Let . Then each of the following four assertions is equivalent to the measurability of .

## (Outer Approximation by Open Sets and Sets)

(i) For each , there is an open set containing for which .

(ii) There is a set containing for which .

## (Inner Approximation by Closed Sets and Sets)

(iii) For each , there is a closed set contained in for which .

(iv) There is an set contained in for which .

Proof:

( measurable implies (i)):

Assume is measurable. Let . First we consider the case where . By the definition of outer measure, there is a countable collection of open intervals which covers and satisfies

Define . Then is an open set containing . By definition of the outer measure of ,

Since is measureable and has finite outer measure, by the excision property,

Now consider the case that . Since is -finite, may be expressed as the disjoint union of a countable collection of measurable sets, each of which has finite outer measure.

By the finite measure case, for each , there is an open set containing for which . The set is open, it contains and

Therefore

Thus property (i) holds for .

((i) implies (ii)):

Assume property (i) holds for . For each , choose an open set that contains such that . Define . Then is a set that contains . Note that for each ,

By monotonicity of outer measure,

Thus and hence (ii) holds.

((ii) is measurable):

Now assume property (ii) holds for . Since a set of measure zero is measurable, is measurable. is a set and thus measurable. Since measurable sets form a -algebra, is measurable.

((i)(iii)):

Assume condition (i) holds. Note that is measurable iff is measurable. Thus there exists an open set such that .

Define which is closed. Note that , and .

((iii)(i)):

Similar.

((ii)(iv)):

Similar idea. Note that a set is iff its complement is .