[proofplan] The symmetry and normalization are direct consequences of the [mixed volume](/page/Mixed%20Volume) definition as the symmetric polarization coefficient in the [Minkowski sum](/page/Minkowski%20Sum) volume polynomial. Multilinearity is proved by comparing one monomial coefficient in two expansions of the same volume polynomial, after noting that Minkowski sums and non-negative dilations of convex bodies remain convex bodies. Translation invariance follows because translating each summand translates the whole Minkowski sum by one vector, and Lebesgue volume is translation invariant. Monotonicity is the only order-theoretic point: for $n=1$ it is ordinary interval-length monotonicity, while for $n \ge 2$ we use the mixed surface area measure representation of mixed volume, whose positive measure immediately converts inclusion of convex bodies into an inequality of [support functions](/page/Support%20Function). Throughout, $\mathcal{K}^n$ denotes the class of nonempty compact convex subsets of $\mathbb{R}^n$, $\mathbb{R}_{+}^{n}$ denotes $[0,\infty)^n$, and $S^{n-1}$ denotes the Euclidean unit sphere $\{u \in \mathbb{R}^n : |u|=1\}$. [/proofplan] [step:Read symmetry from the symmetric polarization definition] Fix $K_1,\dots,K_n \in \mathcal{K}^n$. Let $\sigma$ be a permutation of $\{1,\dots,n\}$. We use the definition of [mixed volume](/page/Mixed%20Volume) as the symmetric polarization of the homogeneous degree-$n$ polynomial $C \mapsto \operatorname{Vol}_n(C)$ on the cone $\mathcal{K}^n$ under [Minkowski addition](/page/Minkowski%20Sum) and non-negative dilation. Equivalently, for convex bodies $C_1,\dots,C_m \in \mathcal{K}^n$ and nonnegative scalars $t_1,\dots,t_m \in [0,\infty)$, \begin{align*} \operatorname{Vol}_n(t_1C_1+\cdots+t_mC_m) = \sum_{i_1=1}^{m}\cdots\sum_{i_n=1}^{m} V(C_{i_1},\dots,C_{i_n}) t_{i_1}\cdots t_{i_n}, \end{align*} where the polarization defining $V:(\mathcal{K}^n)^n \to [0,\infty)$ is symmetric in its $n$ arguments. This is an input from the definition of mixed volume, not a consequence of the displayed polynomial identity alone. Applying that defining symmetry to the ordered $n$-tuple $(K_1,\dots,K_n)$ gives \begin{align*} V(K_{\sigma(1)},\dots,K_{\sigma(n)})=V(K_1,\dots,K_n). \end{align*} This proves symmetry. [/step] [step:Compare the coefficient of $s_1s_2\cdots s_n$ to prove Minkowski linearity] Let $a,b \ge 0$. Since $L,M \in \mathcal{K}^n$ and $\mathcal{K}^n$ is closed under [Minkowski addition](/page/Minkowski%20Sum) and non-negative dilation, the body $aL+bM$ also belongs to $\mathcal{K}^n$. Define \begin{align*} A: \mathbb{R}_{+}^{n} &\to [0,\infty) \\ (s_1,\dots,s_n) &\mapsto \operatorname{Vol}_n\bigl(s_1(aL+bM)+s_2K_2+\cdots+s_nK_n\bigr). \end{align*} By the defining Minkowski polynomial applied to the $n$ convex bodies $aL+bM,K_2,\dots,K_n$, the coefficient of $s_1s_2\cdots s_n$ in $A$ is \begin{align*} n! \, V(aL+bM,K_2,\dots,K_n). \end{align*} Now rewrite the same Minkowski sum as \begin{align*} s_1(aL+bM)+s_2K_2+\cdots+s_nK_n = (as_1)L+(bs_1)M+s_2K_2+\cdots+s_nK_n. \end{align*} Applying the defining Minkowski polynomial to the $n+1$ convex bodies $L,M,K_2,\dots,K_n$, the coefficient of $s_1s_2\cdots s_n$ comes only from choosing either $L$ once and each $K_j$ once, or $M$ once and each $K_j$ once. Using symmetry already proved, that coefficient is \begin{align*} n! \, a V(L,K_2,\dots,K_n) + n! \, b V(M,K_2,\dots,K_n). \end{align*} Since both expressions are the coefficient of the same monomial in the same polynomial $A$, division by $n!$ gives \begin{align*} V(aL+bM,K_2,\dots,K_n) = aV(L,K_2,\dots,K_n)+bV(M,K_2,\dots,K_n). \end{align*} By symmetry, the same argument applies to any argument of $V$. [/step] [step:Use the support function representation to prove monotonicity] First consider the edge case $n=1$. A convex body $C \in \mathcal{K}^1$ is a compact interval $[\alpha_C,\beta_C]$ with $\alpha_C,\beta_C \in \mathbb{R}$ and $\alpha_C \le \beta_C$. Applying the defining polynomial to the single body $C$ gives \begin{align*} \operatorname{Vol}_1(tC)=tV(C) \end{align*} for $t \ge 0$, while one-dimensional Lebesgue volume gives \begin{align*} \operatorname{Vol}_1(tC)=t(\beta_C-\alpha_C). \end{align*} Hence $V(C)=\beta_C-\alpha_C$. If $L \subset M$, interval length is monotone under inclusion, so $V(L) \le V(M)$. Assume now that $n \ge 2$. For a convex body $C \in \mathcal{K}^n$, define its [support function](/page/Support%20Function) \begin{align*} h_C : S^{n-1} &\to \mathbb{R} \\ u &\mapsto \sup_{x \in C} x \cdot u. \end{align*} We use the [Mixed Surface Area Measure Formula](/page/Mixed%20Surface%20Area%20Measure): for fixed nonempty compact convex bodies $K_2,\dots,K_n \in \mathcal{K}^n$, there exists a finite positive Borel measure $S(K_2,\dots,K_n;\cdot)$ on $S^{n-1}$ such that, for every nonempty compact convex body $C \in \mathcal{K}^n$, \begin{align*} V(C,K_2,\dots,K_n) = \frac{1}{n} \int_{S^{n-1}} h_C(u)\, dS(K_2,\dots,K_n;u). \end{align*} The hypotheses of this representation are satisfied because $K_2,\dots,K_n$ and $C$ are convex bodies, hence nonempty compact convex subsets of $\mathbb{R}^n$; compactness makes each support function finite, and convex-body regularity gives continuity on the compact sphere $S^{n-1}$. The measure $S(K_2,\dots,K_n;\cdot)$ is positive by the cited formula. Assume $L \subset M$. For every $u \in S^{n-1}$, taking the supremum over the smaller set gives \begin{align*} h_L(u)=\sup_{x \in L} x\cdot u \le \sup_{x \in M} x\cdot u=h_M(u). \end{align*} Since $S(K_2,\dots,K_n;\cdot)$ is a positive finite Borel measure, integration preserves pointwise inequalities of continuous functions. Therefore \begin{align*} V(L,K_2,\dots,K_n) &= \frac{1}{n} \int_{S^{n-1}} h_L(u)\, dS(K_2,\dots,K_n;u) \\ &\le \frac{1}{n} \int_{S^{n-1}} h_M(u)\, dS(K_2,\dots,K_n;u) \\ &= V(M,K_2,\dots,K_n). \end{align*} By symmetry, monotonicity holds in every argument. [/step] [step:Translate each summand and use translation invariance of volume] Let $x_1,\dots,x_n \in \mathbb{R}^n$. For every $t_1,\dots,t_n \ge 0$, the Minkowski sum of translated bodies satisfies \begin{align*} t_1(K_1+x_1)+\cdots+t_n(K_n+x_n) = t_1K_1+\cdots+t_nK_n+\sum_{j=1}^{n} t_jx_j. \end{align*} The vector $\sum_{j=1}^{n} t_jx_j \in \mathbb{R}^n$ is a single translation vector. Since $n$-dimensional Lebesgue volume is translation invariant, \begin{align*} \operatorname{Vol}_n\bigl(t_1(K_1+x_1)+\cdots+t_n(K_n+x_n)\bigr) = \operatorname{Vol}_n(t_1K_1+\cdots+t_nK_n). \end{align*} Expanding both sides by the defining Minkowski polynomial and comparing the coefficient of $t_1\cdots t_n$ gives \begin{align*} n! \, V(K_1+x_1,\dots,K_n+x_n) = n! \, V(K_1,\dots,K_n). \end{align*} Dividing by $n!$ proves translation invariance. [/step] [step:Recover ordinary volume on the diagonal] Let $K \in \mathcal{K}^n$. For $t \ge 0$, the defining polynomial applied to the single body $K$ gives \begin{align*} \operatorname{Vol}_n(tK)=t^n V(K,\dots,K). \end{align*} On the other hand, the homogeneity of $n$-dimensional Lebesgue volume under dilation gives \begin{align*} \operatorname{Vol}_n(tK)=t^n\operatorname{Vol}_n(K). \end{align*} Comparing the coefficients of $t^n$ yields \begin{align*} V(K,\dots,K)=\operatorname{Vol}_n(K). \end{align*} This proves the normalization and completes the proof of all stated properties. [/step]