Existence of efficient envy-free allocations of a heterogeneous divisible commodity with nonadditive utilities

Abstract

This paper studies the existence of Pareto optimal, envy-free allocations of a heterogeneous, divisible commodity for a finite number of individuals. We model the commodity as a measurable space and make no convexity assumptions on the preferences of individuals. We show that if the utility function of each individual is uniformly continuous and strictly monotonic with respect to set inclusion, and if the partition matrix range of the utility functions is closed, a Pareto optimal envy-free partition exists. This result follows from the existence of Pareto optimal envy-free allocations in an extended version of the original allocation problem.

Appendix

1.1 Uniform topologies on $L^\infty$

A binary relation $U$ on $L^\infty (\Omega ,\mathcal F ,\mu )$ is a subset of $[L^\infty (\Omega ,\mathcal F ,\mu )]^2$. Its composition with itself, $U\circ U,$ is defined by:

$$\begin{aligned} U\circ U=\left\{ (f,h)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\,\Bigl | \begin{array}{l} \exists g\in L^\infty (\Omega ,\mathcal F ,\mu ): \\ (f,g)\in U \text{ and} (g,h)\in U \end{array} \right\} . \end{aligned}$$

The inverse relation $U^{-1}$ of $U$ is defined by:

$$\begin{aligned} U^{-1}=\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid (g,f)\in U \}. \end{aligned}$$

The diagonal $\varDelta =\{ (f,f)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2 \}$ is the identity relation on $L^\infty (\Omega ,\mathcal F ,\mu )$.

Definition 5.1

A family $\mathcal U$ of subsets of $[L^\infty (\Omega ,\mathcal F ,\mu )]^2$ is a uniformity for $L^\infty (\Omega ,\mathcal F ,\mu )$ if it satisfies the following conditions.

• (i) $\varDelta \subset U$ for every $U\in \mathcal U$.

• (ii) $U^{-1}\in \mathcal U$ for every $U\in \mathcal U$.

• (iii) For every $U\in \mathcal U$ there exists $V\in \mathcal U$ such that $V\circ V\subset U$.

• (iv) $U\cap V\in \mathcal U$ for every $U,V\in \mathcal U$.

• (v) $U\in \mathcal U$ and $U\subset V$ imply $V\in \mathcal U$.

The pair $(L^\infty (\Omega ,\mathcal F ,\mu ),\mathcal U )$ is a uniform space. A subfamily $\mathcal B$ of a uniformity $\mathcal U$ is a base for $\mathcal U$ if for every $U\in \mathcal U$ there exists $V\in \mathcal B$ such that $V\subset U$. For $U\subset [L^\infty (\Omega ,\mathcal F ,\mu )]^2$ and $f\in L^\infty (\Omega ,\mathcal F ,\mu )$, define $U(f)=\{ g\in L^\infty (\Omega ,\mathcal F ,\mu )\mid (f,g)\in U \}$. The uniform topology $\tau _\mathcal{U }$ of $L^\infty (\Omega ,\mathcal F ,\mu )$ is the family of subsets of $L^\infty (\Omega ,\mathcal F ,\mu )$ given by:

$$\begin{aligned} \tau _\mathcal{U }=\{ O\subset L^\infty (\Omega ,\mathcal F ,\mu )\mid \forall f\in O\,\exists U\in \mathcal U : U(f)\subset O \}, \end{aligned}$$

where the neighborhood base of $f$ is the family $\{ U(f)\mid U\in \mathcal B \}$. Let $X$ be a subset of $L^\infty (\Omega ,\mathcal F ,\mu )$. The relative uniformity $\mathcal U _X$ for $X$ is the family:

$$\begin{aligned} \mathcal U _X=\{ U\cap (X\times X)\subset [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid U\in \mathcal U \}. \end{aligned}$$

It induces a uniform topology $\tau _\mathcal{U _X}$ on $X$ that coincides with the topology induced by $\tau _\mathcal{U }$.

We present the following result from (Kelley (1955), Theorem 6.26) as adapted to our context.

