Lebesgue Measure on ℝ^{m+n} as the Product of Lebesgue Measures on ℝ^m and ℝ^n

In the setting of Product Measures, Lebesgue Measure on ℝⁿ and Image Measures: Notation, assume the axiom of dependent choice, and let $m,n\in\mathbb{N}$.

(i) Dimension one. For every $B\in\mathcal{B}(\mathbb{R})$, $\tau[B]\in\mathcal{B}(\mathbb{R}^{1})$ and

$$\lambda_{1}(\tau[B])=\lambda(B).$$

(ii) Concatenation. The image of $\lambda_{m}\otimes\lambda_{n}$ under $\iota$ is $\lambda_{m+n}$ ($\iota$ is measurable by Borel Sets of ℝ^{m+n} and the Product σ-Algebra of ℝ^m and ℝ^n (i); $\lambda_{m}$ and $\lambda_{n}$ are $\sigma$-finite by Lebesgue Measure on ℝⁿ: σ-Finiteness, Bounded Sets, Open Sets and Coordinate Hyperplanes (i)):

$$(\lambda_{m}\otimes\lambda_{n})\circ\iota^{-1}=\lambda_{m+n}.$$

(iii) Images of product-measurable sets. For every $E\in\mathcal{B}(\mathbb{R}^{m})\otimes\mathcal{B}(\mathbb{R}^{n})$, $\iota[E]\in\mathcal{B}(\mathbb{R}^{m+n})$ and

$$\lambda_{m+n}(\iota[E])=(\lambda_{m}\otimes\lambda_{n})(E).$$

(iv) Tonelli's formula. Let $f\in\mathcal{M}^{+}(\mathbb{R}^{m+n},\mathcal{B}(\mathbb{R}^{m+n}))$. For every $\xi\in\mathbb{R}^{m}$ the function $\eta\mapsto f(\iota(\xi,\eta))$ lies in $\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, the function $\xi\mapsto\int_{\mathbb{R}^{n}}f(\iota(\xi,\eta))\,d\lambda_{n}(\eta)$ lies in $\mathcal{M}^{+}(\mathbb{R}^{m},\mathcal{B}(\mathbb{R}^{m}))$, and

$$\int_{\mathbb{R}^{m+n}}f\,d\lambda_{m+n}=\int_{\mathbb{R}^{m}}\Bigl(\int_{\mathbb{R}^{n}}f(\iota(\xi,\eta))\,d\lambda_{n}(\eta)\Bigr)d\lambda_{m}(\xi).$$

(v) Fubini's formula. Let $f\in\mathcal{L}^{1}(\lambda_{m+n})$. The set

$$N=\{\xi\in\mathbb{R}^{m}:\text{the function }\eta\mapsto f(\iota(\xi,\eta))\text{ does not lie in }\mathcal{L}^{1}(\lambda_{n})\}$$

lies in $\mathcal{B}(\mathbb{R}^{m})$ and $\lambda_{m}(N)=0$; the function $g:\mathbb{R}^{m}\to\mathbb{R}$ with $g(\xi)=\int_{\mathbb{R}^{n}}f(\iota(\xi,\eta))\,d\lambda_{n}(\eta)$ for $\xi\in\mathbb{R}^{m}\setminus N$ and $g(\xi)=0$ for $\xi\in N$ lies in $\mathcal{L}^{1}(\lambda_{m})$; and

$$\int_{\mathbb{R}^{m+n}}f\,d\lambda_{m+n}=\int_{\mathbb{R}^{m}}g\,d\lambda_{m}.$$