{X}=\{\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_n\}$$

要证 $\mathbf{X}$ 中所有束可排序，$\mathbf{X}=\{\mathbf{x}_{(1)},\mathbf{x}_{(2)},\dots,\mathbf{x}_{(n)}\}$，其中 $\mathbf{x}_{(i)}\precsim\mathbf{x}_{(j)}$ 若 $i\le j$。用归纳法证明。初始只有第一个束 $\mathbf{x}_1$，$\mathbf{X}^1=\{\mathbf{x}_1\}=\{\mathbf{x}_{(1)}\}$（仅一束无差别）。设 $n=k$ 时排序可行，$\mathbf{X}^k=\{\mathbf{x}_1,\dots,\mathbf{x}_k\}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(k)}\}$。加入第 $k+1$ 个束 $\mathbf{x}_{k+1}$ 得 $\mathbf{X}^{k+1}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(k)},\mathbf{x}_{k+1}\}$。把 $\mathbf{x}_{k+1}$ 与其左侧束 $\mathbf{x}_{(i)}$ 比较：由 $A_1$（完备性），要么 $\mathbf{x}_{k+1}\succsim\mathbf{x}_{(i)}$，要么 $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$；若非 $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$ 则交换位置，重复此操作直至 $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$。然后重排下标，写作 $\{\mathbf{x}_{(1)},\mathbf{x}_{(2)},\dots,\mathbf{x}_{(k+1)}\}$。设 $\mathbf{x}_{k+1}$ 现位于第 $j$ 位（$\mathbf{x}_{k+1}=\mathbf{x}_{(j)}$）。该操作不改变前 $k$ 个束的相对位置。由传递性 $A_2$，$\mathbf{x}_{(s)}\precsim\mathbf{x}_{(j)}$ 对 $\forall s\le j$、$\mathbf{x}_{(j)}\precsim\mathbf{x}_{(l)}$ 对 $\forall l\ge j$。故 $\mathbf{X}^{k+1}$ 满足排序要求。由归纳法，排序对 $k=n$ 可行，于是 $\mathbf{X}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(n)}\}$。考虑效用函数 $u(\mathbf{x}_{(1)})=1$，并对 $i=1,\dots,n-1$

We want to show all bundles in $\mathbf{X}$ can be ranked, $\mathbf{X}=\{\mathbf{x}_{(1)},\mathbf{x}_{(2)},\dots,\mathbf{x}_{(n)}\}$ where $\mathbf{x}_{(i)}\precsim\mathbf{x}_{(j)}$ if $i\le j$. We show this by induction. Initially only the first bundle $\mathbf{x}_1$, and $\mathbf{X}^1=\{\mathbf{x}_1\}=\{\mathbf{x}_{(1)}\}$ (one bundle, no difference). Suppose ranking is feasible for $n=k$, $\mathbf{X}^k=\{\mathbf{x}_1,\dots,\mathbf{x}_k\}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(k)}\}$. Adding the $k+1$th bundle $\mathbf{x}_{k+1}$ gives $\mathbf{X}^{k+1}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(k)},\mathbf{x}_{k+1}\}$. Compare $\mathbf{x}_{k+1}$ with the bundle on its left $\mathbf{x}_{(i)}$: by $A_1$ (completeness), either $\mathbf{x}_{k+1}\succsim\mathbf{x}_{(i)}$ or $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$; switch positions if not $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$ and repeat until $\mathbf{x}_{k+1}\precsim\mathbf{x}_{(i)}$. Then renew the subscripts, writing $\{\mathbf{x}_{(1)},\mathbf{x}_{(2)},\dots,\mathbf{x}_{(k+1)}\}$. Suppose $\mathbf{x}_{k+1}$ is now in the $j$th position ($\mathbf{x}_{k+1}=\mathbf{x}_{(j)}$). The operation does not change the relative positions of the previous $k$ bundles. By transitivity $A_2$, $\mathbf{x}_{(s)}\precsim\mathbf{x}_{(j)}$ for $\forall s\le j$ and $\mathbf{x}_{(j)}\precsim\mathbf{x}_{(l)}$ for $\forall l\ge j$. So $\mathbf{X}^{k+1}$ satisfies the ranking requirements. By induction, ranking is feasible for $k=n$, thus $\mathbf{X}=\{\mathbf{x}_{(1)},\dots,\mathbf{x}_{(n)}\}$. Consider the utility function $u(\mathbf{x}_{(1)})=1$, and for $i=1,\dots,n-1$