Theorem 5.1

Let $(L^\infty (\Omega ,\mathcal F ,\mu ),\mathcal U )$ be a uniform space and $X$ be a subset of $L^\infty (\Omega ,\mathcal F ,\mu )$. If $f:X\rightarrow \mathbb R$ is uniformly continuous, then there exists a unique uniformly continuous extension $\hat{f}:\mathrm cl \,X\rightarrow \mathbb R$ of $f$, where $\mathrm cl \,X$ is the closure of $X$ with respect to the uniform topology for $(L^\infty (\Omega ,\mathcal F ,\mu ),\mathcal U )$.

There are many different uniformities for $L^\infty (\Omega ,\mathcal F ,\mu )$. Our focus here is on the uniformity for $L^\infty (\Omega ,\mathcal F ,\mu )$ that is consistent with the weak* topology of $L^\infty (\Omega ,\mathcal F ,\mu )$. The existence of such uniformity is guaranteed by the following result.

Proposition 5.1

The family $\mathcal B$ of all subsets $U$ of $[L^\infty (\Omega ,\mathcal F ,\mu )]^2$ given as:

$$\begin{aligned} U=\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid |\mu _i(f)-\mu _i(g)|<\varepsilon ,\ i=1,\dots ,m \} \end{aligned}$$

for some $m\in \mathbb N ,\mu _1,\dots ,\mu _m\in ca (\Omega ,\mathcal F ,\mu )$ and $\varepsilon >0$, is a base for a uniformity for $L^\infty (\Omega ,\mathcal F ,\mu )$.

Proof

The family $\mathcal B$ is a base for a uniformity for $L^\infty (\Omega ,\mathcal F ,\mu )$ if and only if it satisfies the following conditions. (See (Kelley 1955, Theorem 6.2).)

• (i) $\varDelta \subset U$ for every $U\in \mathcal B$.

• (ii) For every $U\in \mathcal B$ there exists $V\in \mathcal B$ such that $V\subset U^{-1}$.

• (iii) For every $U\in \mathcal B$ there exists $V\in \mathcal B$ such that $V\circ V\subset U$.

• (iv) For every $U,V\in \mathcal B$ there exists $W\in \mathcal B$ such that $W\subset U\cap V$.

We verify that these conditions are satisfied for $\mathcal B$.

• (i): Obvious.

• (ii): This follows from the symmetry $U^{-1}=U$ for every $U\in \mathcal B$.

• (iii): Let $U\in \mathcal B$ be given by:

$$\begin{aligned} U=\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid |\mu _i(f)-\mu _i(g)|<\varepsilon ,\ i=1,\dots ,m \} \end{aligned}$$

with $m\in \mathbb N ,\mu _1,\dots ,\mu _m\in ca (\Omega ,\mathcal F ,\mu )$ and $\varepsilon >0$. Take $V\in \mathcal B$ such that:

$$\begin{aligned} V=\left\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid |\mu _i(f)-\mu _i(g)|<\frac{\varepsilon }{2},\ i=1,\dots ,m \right\} . \end{aligned}$$

If $(f,h)\in V\circ V$, then there exists $g\in L^\infty (\Omega ,\mathcal F ,\mu )$ such that $(f,g)\in V$ and $(g,h)\in V$. Thus, $|\mu _i(f)-\mu _i(h)|\le |\mu _i(f)-\mu _i(g)|+|\mu _i(g)-\mu _i(h)|<\varepsilon /2+\varepsilon /2=\varepsilon$ for each $i=1,\dots ,m$. This shows that $(f,h)\in U$. Hence $V\circ V\subset U$.

• (iv): Let $U,V\in \mathcal B$ have the form:

$$\begin{aligned} U=\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid |\mu _i(f)-\mu _i(g)|<\varepsilon ,\ i=1,\dots ,m \} \end{aligned}$$

with $m\in \mathbb N ,\mu _1,\dots ,\mu _m\in ca (\Omega ,\mathcal F ,\mu )$ and $\varepsilon >0$, and

$$\begin{aligned} V=\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\mid |\lambda _j(f)-\lambda _j(g)|<\varepsilon ^{\prime },\ j=1,\dots ,n \} \end{aligned}$$

with $n\in \mathbb N ,\,\lambda _1,\dots ,\lambda _n\in ca (\Omega ,\mathcal F ,\mu )$ and $\varepsilon ^{\prime }>0$. Define the $(m+n)$-dimensional vector measure by $\overrightarrow{\theta }=(\mu _1,\dots ,\mu _m,\lambda _1,\dots ,\lambda _n)$. Then, the intersection $U\cap V$ contains the set $W\in \mathcal B$ given by:

$$\begin{aligned} W=\left\{ (f,g)\in [L^\infty (\Omega ,\mathcal F ,\mu )]^2\,\Bigl | \begin{array}{l} |\theta _k(f)-\theta _k(g)|<\min \{ \varepsilon ,\varepsilon ^{\prime } \} \\ k=1,\dots ,m+n \end{array} \right\} , \end{aligned}$$

where $\theta _k$ are component measures of $\overrightarrow{\theta }$. $\square$

