 Solution

A measure $\mu$ on $(\Omega ,\mathcal{F})$ is $\sigma$-finite whenThe $\sigma$-finite uniqueness theorem says that two measures agreeing on a generating $\pi$-system agree on the generated $\sigma$-algebra, provided the space is covered by countably many sets in that system having finite common measure. Equivalently, a $\sigma$-finite premeasure has at most one extension to the generated $\sigma$-algebra.