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 $𝐾 \subseteq 𝑋$ be compact. Then $𝐾$ has finite measure. #m/thm/measure

Proof

For each $𝑥 \in 𝐾$ , let $𝑈 𝑥$ be a neighbourhood with finite measure. Since ${𝑈 𝑥} 𝑥 \in 𝐾$ forms an open cover of $𝐾$ it admits a finite subcover ${𝑈 𝑥 𝑖} 𝑖 \in 𝑋$ with ${𝑥 𝑖} 𝑛 𝑖 = 1 \subseteq 𝐾$ . By monotonicity and countable subadditivity of $𝜇$ ,

𝜇 (𝐾) \leq 𝜇 (𝑛 ⋃ 𝑖 = 1 𝑈 𝑖) \leq 𝑛 \sum 𝑖 = 1 𝜇 (𝑈 𝑖) < \infty

hence $𝜇 (𝐾)$ 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 $𝑥 \in 𝑋$ has a compact neighbourhood which in turn has finite measure, $𝑋$ has locally finite measure.

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