Let $(L^\infty (\Omega ,\mathcal F ,\mu ),\mathcal U )$ be the uniform space with the base $\mathcal B$ for $\mathcal U$ given in Proposition 5.1. By construction, the sets of the form:

$$\begin{aligned} U(f)=\{ g\in L^\infty (\Omega ,\mathcal F ,\mu )\mid |\mu _i(f)-\mu _i(g)|<\varepsilon ,\ i=1,\dots ,m \} \end{aligned}$$

with $m\in \mathbb N ,\,\mu _1,\dots ,\mu _m\in ca (\Omega ,\mathcal F ,\mu )$ and $\varepsilon >0$, constitute a neighborhood base of $f$ for the weak* topology of $L^\infty (\Omega ,\mathcal F ,\mu )$. (See Subsection 2.2.) Therefore, the uniform topology and the weak* topology coincide and the relative uniform topology of $X\subset L^\infty (\Omega ,\mathcal F ,\mu )$ coincides with the relative weak* topology of $X$.

1.2 The structure of the Pareto frontier

The next lemma is an immediate consequence of the Banach–Alaoglu theorem (see (Dunford and Schwartz 1958, Corollary V.4.3)).

Lemma 5.1

$\mathcal A$ is weakly* compact in $[L^\infty (\Omega ,\mathcal F ,\mu )]^n$.

Lemma 5.2

Suppose that $\nu _i$ is uniformly continuous and strictly $\mu _i$-monotone for each $i=1,\dots ,n$. Then, an allocation is Pareto optimal if and only if it is weakly Pareto optimal.

Proof

It is evident that Pareto optimality implies weak Pareto optimality. We show the converse implication. Let $(f_1,\dots ,f_n)$ be an allocation for $\widehat{\mathcal{E }}$ that is not Pareto optimal. Then, there is an allocation $(g_1,\dots ,g_n)$ in $\mathcal A$ such that $\hat{\nu }_i(f_i)\le \hat{\nu }_i(g_i)$ for each $i$ and $\hat{\nu }_j(f_j)<\hat{\nu }_j(g_j)$ for some $j$. As $\hat{\nu }_j$ is strictly $\mu _j$-monotone by Theorem 2.2, there exists $A\in \mathcal F$ with $\mu (A)>0$ on which $g_j$ is positive. The mutual absolute continuity of $\mu _1,\dots ,\mu _n$ yields $\mu _i(A)>0$ for each $i$. By the weak* continuity of $\hat{\nu }_j$ established in Theorem 2.1, there is $\varepsilon \in (0,1)$ such that $\hat{\nu }_j(f_j)<\hat{\nu }_j((1-\varepsilon )g_j)$. Define $h_i\in L^\infty (\Omega ,\mathcal F ,\mu )$ by

$$\begin{aligned} h_i= {\left\{ \begin{array}{ll} g_i+\frac{\varepsilon }{n-1}g_j&\text{ if} i\ne j, \\ (1-\varepsilon )g_j&\text{ otherwise}. \end{array}\right.} \end{aligned}$$

It is easy to see that $0\le h_i\le 1$ for each $i,h_i\ge g_i$ and $\mu _i(h_i)=\mu _i(g_i)+\varepsilon \mu _i(g_j)/ (n-1)>\mu _i(g_i)$ for $i\ne j$. By the strict $\mu _i$-monotonicity of $\hat{\nu }_i$ established in Theorem 2.2, the resulting allocation $(h_1,\dots ,h_n)$ satisfies $\hat{\nu }_i(f_i)<\hat{\nu }_i(h_i)$ for each $i$. Thus, allocation $(f_1,\ldots ,f_n)$ is not weakly Pareto optimal. $\square$

For every Pareto optimal allocation in $\widehat{\mathcal{E }}$, an individual exists that no one envies. This is a variant of the simple observation by Varian (1974), which plays an important role in proving the existence of a Pareto optimal envy-free allocation.

