3
$\begingroup$

Let $\mu$ be a $\sigma$-finite measure on a $\sigma$-Algebra $\mathcal A$ and $A_i\in\mathcal A$ ($i\in I$) subsets with $A_i\cap A_j=\emptyset$ for $i\neq j$. Then $\mu(A_i)>0$ for at most countably many $i\in I$.

I know $\sigma$-finite means that there exists a sequence $(A_n)\subset\mathcal A$ with $\Omega=\sum_{i=1}^\infty A_i$ such that $A_i\cap A_j=\emptyset$ for $i\neq j$ and $\mu(A_n)<\infty$ for all $n$. But I don't see how I can show the above using this.

$\endgroup$

1 Answer 1

2
$\begingroup$

Let $(B_n)_{n\geqslant 1}$ be an increasing sequence sets of finite measure which covers $\Omega$.

Fix integers $n$ and $p\geqslant 1$. The set $J_{n,p}:=\{i\in I\mid\mu(A_i\cap B_n)\geqslant p^{-1}\}$ is finite. Hence the set $\bigcup\limits_{n,p\geqslant 1}J_{n,p}$ is at most countable.

$\endgroup$
2
  • $\begingroup$ Thanks! Is $J_{n,p}$ finite because $(B_n)$ is an increasing sequence? $\endgroup$
    – dinosaur
    Nov 7, 2013 at 15:33
  • 1
    $\begingroup$ No ($n$ is fixed). This is because $B_n$ has finite measure and the $A_i\cap B_n$ are disjoint. $\endgroup$ Nov 7, 2013 at 15:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .