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Josia Pietsch 2024-02-06 15:16:20 +01:00
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@ -110,11 +110,11 @@
Whenever $B \subseteq X$ is Borel,
we have that $b^{-1}(B)$ is Borel,
since $b$ is continuous.
For $A \subseteq \cN$ is Borel,
we have that $b$ with respect to $b(A)$
For $A \subseteq D$ Borel
be get by \yaref{thm:lusinsouslin},
that $b$ with respect to $b(A)$
is Borel,
since $b\defon{A}$ is injective,
by \yaref{thm:lusinsouslin}.
since $b\defon{A}$ is injective.
Hence \yaref{thm:bsb} can be applied.
\end{proof}