$$u(\mathbf{x}_{(i+1)})=\begin{cases}u(\mathbf{x}_{(i)})&\text{if }\mathbf{x}_{(i+1)}\sim\mathbf{x}_{(i)}\\u(\mathbf{x}_{(i)})+1&\text{otherwise}\end{cases}$$

显然 $u(\mathbf{x}_{(i)})\ge u(\mathbf{x}_{(j)})\Leftrightarrow\mathbf{x}_{(i)}\succsim\mathbf{x}_{(j)}$，故此 $u$ 表示 $\succsim$。$\blacksquare$

Clearly $u(\mathbf{x}_{(i)})\ge u(\mathbf{x}_{(j)})\Leftrightarrow\mathbf{x}_{(i)}\succsim\mathbf{x}_{(j)}$. So such utility function represents $\succsim$. $\blacksquare$

5.2.2 Infinite consumption set

然而，当 $\mathbf{X}$ 非有限时，满足 $A_1,A_2$ 的二元关系不一定能被效用函数表示。下面是一个反例。

5.2.2 Infinite consumption set

However, a binary relation satisfying $A_1$ and $A_2$ is not necessarily representable by a utility function when $\mathbf{X}$ is not finite. Below is a counter-example.

Important

Example 5.1（Lexicographic preferences / 字典序偏好）设 $\mathbf{X}=\mathbb{R}_+^2$。称二元关系 $\succsim_L$ 为字典序偏好，若Suppose $\mathbf{X}=\mathbb{R}_+^2$. The binary relation $\succsim_L$ is called lexicographic preferences if

$$ \begin{aligned} \mathbf{x}^1=(x_1^1,x_2^1)\succsim_L\mathbf{x}^2=(x_1^2,x_2^2)\ \Leftrightarrow\ &x_1^1>x_1^2\\ \text{or }\ &x_1^1=x_1^2,\ x_2^1>x_2^2\\ \text{or }\ &\mathbf{x}^1=\mathbf{x}^2 \end{aligned} $$

易证 $\succsim_L$ 满足 $A_1,A_2$，但 $\succsim_L$ 不可能被效用函数表示。

It is easy to show that $\succsim_L$ satisfies $A_1$ and $A_2$, but $\succsim_L$ is impossible to be represented by a utility function.

Note

证明 / Proof (Example 5.1)

反证。设 $u$ 表示 $\succsim_L$。定义 $r:\mathbb{R}\to\mathbb{Q}$，使对任意 $x_1\in\mathbb{R}$，$u(x_1,2)>r(x_1)>u(x_1,1)$。则

By contradiction. Suppose $u$ represents $\succsim_L$. Define $r:\mathbb{R}\to\mathbb{Q}$ such that for any $x_1\in\mathbb{R}$, $u(x_1,2)>r(x_1)>u(x_1,1)$. Then

$$ > \begin{aligned} > x_1&>x_0\\ > \Rightarrow r(x_1)&>u(x_1,1)>u(x_0,2)>r(x_0)\\ > \Rightarrow r(x_1)&>r(x_0) > \end{aligned} > $$

这意味着 $x_1\ne x_0\Rightarrow r(x_1)\ne r(x_0)$，即 $|\mathbb{Q}|\ge|\mathbb{R}|$。而 $|\mathbb{Q}|<|\mathbb{R}|$，矛盾。故这样的 $u$ 不存在。$\blacksquare$

which means $x_1\ne x_0\Rightarrow r(x_1)\ne r(x_0)$, i.e. $|\mathbb{Q}|\ge|\mathbb{R}|$. This is not true since $|\mathbb{Q}|<|\mathbb{R}|$. So such utility function $u$ doesn't exist. $\blacksquare$

为用效用函数表示偏好关系，需如下连续性公理 $A_3$。

To represent a preference relation with a utility function, we need the following continuity axiom $A_3$.

