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Minor typo fix.
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Bokuan Li
2026-03-17 15:32:44 -04:00
parent dd49a42b4a
commit c26e3fdfcb
1 changed files with 1 additions and 1 deletions
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src/measure/bochner-integral/bochner.tex
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@@ -30,7 +30,7 @@
I\phi + I\psi &= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} + \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z} \\ I\phi + I\psi &= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} + \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z} \\
&= \sum_{y \in \phi(X)} \sum_{z \in \psi(X)} (y + z) \cdot \mu\bracs{\phi = y, \psi = z} \\ &= \sum_{y \in \phi(X)} \sum_{z \in \psi(X)} (y + z) \cdot \mu\bracs{\phi = y, \psi = z} \\
&= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'} \\ &= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'} \\
&= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu(\bracs{\phi + \psi = y}) = I\phi + I\psi &= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu\bracs{\phi + \psi = y} = I\phi + I\psi
\end{align*} \end{align*}
so $I$ is a linear operator on $\Sigma(X; E) \cap L^1(X; E)$ that satisfies (1). so $I$ is a linear operator on $\Sigma(X; E) \cap L^1(X; E)$ that satisfies (1).