- #1
Tom555
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I've just started self-studying measure theory by reading Pugh's Mathematical Analysis. I'm trying to understand his argument for why the exclusion of a zero set does not change the outer measure: $m^*(E\setminus Z)=m^*(E)$:
(Pugh's arugment): Let $Z$ be a zero set, $E\subseteq\mathbb{R}$, and $m^*$ be the Lebesgue outer measure. Since $m^*(E)=m^*(E\cup Z)$, applying this to the set $E\setminus Z$ gives $m^*(E\setminus Z)=m^*((E\setminus Z)\cup(E\cap Z))=m^*(E).$ QED
My question is where does the $E\cap Z$ come from in the second equality above? If you're using $m^*(E)=m^*(E\cup Z)$ and making the substitution $E\to E\setminus Z$, why isn't it $m^*(E\setminus Z)=m^*((E\setminus Z)\cup Z)?$
Also, this is my first post on this site, so I apologize if something isn't formatted correctly.
(Pugh's arugment): Let $Z$ be a zero set, $E\subseteq\mathbb{R}$, and $m^*$ be the Lebesgue outer measure. Since $m^*(E)=m^*(E\cup Z)$, applying this to the set $E\setminus Z$ gives $m^*(E\setminus Z)=m^*((E\setminus Z)\cup(E\cap Z))=m^*(E).$ QED
My question is where does the $E\cap Z$ come from in the second equality above? If you're using $m^*(E)=m^*(E\cup Z)$ and making the substitution $E\to E\setminus Z$, why isn't it $m^*(E\setminus Z)=m^*((E\setminus Z)\cup Z)?$
Also, this is my first post on this site, so I apologize if something isn't formatted correctly.