Proposition 5.1

For every Pareto optimal allocation $(f_1,\dots ,f_n)$ there exists $j$ such that $\hat{\nu }_i(f_j)\le \hat{\nu }_i(f_i)$ for each $i=1,\dots ,n$.

Proof

Take an arbitrary Pareto optimal allocation $(f_1,\dots ,f_n)$. Suppose, to the contrary, that for each $j$ there exists $\pi (j)\in \{ 1,\dots ,n \}$ such that $\hat{\nu }_{\pi (j)}(f_{\pi (j)})<\hat{\nu }_{\pi (j)}(f_j)$. Then, the map $\pi$ from $\{ 1,\dots ,n \}$ into itself defined by $j\mapsto \pi (j)$ satisfies $\pi (j)\ne j$ for each $j$. Thus, we have $\pi ^s(j)\ne \pi ^{s+1}(j)$ and $\hat{\nu }_{\pi ^{s+1}(j)}(f_{\pi ^{s+1}(j)})<\hat{\nu }_{\pi ^{s+1}(j)}(f_{\pi ^s(j)})$ for every $s=0,1,\dots ,$ where $\pi ^s$ is the $s$-th iteration of $\pi$ with $\pi ^0$ the identity map on $\{ 1,\dots ,n \}$. As the sequence $\{ \pi ^s(j) \}_{s=0}^\infty$ is contained in $\{ 1,\dots ,n \}$, and hence finite, we have $\pi ^s(j)=\pi (j)^{s-t}$ for some integers $s>t\ge 0$. Let $i_0=\pi ^s(j),i_1=\pi ^{s-1}(j),\dots ,i_t=\pi ^{s-t}(j)$ and $I=\{ i_0,\dots ,i_t \}$. It is evident that $\hat{\nu }_{i_0}(f_{i_0})<\hat{\nu }_{i_0}(f_{i_1}),\dots ,\hat{\nu }_{i_{t-1}}(f_{i_{t-1}})<\hat{\nu }_{i_{t-1}}(f_{i_t})$, and $\hat{\nu }_{i_t}(f_{i_t})<\hat{\nu }_{i_t}(f_{i_0})$. Define the allocation $(g_1,\dots ,g_n)$ by:

$$\begin{aligned} g_i= {\left\{ \begin{array}{ll} f_{i_{k+1}}&\text{ if} i=i_k \text{ for} 1\le k \le t-1, \\ f_{i_0}&\text{ if} i=i_t,\\ f_i&\text{ if} i\not \in I. \end{array}\right.} \end{aligned}$$

It is obvious that the resulting allocation $(g_1,\dots ,g_n)$ satisfies $\hat{\nu }_i(f_i)<\hat{\nu }_i(g_i)$ for each $i\in I$ and $\hat{\nu }_i(g_i)=\hat{\nu }_i(f_i)$ for each $i\not \in I.$ This contradicts the Pareto optimality of $(f_1,\dots ,f_n)$. $\square$

The weak* continuity of $\hat{\nu }_i$ by Theorem 2.1 and the weak* compactness of $\mathcal A$ by Lemma 5.1 guarantee that $\varGamma$ is compact in $\mathbb R ^n$ and that $\varGamma ^\mathrm{P }$ is nonempty and closed by Lemma 5.2. It follows from the strict $\mu _i$-monotonicity of $\hat{\nu }_i$ that $\varGamma ^\mathrm{P }$ is included in the boundary of $\varGamma$. Note also that $\varGamma$ is comprehensive from below. That is, $(x_1,\dots ,x_n)\in \varGamma$ and $0\le (y_1,\dots ,y_n)\le (x_1,\dots ,x_n)$ imply $(y_1,\dots ,y_n)\in \varGamma$.

Under our hypotheses, the Pareto frontier is homeomorphic to the unit simplex. The following technique to demonstrate this significant property is based on the argument developed by Hüsseinov (2009), Mas-Colell (1986), Sagara (2008).

