Locally finite measure

Local finite measure of a compact set is finite

Let 𝑋 be a hausdorff Topological space equipped with a Locally finite measure 𝜇, and let 𝐾 ⊆𝑋 be compact. Then 𝐾 has finite measure. measure

Proof

For each 𝑥 ∈𝐾, let 𝑈𝑥 be a neighbourhood with finite measure. Since {𝑈𝑥}𝑥∈𝐾 forms an open cover of 𝐾 it admits a finite subcover {𝑈𝑥𝑖}𝑖∈𝑋 with {𝑥𝑖}𝑛𝑖=1 ⊆𝐾. By monotonicity and countable subadditivity of 𝜇,

𝜇(𝐾)≤𝜇(𝑛⋃𝑖=1𝑈𝑖)≤𝑛∑𝑖=1𝜇(𝑈𝑖)<∞

hence 𝜇(𝐾) is finite.

Corollary

A locally compact measure space has Locally finite measure iff the measure of every compact set is finite. measure

Proof

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

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