signma-finite measure satisfies the finite set property

Question

If $\mu$ is a sigma-finite measure then $\mu$ satisfies the finite set property.

Note, we sat $\mu$ satisfies the finite set property if for all $A$ in the sigma-algebra, $\mu(A) = \infty$ there exists $B$ in the sigma algebra so $B \subset A$ and $\mu(B) \in (0, \infty)$

Proof:

I'm not really sure how to begin... We have

Suppose $\mu$ satisfies the sigma-finite measure, then $\bigcup_{n \in \mathbb{N}}A_{n} = X$. Do I then take $B = A_{n}$ for some fixed $n$?

Answer 1

Actually I think there is a consequence stronger than your finite set property on $\sigma$-finite measures. It is called the semifinite measures.

Semifinite measures:

A measure $\mu$ is semifinite if for all measurable $A$ with $\mu(A) = \infty$, there is a subset $B$ of $A$ with arbitrarily larger finite measure, i.e. for all $c > 0$, there is $B$ such that $B \subset A$ and $\mu(B) > c$.

Since in you question only ask for one $B$ with finite measure, showing that $\mu$ is semifinite should be a valid answer.

Proof: As $\mu$ is $\sigma$-finite, let $\{A_n\}_{n = 1} ^\infty$ be a disjoint collection of measurable sets such that $$ \bigcup _{n = 1} ^\infty A_n = X \quad \text{and} \quad \mu(A_n) < \infty \quad \text{for all } n \, . $$ Suppose $A$ is a measurable set with $\mu(A) = \infty$. By the countable additivity, $$ \infty = \mu(A) = \mu \left( \bigcup _{n = 1} ^\infty [A_n \cap A] \right) = \sum _{n = 1} ^\infty \mu(A_n \cap A) \, . $$ As the series above diverges to $\infty$, for every $c > 0$, there is $N$ such that $$ \sum _{n = 1} ^N \mu(A_n \cap A) > c \, . $$ The above is a finite sum, meaning the value is still finite. Thus $\bigcup _{n=1}^N [A_n \cap A]$ is a measurable subset of $A$ with finite measure greater than $c$.