$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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The Volume of a Box is Equal to the Sum of the Volume of its Partitions

Let $I\subseteq\mathbb{R}^n$ be a box and let $P$ be a partition with indexing set $J$ and subboxes $\{I^{(\vec\alpha)}:\vec{\alpha}\in J\}$. Then $$\begin{align*} \mu(I)=\sum_{\vec{\alpha}\in J}\\ \mu(I^{\vec{\alpha}})=\prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}), \end{align*}$$ for each $k\in\{1, \dots, n \}$, where $$\begin{align*} (b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)}). \end{align*}$$ Then using a combination of expanding, rearranging, and subsituting, we get $$\begin{align*} \mu(I)&=\prod_{k=1}^{n}\sum_{i=1}^{l_k}(x_{k}^{(i)}-x_k^{(i-1)})\\ &=[\sum_{i=1}^{l_1}(x_k^{(\alpha_1)}-x_k^{(\alpha_1-1)})]\dots [\sum_{j=1}^{l_n}(x_k^{(\alpha_n)}-x_k^{(\alpha_n-1)})] \\ &=\sum_{\alpha_1=1}^{l_1}\dots\sum_{\alpha_n=1}^{l_n}\prod_{k=1}^n (x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}) \\ &=\sum_{\alpha\in J} \prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)})\\ &=\sum_{\alpha\in J}\mu(I^{\vec{\alpha}}). \end{align*}$$

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Proof

By definition, $$\begin{align*} \mu(I)&=\prod_{k=1}^{n} (b_k-a_k)\\ \mu(I^{(\vec{\alpha})})&=\prod_{k=1}^n (x_k^{\alpha_k}-x_k^{(\alpha_k-1)}) \end{align*}$$ for each $k\in\{ 1, \dots, n\}$, where $(b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)})$. Then we get $$\begin{align*} \end{align*}$$

The Volume of a Box is Equal to the Sum of the Volume of its Partitions

Let $I\subseteq\mathbb{R}^n$ be a box and let $P$ be a partition with indexing set $J$ and subboxes $\{I^{(\vec\alpha)}:\vec{\alpha}\in J\}$. Then $$\begin{align*} \mu(I)=\sum_{\vec{\alpha}\in J}\\ \mu(I^{\vec{\alpha}})=\prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}), \end{align*}$$ for each $k\in\{1, \dots, n \}$, where $$\begin{align*} (b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)}). \end{align*}$$ Then using a combination of expanding, rearranging, and subsituting, we get $$\begin{align*} \mu(I)&=\prod_{k=1}^{n}\sum_{i=1}^{l_k}(x_{k}^{(i)}-x_k^{(i-1)})\\ &=[\sum_{i=1}^{l_1}(x_k^{(\alpha_1)}-x_k^{(\alpha_1-1)})]\dots [\sum_{j=1}^{l_n}(x_k^{(\alpha_n)}-x_k^{(\alpha_n-1)})] \\ &=\sum_{\alpha_1=1}^{l_1}\dots\sum_{\alpha_n=1}^{l_n}\prod_{k=1}^n (x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}) \\ &=\sum_{\alpha\in J} \prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)})\\ &=\sum_{\alpha\in J}\mu(I^{\vec{\alpha}}). \end{align*}$$

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Proof

By definition, $$\begin{align*} \mu(I)&=\prod_{k=1}^{n} (b_k-a_k)\\ \mu(I^{(\vec{\alpha})})&=\prod_{k=1}^n (x_k^{\alpha_k}-x_k^{(\alpha_k-1)}) \end{align*}$$ for each $k\in\{ 1, \dots, n\}$, where $(b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)})$. Then we get $$\begin{align*} \end{align*}$$
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concepts
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proof

The Volume of a Box is Equal to the Sum of the Volume of its Partitions

Let $I\subseteq\mathbb{R}^n$ be a box and let $P$ be a partition with indexing set $J$ and subboxes $\{I^{(\vec\alpha)}:\vec{\alpha}\in J\}$. Then $$\begin{align*} \mu(I)=\sum_{\vec{\alpha}\in J}\\ \mu(I^{\vec{\alpha}})=\prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}), \end{align*}$$ for each $k\in\{1, \dots, n \}$, where $$\begin{align*} (b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)}). \end{align*}$$ Then using a combination of expanding, rearranging, and subsituting, we get $$\begin{align*} \mu(I)&=\prod_{k=1}^{n}\sum_{i=1}^{l_k}(x_{k}^{(i)}-x_k^{(i-1)})\\ &=[\sum_{i=1}^{l_1}(x_k^{(\alpha_1)}-x_k^{(\alpha_1-1)})]\dots [\sum_{j=1}^{l_n}(x_k^{(\alpha_n)}-x_k^{(\alpha_n-1)})] \\ &=\sum_{\alpha_1=1}^{l_1}\dots\sum_{\alpha_n=1}^{l_n}\prod_{k=1}^n (x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}) \\ &=\sum_{\alpha\in J} \prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)})\\ &=\sum_{\alpha\in J}\mu(I^{\vec{\alpha}}). \end{align*}$$

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Hypothesis

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Results

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Proof

By definition, $$\begin{align*} \mu(I)&=\prod_{k=1}^{n} (b_k-a_k)\\ \mu(I^{(\vec{\alpha})})&=\prod_{k=1}^n (x_k^{\alpha_k}-x_k^{(\alpha_k-1)}) \end{align*}$$ for each $k\in\{ 1, \dots, n\}$, where $(b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)})$. Then we get $$\begin{align*} \end{align*}$$

The Volume of a Box is Equal to the Sum of the Volume of its Partitions

Let $I\subseteq\mathbb{R}^n$ be a box and let $P$ be a partition with indexing set $J$ and subboxes $\{I^{(\vec\alpha)}:\vec{\alpha}\in J\}$. Then $$\begin{align*} \mu(I)=\sum_{\vec{\alpha}\in J}\\ \mu(I^{\vec{\alpha}})=\prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}), \end{align*}$$ for each $k\in\{1, \dots, n \}$, where $$\begin{align*} (b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)}). \end{align*}$$ Then using a combination of expanding, rearranging, and subsituting, we get $$\begin{align*} \mu(I)&=\prod_{k=1}^{n}\sum_{i=1}^{l_k}(x_{k}^{(i)}-x_k^{(i-1)})\\ &=[\sum_{i=1}^{l_1}(x_k^{(\alpha_1)}-x_k^{(\alpha_1-1)})]\dots [\sum_{j=1}^{l_n}(x_k^{(\alpha_n)}-x_k^{(\alpha_n-1)})] \\ &=\sum_{\alpha_1=1}^{l_1}\dots\sum_{\alpha_n=1}^{l_n}\prod_{k=1}^n (x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)}) \\ &=\sum_{\alpha\in J} \prod_{k=1}^n(x_k^{(\alpha_k)}-x_k^{(\alpha_k-1)})\\ &=\sum_{\alpha\in J}\mu(I^{\vec{\alpha}}). \end{align*}$$

Concepts

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Hypothesis

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Results

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Proof

By definition, $$\begin{align*} \mu(I)&=\prod_{k=1}^{n} (b_k-a_k)\\ \mu(I^{(\vec{\alpha})})&=\prod_{k=1}^n (x_k^{\alpha_k}-x_k^{(\alpha_k-1)}) \end{align*}$$ for each $k\in\{ 1, \dots, n\}$, where $(b_k-a_k)=\sum_{i=1}^{l_k}(x_k^{(i)}-x_k^{(i-1)})$. Then we get $$\begin{align*} \end{align*}$$
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