Local finite measure of a compact set is finite

Let be a hausdorff Topological space equipped with a Locally finite measure , and let be compact. Then has finite measure. #m/thm/measure

Proof

For each , let be a neighbourhood with finite measure. Since forms an open cover of it admits a finite subcover with . By monotonicity and countable subadditivity of ,

hence is finite.

Corollary

A locally compact measure space has Locally finite measure iff the measure of every compact set is finite. #m/thm/measure

Proof

The forward direction is given above. Since every has a compact neighbourhood which in turn has finite measure, has locally finite measure.

#state/tidy | #lang/en | #SemBr