Proposition 5.2

Define the function $\rho :\varDelta ^{n-1}\rightarrow \mathbb R$ by

$$\begin{aligned} \rho (x)=\sup \{ r\ge 0 \mid rx\in \varGamma \}, \end{aligned}$$

and let $h:\varDelta ^{n-1}\rightarrow \mathbb R ^n$ be defined by:

$$\begin{aligned} h(x)=\rho (x)x. \end{aligned}$$

Then, $h$ is a homeomorphism between $\varDelta ^{n-1}$ and $\varGamma ^\mathrm{P }$.

Proof

It follows from the closedness of $\varGamma$ that $h(x)\in \varGamma$. If $h(x)\not \in \varGamma ^\mathrm{P }$, then there exists $y\in \varGamma$ such that $h(x)<y$. This implies that $0\le (\rho (x)+\varepsilon )x<y$ for any sufficiently small $\varepsilon >0$, and hence $(\rho (x)+\varepsilon )x\in \varGamma$. This contradicts the definition of $\rho$. Therefore, $h$ is a mapping from $\varDelta ^{n-1}$ into the compact set $\varGamma ^\mathrm{P }$. By the strict $\mu _i$-monotonicity of $\hat{\nu }_i$, it is evident that $\varGamma$ contains a strictly positive vector. Hence, $\rho (x)>0$ for every $x\in \varDelta ^{n-1}$ because $\varGamma$ is comprehensive from below. It follows easily from this that $h$ is an injection.

We show that $h:\varDelta ^{n-1}\rightarrow \varGamma ^\mathrm{P }$ is a surjection. To this end, choose any $y\in \varGamma ^\mathrm{P }$. Note that $y$ is nonzero by the strict $\mu _i$-monotonicity of $\nu _i$. Define $x_i=y_i/\sum _{k=1}^ny_k$ for each $i$. Then, we have $x\in \varDelta ^{n-1}$ and $y=\sum _{k=1}^ny_kx$. Suppose that $\sum _{k=1}^ny_k\ne \rho (x)$. By the definition of $\rho (x)$ and the fact that $\sum _{k=1}^ny_kx\in \varGamma ^\mathrm{P }$, we must have $\sum _{k=1}^ny_k<\rho (x)$. Thus, $y_i=\sum _{k=1}^ny_kx_i\le \rho (x)x_i$ for each $i$ and $y_j=\sum _{k=1}^ny_kx_j<\rho (x)x_j$ for some $j$ with $x_j>0$. This contradicts the fact that $y\in \varGamma ^\mathrm{P }$ in view of $h(x)=\rho (x)x\in \varGamma ^\mathrm{P }$. Thus we have $\sum _{k=1}^ny_k=\rho (x)$, and hence $h(x)=y$.

Since $\varDelta ^{n-1}$ is compact, to complete the proof it suffices to show that $h$ is continuous. This will follow if we show that $\rho$ is a continuous function. To show the upper semicontinuity of $\rho$, assume, by way of contradiction, that $x^k\in \varDelta ^{n-1}$ and $x^k\rightarrow x$ imply ${\overline{\mathrm{lim}}}_k\rho (x^k)>\rho (x)$. Then, as $\rho$ is bounded, there exists a subsequence $\{ x^{k_m} \}$ of sequence $\{ x^k \}$ such that $\rho (x^{k_m})\rightarrow r_0>\rho (x)$. The closedness of $U$ implies $r_0x\in \varGamma$. But, $r_0>\rho (x)$ contradicts the definition of $\rho$. To demonstrate the lower semicontinuity of $\rho$, assume, by way of contradiction, that $x^k\in \varDelta ^{n-1}$ and $x^k\rightarrow x$ imply ${\underline{\mathrm{lim}}}_k\rho (x^k)<\rho (x)$. Then, there exists a subsequence $\{ x^{k_m^{\prime }} \}$ of $\{ x^k \}$ such that $\rho (x^{k_m^{\prime }})\rightarrow r_0^{\prime }<\rho (x)$. Thus, $\{ \rho (x^{k_m^{\prime }})x^{k_m^{\prime }} \}$ is a sequence in $\varGamma ^\mathrm{P }$ with the limit $r_0^{\prime }x$ not in $\varGamma ^\mathrm{P }$. This contradicts the closedness of $\varGamma$. $\square$