$ \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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Fubini's Theorem

Let $A\subseteq\mathbb{R}^n$ and $B\subseteq\mathbb{R}^m$ be two boxes, and let $f:A\times B\to\mathbb{R}$ be a bounded and integrable function on $A\times B$. If for each $\vec{x}\in A$, the function $f(\vec{x}, \cdot)$ is integrable on $B$, then $\int_Bf(\cdot, \vec{y})d\vec{y}$ is integrable on $A$ and $$\begin{align*} \int_{A\times B}f(\vec{x}, \vec{y}) d(\vec{x}, \vec{y})=\int_A[\int_B f(\vec{x}, \vec{y}) d\vec{y}] d\vec{x} \end{align*}$$

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Fubini's Theorem

Let $A\subseteq\mathbb{R}^n$ and $B\subseteq\mathbb{R}^m$ be two boxes, and let $f:A\times B\to\mathbb{R}$ be a bounded and integrable function on $A\times B$. If for each $\vec{x}\in A$, the function $f(\vec{x}, \cdot)$ is integrable on $B$, then $\int_Bf(\cdot, \vec{y})d\vec{y}$ is integrable on $A$ and $$\begin{align*} \int_{A\times B}f(\vec{x}, \vec{y}) d(\vec{x}, \vec{y})=\int_A[\int_B f(\vec{x}, \vec{y}) d\vec{y}] d\vec{x} \end{align*}$$

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FullPage
result
concepts
hypothesis
implications
proof

Fubini's Theorem

Let $A\subseteq\mathbb{R}^n$ and $B\subseteq\mathbb{R}^m$ be two boxes, and let $f:A\times B\to\mathbb{R}$ be a bounded and integrable function on $A\times B$. If for each $\vec{x}\in A$, the function $f(\vec{x}, \cdot)$ is integrable on $B$, then $\int_Bf(\cdot, \vec{y})d\vec{y}$ is integrable on $A$ and $$\begin{align*} \int_{A\times B}f(\vec{x}, \vec{y}) d(\vec{x}, \vec{y})=\int_A[\int_B f(\vec{x}, \vec{y}) d\vec{y}] d\vec{x} \end{align*}$$

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Fubini's Theorem

Let $A\subseteq\mathbb{R}^n$ and $B\subseteq\mathbb{R}^m$ be two boxes, and let $f:A\times B\to\mathbb{R}$ be a bounded and integrable function on $A\times B$. If for each $\vec{x}\in A$, the function $f(\vec{x}, \cdot)$ is integrable on $B$, then $\int_Bf(\cdot, \vec{y})d\vec{y}$ is integrable on $A$ and $$\begin{align*} \int_{A\times B}f(\vec{x}, \vec{y}) d(\vec{x}, \vec{y})=\int_A[\int_B f(\vec{x}, \vec{y}) d\vec{y}] d\vec{x} \end{align*}$$

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