Important

$A_3$（Continuity / 连续性）对 $\forall\mathbf{x}\in\mathbf{X}$，上等高集 $\succsim(\mathbf{x})$ 与不优于集 $\precsim(\mathbf{x})$ 都是 $\mathbb{R}_+^n$ 中的闭集。闭集是包含其中所有序列极限点的集合，例如 $\succsim(\mathbf{x})$ 闭，意为若 $\mathbf{y}^k\succsim\mathbf{x}$（$k=1,2,\dots$）且 $\mathbf{y}^k\to\mathbf{y}$，则 $\mathbf{y}\succsim\mathbf{x}$。For $\forall\mathbf{x}\in\mathbf{X}$, both $\succsim(\mathbf{x})$ and $\precsim(\mathbf{x})$ are closed sets in $\mathbb{R}_+^n$. A closed set contains the limit points of all sequences in that set; e.g. $\succsim(\mathbf{x})$ is closed if $\mathbf{y}^k\succsim\mathbf{x}$ ($k=1,2,\dots$) and $\mathbf{y}^k\to\mathbf{y}$, then $\mathbf{y}\succsim\mathbf{x}$.

Tip

注记 5.1 / Remark 5.1 $\succsim(\mathbf{x})$（$\precsim(\mathbf{x})$）为闭集意味着：若一列商品束都不比 $\mathbf{x}$ 差（不比 $\mathbf{x}$ 好），则取极限后它不能突然变得比 $\mathbf{x}$ 差（好）。换言之，若两个商品束彼此足够接近，它们就不能对第三个束有相反的偏好关系——这正是偏好连续性的含义。$\succsim(\mathbf{x})$ ($\precsim(\mathbf{x})$) being closed means that if a sequence of bundles is no worse (no better) than $\mathbf{x}$, then it cannot suddenly jump to be worse (better) than $\mathbf{x}$ by taking the limits. In other words, if two bundles are close enough, they cannot have flipped preference relations with a third bundle, which is exactly the meaning of continuity of preference.

Important

Theorem 5.2 设 $\mathbf{X}=\mathbb{R}_+^n$。若二元关系 $\succsim$ 满足 $A_1,A_2,A_3$，则 $\succsim$ 可被一个连续效用函数表示。Let $\mathbf{X}=\mathbb{R}_+^n$. If the binary relation $\succsim$ satisfies $A_1,A_2,A_3$, then $\succsim$ can be represented by a continuous utility function.

注意 $A_1,A_2,A_3$ 足以证明存在表示 $\succsim$ 的效用函数。下面的附加假设仅为使证明更易，且在多数情形下合理，故加入无妨。

Note $A_1,A_2,A_3$ are sufficient to prove a utility function exists representing $\succsim$. The following assumptions are proposed only to make the proof easier, and are reasonable in most cases, so it is fine to add them.

Important

附加公理 / Additional axioms $A_4$（严格单调性，Strict monotonicity）：对所有 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，若 $\mathbf{x}^1\ge\mathbf{x}^2$（$\mathbf{x}^1$ 每个分量都不小于 $\mathbf{x}^2$ 对应分量），则 $\mathbf{x}^1\succsim\mathbf{x}^2$；进而若 $\mathbf{x}^1\gg\mathbf{x}^2$（每个分量严格更大），则 $\mathbf{x}^1\succ\mathbf{x}^2$。
$A_5'$（凸性，Convexity）：对任意 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，若 $\mathbf{x}^1\succsim\mathbf{x}^2$，则 $t\mathbf{x}^1+(1-t)\mathbf{x}^2\succsim\mathbf{x}^2$，$\forall t\in[0,1]$。
$A_5$（严格凸性，Strict convexity）：对任意 $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$，若 $\mathbf{x}^1\ne\mathbf{x}^2$ 且 $\mathbf{x}^1\succsim\mathbf{x}^2$，则 $t\mathbf{x}^1+(1-t)\mathbf{x}^2\succ\mathbf{x}^2$，$\forall t\in(0,1)$。$A_4$ (Strict monotonicity): for all $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, if $\mathbf{x}^1\ge\mathbf{x}^2$ (each element of $\mathbf{x}^1$ no less than the corresponding element in $\mathbf{x}^2$), then $\mathbf{x}^1\succsim\mathbf{x}^2$; moreover if $\mathbf{x}^1\gg\mathbf{x}^2$ (each element strictly larger), then $\mathbf{x}^1\succ\mathbf{x}^2$.
$A_5'$ (Convexity): for any $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, if $\mathbf{x}^1\succsim\mathbf{x}^2$ then $t\mathbf{x}^1+(1-t)\mathbf{x}^2\succsim\mathbf{x}^2$, $\forall t\in[0,1]$.
$A_5$ (Strict convexity): for any $\mathbf{x}^1,\mathbf{x}^2\in\mathbf{X}$, if $\mathbf{x}^1\ne\mathbf{x}^2$ and $\mathbf{x}^1\succsim\mathbf{x}^2$, then $t\mathbf{x}^1+(1-t)\mathbf{x}^2\succ\mathbf{x}^2$, $\forall t\in(0,1)$.

图 5.1（凸性，$\mathbb{R}_+^2$，已转述）：由于不要求严格，无差异曲线上可有平段。$A_5'$ 也意味着任何束的上等高集是凸的。图中 $\mathbf{x}^1$ 与 $\mathbf{x}^2$ 在同一条无差异曲线上，凸组合 $(1-t)\mathbf{x}^1+t\mathbf{x}^2$ 位于该曲线之上或之上方（即上等高集内）。

图 5.2（严格凸性，$\mathbb{R}_+^2$，已转述）：无差异曲线上不能有任何平段。$A_5$ 也意味着任何束的上等高集是严格凸的。凸组合 $(1-t)\mathbf{x}^1+t\mathbf{x}^2$ 严格优于端点。

现在可用 $A_1,A_2,A_3,A_4$ 证明存在表示 $\succsim$ 的效用函数。

Figure 5.1 (convexity, $\mathbb{R}_+^2$, paraphrased): since we do not require strictness, there can be a flat segment on the indifference curve. $A_5'$ also means the upper-contour set for any bundle is convex. In the figure, $\mathbf{x}^1$ and $\mathbf{x}^2$ are on the same indifference curve, and the convex combination $(1-t)\mathbf{x}^1+t\mathbf{x}^2$ lies on or above that curve (i.e. inside the upper-contour set).

Figure 5.2 (strict convexity, $\mathbb{R}_+^2$, paraphrased): there cannot be any flat segment on the indifference curve. $A_5$ also means the upper-contour set for any bundle is strictly convex. The convex combination $(1-t)\mathbf{x}^1+t\mathbf{x}^2$ is strictly preferred to the endpoints.

Now we can use axioms $A_1,A_2,A_3,A_4$ on $\succsim$ to prove there exists a utility function representing $\succsim$.

Tip

注记 5.2 / Remark 5.2 若 $u$ 表示 $\succsim$，则 $u$ 的任意严格递增变换 $f$，即 $v=f(u)$，也表示 $\succsim$。因此若有一个表示 $\succsim$ 的效用函数，就有无穷多个，我们只需找出其中一个。If $u$ represents $\succsim$, then any strictly increasing transformation $f$ of $u$, i.e. $v=f(u)$, also represents $\succsim$. So if there is one utility function representing $\succsim$, there are infinitely many, and we only need to find one of them.

