Compact Subsets of the Real Line have Finite Measures

Question

I just want to check my reasoning on this one question:

Prove that if K is a compact subset of the real line, the measure of K is finite.

I am assuming this is simply because compact sets are closed and closed sets have finite outer measures. Is this correct or am I oversimplifying?

Answer 1

You’re not oversimplifying; you’re simplifying in the wrong direction. Compact sets are closed and bounded but it is the boundedness you need here. (The closedness happens to be irrelevant.) Boundedness means there is an $M$ such that $K\subset[-M,M]$. By the monotonicity of measures, $$\mu(K)\leq\mu([-M,M])=2M<\infty.$$

Answer 2

Any point of $\Bbb R$ has an open neighbourhood of finite measure. (this is trivial as $\mu((x-1,x+1)) = 2$, e.g. but is a property that many measures ons paces have (it's called a locally finite measure).

A compact subset $K$ has a cover by such neighbourhoods thus a finite such cover too. Then (finite)subadditivity and monotonicity of the measure do the rest and $\mu(K) < +\infty$.