In the setting of Product Measures, Lebesgue Measure on ℝⁿ and Image Measures: Notation , assume the axiom of dependent choice . Let n ∈ N n\in\mathbb{N} n ∈ N , let A ∈ M n ( R ) A\in\mathcal{M}_{n}(\mathbb{R}) A ∈ M n ( R ) be invertible , let b ∈ R n b\in\mathbb{R}^{n} b ∈ R n , and let T : R n → R n T:\mathbb{R}^{n}\to\mathbb{R}^{n} T : R n → R n be the function
T ( x ) = A x + b . T(x)=Ax+b . T ( x ) = A x + b .
(i) Images of Borel sets. For every B ∈ B ( R n ) B\in\mathcal{B}(\mathbb{R}^{n}) B ∈ B ( R n ) , T [ B ] ∈ B ( R n ) T[B]\in\mathcal{B}(\mathbb{R}^{n}) T [ B ] ∈ B ( R n ) and
λ n ( T [ B ] ) = ∣ det A ∣ λ n ( B ) . \lambda_{n}(T[B])=|\det A|\,\lambda_{n}(B). λ n ( T [ B ]) = ∣ det A ∣ λ n ( B ) .
(ii) The image measure. T T T is B ( R n ) / B ( R n ) \mathcal{B}(\mathbb{R}^{n})/\mathcal{B}(\mathbb{R}^{n}) B ( R n ) / B ( R n ) -measurable, and its image measure is
( λ n ∘ T − 1 ) ( B ) = ∣ det A ∣ − 1 λ n ( B ) ( B ∈ B ( R n ) ) . (\lambda_{n}\circ T^{-1})(B)=|\det A|^{-1}\lambda_{n}(B)\qquad(B\in\mathcal{B}(\mathbb{R}^{n})). ( λ n ∘ T − 1 ) ( B ) = ∣ det A ∣ − 1 λ n ( B ) ( B ∈ B ( R n )) .
(iii) Nonnegative functions. For every f ∈ M + ( R n , B ( R n ) ) f\in\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n})) f ∈ M + ( R n , B ( R n )) , f ∘ T ∈ M + ( R n , B ( R n ) ) f\circ T\in\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n})) f ∘ T ∈ M + ( R n , B ( R n )) and
∫ f ∘ T d λ n = ∣ det A ∣ − 1 ∫ f d λ n . \int f\circ T\,d\lambda_{n}=|\det A|^{-1}\int f\,d\lambda_{n}. ∫ f ∘ T d λ n = ∣ det A ∣ − 1 ∫ f d λ n .
(iv) Integrable functions. Let f : R n → R f:\mathbb{R}^{n}\to\mathbb{R} f : R n → R be measurable. Then f ∘ T f\circ T f ∘ T is measurable, f ∘ T ∈ L 1 ( λ n ) f\circ T\in\mathcal{L}^{1}(\lambda_{n}) f ∘ T ∈ L 1 ( λ n ) if and only if f ∈ L 1 ( λ n ) f\in\mathcal{L}^{1}(\lambda_{n}) f ∈ L 1 ( λ n ) , and in that case
∫ f ∘ T d λ n = ∣ det A ∣ − 1 ∫ f d λ n . \int f\circ T\,d\lambda_{n}=|\det A|^{-1}\int f\,d\lambda_{n}. ∫ f ∘ T d λ n = ∣ det A ∣ − 1 ∫ f d λ n .