Let $A\subset \mathbb R^n$ be an arbitrary subset(not necessarily Lebesgue measurable). Denote the Lebesgue outer measure of $A$ by $m^{\ast}(A)$. Show that there exists a Lebesgue measurable set $B\subset \mathbb R^n$ such that $A\subset B$ and $m(B)=m^{\ast}(A)$.

How should I deal with this problem? What I know is just the definition of outer measure $\mu^{\ast}(A)= \mathrm{inf}\{\sum_{i=1}^\infty l(U_i): A\subset\cup_i U_i\}$ where $U_i$'s are open sets in $\mathbb R^n$. I have no idea how to proceed. What should I do?

Thank you for your help!


1 Answer 1


For each $m\in\mathbb N$, you know that there is a family $(U_{k,m})_{k\in\mathbb N}$ of open sets in $\mathbb{R}^n$ such that $A\subset\bigcup_{k\in\mathbb N}U_{k,m}$ and that $\sum_{k=1}^\infty l(U_{k,m})<\mu^*(A)+\frac1m$. Define $B_m=\bigcup_{k\in\mathbb N}U_{k,m}$. Then $A\subset B_m$ and$$\mu^*(A)\leqslant\mu(B_m)<\mu^*(A)+\frac1m.$$ Now, define, $B=\bigcap_{m\in\mathbb N}B_m$. Then $B$ is Borelian (and therefore measurable) and $\mu(B)=\mu^*(A)$.

  • $\begingroup$ it appears $n$ has several meanings $\endgroup$
    – zhw.
    Jul 30, 2017 at 14:43
  • $\begingroup$ @zhw. A capital error, if there is one. :-) Edited. $\endgroup$ Jul 30, 2017 at 14:49

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