Countable union of positive sets

Question

Let $\mu$ be a finite signed measure on a measurable space $(E, \mathcal{E})$. Let $(A_n)_{n\geq 1}$ be a sequence of $\mu$-positive sets. Show that $B:= \bigcup_{n\geq 1} A_n$ is $\mu$-positive with $$\mu(B) \geq \sup_{n\geq 1} \mu(A_n).$$

I already showed that $B$ is $\mu$-positive, but I have some difficulties to prove the inequality for the measure.

Any suggestions? Thanks in advance!

Answer 1

Since you already know that $B$ is $\mu$-positive, you know that $\mu$ behaves as a positive measure when restricted to $B$. By monotonicity of positive measures, you then have

$$\mu(B)=\mu|_B(B)=\mu|_B\left(\bigcup_{n=1}^\infty A_n\right)\ge\mu|_B(A_k)=\mu(A_k)$$

for all $k\ge1$. Taking the supremum over all $k\ge1$ on both sides yields your claim.