Important

Theorem 5.3 设 $\mathbf{X}=\mathbb{R}_+^n$。若二元关系 $\succsim$ 满足 $A_1,A_2,A_3,A_4$，则 $\succsim$ 可被一个连续效用函数 $u:\mathbb{R}_+^n\to\mathbb{R}$ 表示。Let $\mathbf{X}=\mathbb{R}_+^n$. If the binary relation $\succsim$ satisfies $A_1,A_2,A_3,A_4$, then $\succsim$ can be represented by a continuous utility function $u:\mathbb{R}_+^n\to\mathbb{R}$.

Note

证明 / Proof (Theorem 5.3)

令 $\mathbf{e}\equiv(1,1,\dots,1)\in\mathbb{R}_+^n$。定义 $u:\mathbb{R}_+^n\to\mathbb{R}_+$，使对任意 $\mathbf{x}\in\mathbb{R}_+^n$，$u(\mathbf{x})\mathbf{e}\sim\mathbf{x}$。显然，若这样的 $u(\mathbf{x})$ 被唯一钉定，则在 $A_1,A_2,A_3,A_4$ 下它满足

Let $\mathbf{e}\equiv(1,1,\dots,1)\in\mathbb{R}_+^n$. Define $u:\mathbb{R}_+^n\to\mathbb{R}_+$ such that for any $\mathbf{x}\in\mathbb{R}_+^n$, $u(\mathbf{x})\mathbf{e}\sim\mathbf{x}$. Clearly, if such $u(\mathbf{x})$ is uniquely pinned down, then with $A_1,A_2,A_3,A_4$ it satisfies

$$\mathbf{x}^1\succsim\mathbf{x}^2\Leftrightarrow u(\mathbf{x}^1)\mathbf{e}\succsim u(\mathbf{x}^2)\mathbf{e}\Leftrightarrow u(\mathbf{x}^1)\ge u(\mathbf{x}^2)$$

即 $u(\mathbf{x})$ 表示 $\succsim$。所以须证：对每个 $\mathbf{x}\in\mathbb{R}_+^n$ 恰有一个数作为 $u(\mathbf{x})$ 使 $u(\mathbf{x})\mathbf{e}\sim\mathbf{x}$。先证存在性，再证唯一性。
存在性：固定任意 $\mathbf{x}\in\mathbb{R}_+^n$，考虑两个非负实数集 $A\equiv\{t\ge0:t\mathbf{e}\succsim\mathbf{x}\}$、$B\equiv\{t\ge0:t\mathbf{e}\precsim\mathbf{x}\}$。只需证 $A\cap B\ne\emptyset$，即 $\exists t^\star\in A\cap B$。

i.e. $u(\mathbf{x})$ represents $\succsim$. So we must show there is precisely one number as $u(\mathbf{x})$ for every $\mathbf{x}\in\mathbb{R}_+^n$ such that $u(\mathbf{x})\mathbf{e}\sim\mathbf{x}$. First show existence, then uniqueness.
Existence: fix any $\mathbf{x}\in\mathbb{R}_+^n$, and consider two sets of non-negative reals $A\equiv\{t\ge0:t\mathbf{e}\succsim\mathbf{x}\}$, $B\equiv\{t\ge0:t\mathbf{e}\precsim\mathbf{x}\}$. It suffices to show $A\cap B\ne\emptyset$, i.e. $\exists t^\star\in A\cap B$.

Important

Lemma 5.1 $A$ 与 $B$ 都是闭集。$A$ and $B$ are both closed sets.

证明：设 $t_1,t_2,\dots\in A$，若 $t_n\to\tilde t$，则 $t_n\mathbf{e}\to\tilde t\mathbf{e}$。因 $t_n\mathbf{e}\succsim\mathbf{x}$，由 $A_3$（连续性），$\tilde t\mathbf{e}\succsim\mathbf{x}$，即 $\tilde t\in A$。故 $A$ 闭。同理 $B$ 闭。$\blacksquare$

