Let $ (X, \Sigma, \mu) $ it will be a space with measure.
$\mu$ is $\sigma$-finite measure if it exist sequence of sets $X_{i} \in \Sigma $ and $\cup_{i=1}^{\infty}X_{i}=X$ and $\mu(X_{i})<\infty$ for all i
$\mu$ is semi-finite measure if for all $G \in \Sigma $ and $\mu (G)=\infty$ it exist $H \in \Sigma$ and $H \subset G$ and $0<\mu(H)<\infty$
Show that if $\mu$ is $\sigma$-finite measure then $\mu$ is semi-finite measure
$\endgroup$ 32 Answers
$\begingroup$Let $G\in \Sigma$ with $\mu(G) = \infty$. Choose $X_i \in \Sigma$ with $\cup_{i=1}^\infty X_i = X$ and $\mu(X_i) < \infty$ for all $i$ by the definition of $\sigma$-finite. Without loss of generality assume the $X_i$ are increasing (replace $X_i$ with $\cup_{j\leq i} X_i$) Then$$ G = \cup_{i=1}^\infty (G \cap X_i) $$so$$ \infty = \mu(G) = \lim_{i} \mu(G \cap X_i). $$All the terms on the right side are finite by definition of $X_i$. So we have a sequence of finite numbers approaching $\infty$. In particular, they are not all $0$, because $\lim_i 0 = 0 \neq \infty$. Hence one of them must be strictly positive, i.e. there is some $i$ for which $\mu(G \cap X_i) > 0$. Thus $H:= G \cap X_i$ suffices.
$\endgroup$ $\begingroup$Hint: $G=G\cap X=G\cap(\bigcup_i X_i)=\bigcup_i(G\cap X_i)$.
$\endgroup$ 4
