Let $(X,m)$ be a outer measure space such that $m(X)$ is finite and for any $A \subseteq X$ , there is a Caratheodory measurable (w.r.t. outer measure $m$) subset $B $ containing $A$ such that $m(A)=m(B)$ . Now if $E \subseteq X$ such that $m(X)=m(E)+m(X \setminus E)$ , then is it true that $E$ is Caratheodry measurable i.e. that $m(A)=m(A \cap E)+m(A \setminus E) ,\forall A \subseteq X$ ?
$(X,m)$ be a regular outer measure space , $m(X)$ finite , if $m(X)=m(E)+m(X \setminus E)$ , then is $E$ Caratheodory measurable?
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measure-theory
outer-measure
1 Answers
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Thanks to regularity, there is some measurable $F\supseteq E$ with $m(F) = m(E)$ and $G\supseteq X\setminus E$ with $m(G) = m(X\setminus E)$.
It is easy to deduce from $m(E) + m(X\setminus E) = m(X)$ that $m(F\cap G)=0$.
Then for any $A\subseteq X$, $$m(A\cap E) + m(A\setminus E)\le m(A\cap F) + m(A\cap G)\\ = m(A\cap F) + m(A\cap G\cap F) + m(A\cap G\setminus F)\\\le m(A\cap F) + m(G\cap F) + m(A\setminus F) = m(A\cap F) + m(A\setminus F) = m(A);$$ the opposite inequality follows from the definition of outer measure. Therefore, $E$ indeed is measurable.