Remark 1.4.15 (Measurable induction)

Question

Let $X$ be a set, and let $\mathcal{F}$ be a family of sets in $X$ and $P(E)$ a property of sets $E \subseteq X$ which obeys the following axioms:
\begin{enumerate}[label=(\roman*)]
\item $P(\emptyset)$ is true.
\item If $P(E)$ is true for some $E\subseteq X$, then $P(X\setminus E)$ is true also.
\item If $E_1,E_2,\ldots \subseteq X$ are such that $P(E_n)$ is true for all $n \in \mathbf{N}$, then $P\left(\bigcup_{n=1}^{\infty} E_n\right)$ is true also.
\item $P(E)$ is true for all $E \in \mathcal{F}$.
\end{enumerate}
Then one can conclude that $P(E)$ is true for all $E \in \left\langle F \right\rangle$.

Elu Thingol · Accepted Answer

Consider the set
$$\mathcal{A} :=\left\{E \subseteq X \mid P(E) 	ext{ is true}ight\}.$$

It is easy to verify that this set defines a $\sigma$-algebra.
\begin{enumerate}[label=(oman*)]
\item $\emptyset \in \mathcal{A}$ since $P(\emptyset)$ is true by the first assumption.
\item Let $E\in \mathcal{A}$ be arbitrary. Then $P(X \setminus E)$ is true as well by the second assumption; thus, $X \setminus E \in \mathcal{A}$.
\item Similarly, by the third assumption, $P\left(\bigcup E_night)$ is true and therefore $\bigcup_{n\in\mathbf{N}} E_n \in X$.
\end{enumerate}

By the fourth assumption, $\mathcal{F}$ is contained in $\mathcal{A}$. Since $\mathcal{A}$ is itself a $\sigma$-algebra, we conclude that $\langle \mathcal{F} angle \subseteq \mathcal{A}$. In other words, property $P$ must hold for every $E \in \mathcal{F}$, as desired.

Remark 1.4.15 (Measurable induction)

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Remark 1.4.15 (Measurable induction)

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