Suppose that $\mu$ is a positive $\sigma$-finite Borel measure on $\mathbb{R}^n$.
Does it imply that $\mu(K)<\infty$ for every compact set $K \subset \mathbb{R}^n$ ?
I think it does not. However, I do not know how prove it. Could you please help? Thank you.
Say $\delta_x$ is a point mass at $x$: $$\delta_x(E)=\begin{cases} 1,&(x\in E), \\0,&(x\notin E).\end{cases}$$Let $$\mu=\sum_{n=1}^\infty\delta_{1/n}.$$Then $\mu$ is a $\sigma$-finite Borel measure on $\Bbb R$, but $\mu([0,1])=\infty$.
Note: The right answer to the question depends on the context. If, say, it was a homework problem to determine whether a $\sigma$-finite Borel measure must be finite on compact sets, there's a counterexample. But another possibility is that you're reading something where the author seems to take this result for granted, and you're wondering about that. If so the answer is that people often say just "Borel measure" when they mean "regular Borel measure", or "Borel Radon measure" or some such.