\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Measures
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Section Measures

Why

Toward a theory of iterated integrals, we need to know the function measuring a section is integrable.

Results

Let $(X, \mathcal{A} , \mu )$ and $(Y, \mathcal{B} , \nu )$ be sigma-finite measurable spaces. Let $E \in \mathcal{A} \times \mathcal{B} $. The function $x \mapsto \nu (E_x)$ is $\mathcal{A} $-measurable and the function $y \mapsto \mu (E^y)$ is $\mathcal{B} $-measurable.
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