Inner and Outer Approximation of Lebesgue Measurable Sets

Let E\subseteq\mathbb{R}. Then each of the following four assertions is equivalent to the measurability of E.

(Outer Approximation by Open Sets and G_\delta Sets)

(i) For each \epsilon>0, there is an open set G containing E for which m^*(G\setminus E)<\epsilon.

(ii) There is a G_\delta set G containing E for which m^*(G\setminus E)=0.

(Inner Approximation by Closed Sets and F_\sigma Sets)

(iii) For each \epsilon>0, there is a closed set F contained in E for which m^*(E\setminus F)<\epsilon.

(iv) There is an F_\sigma set F contained in E for which m^*(E\setminus F)=0.

(E measurable implies (i)):

Assume E is measurable. Let \epsilon>0. First we consider the case where m^*(E)<\infty. By the definition of outer measure, there is a countable collection of open intervals \{I_k\}_{k=1}^\infty which covers E and satisfies \displaystyle \sum_{k=1}^\infty l(I_k)<m^*(E)+\epsilon.

Define G=\bigcup_{k=1}^\infty I_k. Then G is an open set containing E. By definition of the outer measure of G, \displaystyle m^*(G)\leq\sum_{k=1}^\infty l(I_k)<m^*(E)+\epsilon.

Since E is measureable and has finite outer measure, by the excision property, \displaystyle m^*(G\setminus E)=m^*(G)-m^*(E)<\epsilon.

Now consider the case that m^*(E)=\infty. Since \mathbb{R} is \sigma-finite, E may be expressed as the disjoint union of a countable collection \{E_k\}_{k=1}^\infty of measurable sets, each of which has finite outer measure.

By the finite measure case, for each k\in\mathbb{N}, there is an open set G_k containing E_k for which m^*(G_k\setminus E_k)<\epsilon/2^k. The set G=\bigcup_{k=1}^\infty G_k is open, it contains E and \displaystyle G\setminus E=(\bigcup_{k=1}^\infty G_k)\setminus E\subseteq\bigcup_{k=1}^\infty (G_k\setminus E_k).

\begin{aligned}  m^*(G\setminus E)&\leq\sum_{k=1}^\infty m^*(G_k\setminus E_k)\\  &<\sum_{k=1}^\infty\epsilon/2^k\\  &=\epsilon.  \end{aligned}
Thus property (i) holds for E.

((i) implies (ii)):

Assume property (i) holds for E. For each k\in\mathbb{N}, choose an open set O_k that contains E such that m^*(O_k\setminus E)<1/k. Define G=\bigcap_{k=1}^\infty O_k. Then G is a G_\delta set that contains E. Note that for each k, \displaystyle G\setminus E\subseteq O_k\setminus E.

By monotonicity of outer measure, \displaystyle m^*(G\setminus E)\leq m^*(O_k\setminus E)<1/k.

Thus m^*(G\setminus E)=0 and hence (ii) holds.

((ii)\implies E is measurable):

Now assume property (ii) holds for E. Since a set of measure zero is measurable, G\setminus E is measurable. G is a G_\delta set and thus measurable. Since measurable sets form a \sigma-algebra, E=G\cap(G\setminus E)^c is measurable.


Assume condition (i) holds. Note that E^c is measurable iff E is measurable. Thus there exists an open set G\supseteq E^c such that m^*(G\setminus E^c)<\epsilon.

Define F=\mathbb{R}\setminus G which is closed. Note that F\subseteq E, and m^*(E\setminus F)=m^*(G\setminus E^c)<\epsilon.




Similar idea. Note that a set is G_\delta iff its complement is F_\sigma.

Author: mathtuition88

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