[proofplan] We prove the three identities separately. Translation invariance of Lebesgue measure gives the first identity. The dilation identity follows from the linear change-of-variables formula for the scalar map $x \mapsto tx$, with the case $t=0$ handled separately. For a general [linear map](/page/Linear%20Map), we first treat the invertible case by change of variables applied to the indicator function of $K$, and then treat the singular case by observing that $TK$ lies in a proper linear subspace and hence has zero $n$-dimensional Lebesgue measure. The convexity assumption is stronger than what the measure-theoretic argument needs; it is used here through the fact that a convex body is compact, hence its affine images are compact and Lebesgue measurable. [/proofplan] [step:Apply translation invariance to the translated convex body] Since $K$ is a convex body in $\mathbb{R}^n$, it is compact, hence Borel measurable and Lebesgue measurable. The set $K+x_0$ is also compact, hence Lebesgue measurable. By translation invariance of $n$-dimensional Lebesgue measure, \begin{align*} |K+x_0| = \mathcal{L}^n(K+x_0) = \mathcal{L}^n(K) = |K|. \end{align*} [/step] [step:Compute the determinant of the dilation map] If $t=0$, then $0K=\{0\}$, so \begin{align*} |0K|=\mathcal{L}^n(\{0\})=0=0^n |K|. \end{align*} Assume now that $t>0$. Define the linear dilation map \begin{align*} S_t: \mathbb{R}^n &\to \mathbb{R}^n\\ x &\mapsto tx. \end{align*} Let $I_n \in \mathbb{R}^{n \times n}$ denote the $n \times n$ identity matrix. Then $S_tK=tK$ and the matrix of $S_t$ in the standard basis is $tI_n$, so \begin{align*} \det S_t=\det(tI_n)=t^n. \end{align*} Since $t>0$, the map $S_t$ is invertible. By the [linear change-of-variables formula](/theorems/???) for Lebesgue measure applied to $S_t$ and to the measurable set $K$, \begin{align*} |tK| = \mathcal{L}^n(S_tK) = |\det S_t|\,\mathcal{L}^n(K) = t^n |K|. \end{align*} [guided] We separate the case $t=0$ because the map $x \mapsto 0x$ is not invertible. In that case the image of every point of $K$ is the origin, so $0K=\{0\}$. A singleton in $\mathbb{R}^n$ has zero $n$-dimensional Lebesgue measure, and therefore \begin{align*} |0K|=\mathcal{L}^n(\{0\})=0=0^n|K|. \end{align*} Now assume $t>0$. Define the dilation as an explicit linear map \begin{align*} S_t: \mathbb{R}^n &\to \mathbb{R}^n\\ x &\mapsto tx. \end{align*} Let $I_n \in \mathbb{R}^{n \times n}$ denote the $n \times n$ identity matrix. The image of $K$ under this map is exactly $S_tK=tK$. In the standard basis, the matrix of $S_t$ is $tI_n$, so its determinant is \begin{align*} \det S_t=\det(tI_n)=t^n. \end{align*} The [linear change-of-variables formula](/theorems/???) says that an invertible linear map scales $n$-dimensional Lebesgue measure by the absolute value of its determinant. Since $t>0$, the map $S_t$ is invertible. Applying the formula to the measurable set $K$ gives \begin{align*} |tK| = \mathcal{L}^n(S_tK) = |\det S_t|\,\mathcal{L}^n(K) = t^n|K|. \end{align*} [/guided] [/step] [step:Use change of variables for an invertible linear map] Assume first that $T$ is invertible. Let \begin{align*} T^{-1}: \mathbb{R}^n &\to \mathbb{R}^n \end{align*} denote the inverse linear map of $T$. Since $T$ is linear, it is continuous; hence $TK$ is compact as the continuous image of the compact set $K$, and therefore $TK$ is Lebesgue measurable. Define the indicator functions \begin{align*} \mathbb{1}_K: \mathbb{R}^n &\to \{0,1\}\\ x &\mapsto \begin{cases} 1, & x \in K,\\ 0, & x \notin K, \end{cases} \end{align*} and \begin{align*} \mathbb{1}_{TK}: \mathbb{R}^n &\to \{0,1\}\\ y &\mapsto \begin{cases} 1, & y \in TK,\\ 0, & y \notin TK. \end{cases} \end{align*} Since $K$ and $TK$ are Lebesgue measurable, $\mathbb{1}_K$ and $\mathbb{1}_{TK}$ are Lebesgue measurable. For every $y \in \mathbb{R}^n$, \begin{align*} \mathbb{1}_{TK}(y)=\mathbb{1}_K(T^{-1}y), \end{align*} because $y \in TK$ if and only if $T^{-1}y \in K$. Hence, using the [linear change-of-variables formula](/theorems/???) with the substitution $y=T(x)$, whose Jacobian determinant is $\det T$, \begin{align*} |TK| &= \int_{\mathbb{R}^n} \mathbb{1}_{TK}(y)\,d\mathcal{L}^n(y)\\ &= \int_{\mathbb{R}^n} \mathbb{1}_K(T^{-1}y)\,d\mathcal{L}^n(y)\\ &= |\det T|\int_{\mathbb{R}^n}\mathbb{1}_K(x)\,d\mathcal{L}^n(x)\\ &= |\det