Proof: suppose $t_1,t_2,\dots\in A$. If $t_n\to\tilde t$, then $t_n\mathbf{e}\to\tilde t\mathbf{e}$. Since $t_n\mathbf{e}\succsim\mathbf{x}$, by $A_3$ (continuity), $\tilde t\mathbf{e}\succsim\mathbf{x}$, i.e. $\tilde t\in A$. So $A$ is closed. By the same argument $B$ is closed. $\blacksquare$

由 $A_4$（严格单调性），若 $t\in A$，则 $\forall t'\ge t$，$t'\in A$。$A$ 已证为闭集，故唯一可能是 $A=[\underline t,\infty)$，某 $\underline t\in\mathbb{R}_+$。类似 $B=[0,\bar t]$，某 $\bar t\in\mathbb{R}_+$。对任意 $t\in\mathbb{R}_+$，由 $A_1$（完备性），要么 $t\mathbf{e}\succsim\mathbf{x}$，要么 $t\mathbf{e}\precsim\mathbf{x}$，故 $A\cup B=\mathbb{R}_+$。因此 $\underline t\le\bar t$，从而 $A\cap B\ne\emptyset$，即 $\exists t^\star\in A\cap B$。
唯一性：反证。设 $\exists t_1^\star,t_2^\star\in A\cap B$ 且 $t_1^\star连续性：只需证对每个 \(\alpha\in\mathbb{R}$，集合 $\{\mathbf{x}\in\mathbb{R}_+^n:u(\mathbf{x})\le\alpha\}$ 与 $\{\mathbf{x}\in\mathbb{R}_+^n:u(\mathbf{x})\ge\alpha\}$ 都是闭集。注意 $\{\mathbf{x}:u(\mathbf{x})\le\alpha\}=\{\mathbf{x}:u(\mathbf{x})\mathbf{e}\precsim\alpha\mathbf{e}\}=\{\mathbf{x}:\mathbf{x}\precsim\alpha\mathbf{e}\}$，由 $A_3$（连续性）$\{\mathbf{x}:\mathbf{x}\precsim\alpha\mathbf{e}\}$ 闭，故 $\{\mathbf{x}:u(\mathbf{x})\le\alpha\}$ 闭。对 $\{\mathbf{x}:u(\mathbf{x})\ge\alpha\}$ 同理。$\blacksquare$

By $A_4$ (strict monotonicity), if $t\in A$, then $\forall t'\ge t$, $t'\in A$. $A$ is proved closed, so the only possibility is $A=[\underline t,\infty)$ for some $\underline t\in\mathbb{R}_+$. Similarly $B=[0,\bar t]$ for some $\bar t\in\mathbb{R}_+$. For any $t\in\mathbb{R}_+$, by $A_1$ (completeness), either $t\mathbf{e}\succsim\mathbf{x}$ or $t\mathbf{e}\precsim\mathbf{x}$, which implies $A\cup B=\mathbb{R}_+$. Therefore $\underline t\le\bar t$, and thus $A\cap B\ne\emptyset$, i.e. $\exists t^\star\in A\cap B$.
Uniqueness: by contradiction. Suppose $\exists t_1^\star,t_2^\star\in A\cap B$ and $t_1^\starContinuity: it suffices to show that for every \(\alpha\in\mathbb{R}$, the sets $\{\mathbf{x}\in\mathbb{R}_+^n:u(\mathbf{x})\le\alpha\}$ and $\{\mathbf{x}\in\mathbb{R}_+^n:u(\mathbf{x})\ge\alpha\}$ are both closed. Note $\{\mathbf{x}:u(\mathbf{x})\le\alpha\}=\{\mathbf{x}:u(\mathbf{x})\mathbf{e}\precsim\alpha\mathbf{e}\}=\{\mathbf{x}:\mathbf{x}\precsim\alpha\mathbf{e}\}$. By $A_3$ (continuity), $\{\mathbf{x}:\mathbf{x}\precsim\alpha\mathbf{e}\}$ is closed, so $\{\mathbf{x}:u(\mathbf{x})\le\alpha\}$ is closed. The same argument applies to $\{\mathbf{x}:u(\mathbf{x})\ge\alpha\}$. $\blacksquare$