# Show that it is at most countable

In a space of finite measure, show that a family of disjoint measurable sets with positive measure is at most countable.

Could you give me some hints what I am supposed to do??

Hint: Let $\{A_i\}_{i\in I}$ be the family of sets with $\mu(A_i)>0$ for all $i$. Now consider the sets with measure greater than $1/n$ for each $n\in\Bbb N$. If the family of sets $\{A_i\}$ were uncountable, what conclusion can you draw (keeping in mind that these all are subsets of a space with finite measure)?
• Do we have the following?? $$\frac{|I|}{n} <\mu(\{A_i\})=\mu(\cup_{i \in I}A_i)\leq \sum_{i \in I}\mu(A_i)$$ – Mary Star Oct 16 '14 at 14:37
• @MaryStar: If $\{A_i\}_{i\in I}$ is uncountable, then there must be infinitely many $A_i$ with the property that $\mu(A_i)\geq\frac{1}{n}$ for some $n\in\Bbb N$. Does this clarify the hint? – Clayton Oct 16 '14 at 16:46
• @MaryStar: $\infty=\sum\frac{1}{n}\leq\sum\mu(A_i)\leq\mu(X)=M<\infty$, where $X$ is the space of finite measure. – Clayton Oct 19 '14 at 0:58
• @MaryStar: if there are uncountable many $A_i$ of positive measure, then there are at least countably many with $\mu(A_i)\geq 1/n$. Since $1/n\leq\mu(A_i)$, we can add up the countably many $A_i$ we chose. Since the $A_i$ are disjoint, we have $\sum\mu(A_i)=\mu(\bigcup A_i)\leq\mu(X).$ – Clayton Oct 25 '14 at 13:50