T|\,|K|. \end{align*} [guided] We first handle the case where $T$ is invertible, because then each point $y \in \mathbb{R}^n$ has a unique preimage under the inverse map. Let \begin{align*} T^{-1}: \mathbb{R}^n &\to \mathbb{R}^n \end{align*} denote the inverse linear map of $T$. Since $T$ is linear, it is continuous. Therefore $TK$ is compact as the continuous image of the compact set $K$, so $TK$ is Borel measurable and hence Lebesgue measurable. Define the indicator function of $K$ by \begin{align*} \mathbb{1}_K: \mathbb{R}^n &\to \{0,1\}\\ x &\mapsto \begin{cases} 1, & x \in K,\\ 0, & x \notin K, \end{cases} \end{align*} and define the indicator function of $TK$ by \begin{align*} \mathbb{1}_{TK}: \mathbb{R}^n &\to \{0,1\}\\ y &\mapsto \begin{cases} 1, & y \in TK,\\ 0, & y \notin TK. \end{cases} \end{align*} These functions are Lebesgue measurable because $K$ and $TK$ are Lebesgue measurable. The key point is to rewrite the indicator of the image set $TK$ in terms of the indicator of $K$. For $y \in \mathbb{R}^n$, we have \begin{align*} y \in TK \iff T^{-1}y \in K, \end{align*} because $T$ is bijective. Therefore \begin{align*} \mathbb{1}_{TK}(y)=\mathbb{1}_K(T^{-1}y). \end{align*} Now compute the volume of $TK$ by integrating its indicator function with respect to $n$-dimensional Lebesgue measure: \begin{align*} |TK| = \int_{\mathbb{R}^n}\mathbb{1}_{TK}(y)\,d\mathcal{L}^n(y) = \int_{\mathbb{R}^n}\mathbb{1}_K(T^{-1}y)\,d\mathcal{L}^n(y). \end{align*} Apply the [linear change-of-variables formula](/theorems/???) with the substitution $y=T(x)$. Under this substitution, Lebesgue measure transforms by \begin{align*} d\mathcal{L}^n(y)=|\det T|\,d\mathcal{L}^n(x). \end{align*} Thus \begin{align*} |TK| &= |\det T|\int_{\mathbb{R}^n}\mathbb{1}_K(x)\,d\mathcal{L}^n(x)\\ &= |\det T|\,\mathcal{L}^n(K)\\ &= |\det T|\,|K|. \end{align*} [/guided] [/step] [step:Show that a singular linear map sends $K$ into a null set] Assume now that $T$ is singular. Define the range of $T$ by \begin{align*} \operatorname{Range}(T) := \{T(x) : x \in \mathbb{R}^n\} \subset \mathbb{R}^n. \end{align*} Then $\det T=0$ and $\operatorname{Range}(T)$ is a proper linear subspace of $\mathbb{R}^n$. Since $T$ is continuous and $K$ is compact, the image $TK$ is compact, hence Lebesgue measurable. Since $TK \subset \operatorname{Range}(T)$ and every proper linear subspace of $\mathbb{R}^n$ has zero $n$-dimensional Lebesgue measure, \begin{align*} |TK| = \mathcal{L}^n(TK) \leq \mathcal{L}^n(\operatorname{Range}(T)) =0. \end{align*} Therefore \begin{align*} |TK|=0=|\det T|\,|K|. \end{align*} [guided] When $T$ is singular, it is not invertible, so the previous change-of-variables argument cannot be applied directly. Instead, singularity gives a geometric replacement: the image of $T$ is lower-dimensional. Since $T$ is singular, \begin{align*} \det T=0 \end{align*} and the rank of $T$ is strictly less than $n$. Define the range of $T$ by \begin{align*} \operatorname{Range}(T) := \{T(x) : x \in \mathbb{R}^n\} \subset \mathbb{R}^n. \end{align*} This range is a proper linear subspace of $\mathbb{R}^n$. Before writing $|TK|$ as a Lebesgue measure, we verify measurability. The map $T$ is linear, hence continuous, and $K$ is compact because it is a convex body. Therefore $TK$ is compact as the continuous image of a compact set. In particular, $TK$ is Borel measurable and hence Lebesgue measurable. The set $TK$ is contained in this range, because every point of $TK$ has the form $T(x)$ for some $x \in K$. Thus \begin{align*} TK \subset \operatorname{Range}(T). \end{align*} A proper linear subspace of $\mathbb{R}^n$ has zero $n$-dimensional Lebesgue measure: it has dimension at most $n-1$, so it occupies no $n$-dimensional volume. Therefore monotonicity of Lebesgue measure gives \begin{align*} |TK| = \mathcal{L}^n(TK) \leq \mathcal{L}^n(\operatorname{Range}(T)) =0. \end{align*} Since Lebesgue measure is nonnegative, this implies $|TK|=0$. Combining this with $\det T=0$, we obtain \begin{align*} |TK|=0=|\det T|\,|K|. \end{align*} [/guided] [/step] [step:Combine the three cases] The translation identity, the dilation identity, and the linear-map identity have been proved under exactly the stated hypotheses. Therefore \begin{align*} |K+x_0| &= |K|,\\ |tK| &= t^n |K|,\\ |TK| &= |\det T|\,|K|. \end{align*} This completes the proof. [/step]