Suppose $\mu$ is a measure in a ring $R$ of a set $X$. So for any mutually disjoint subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, we have that $\mu(\sum_{i=1}^\infty A_i) = \mu(A_1)+\mu(A_2)+\cdots$. (i.e. countable additivity)
I'm wondering: Does that imply countable subadditivity? I.e. for any subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, is it always true that $\mu(\sum_{i=1}^\infty A_i)\le \mu(A_1)+\mu(A_2)+\cdots$?
I believe it should be true, because you can take the mutually disjoint subsets $A_1,A_2-A_1,A_3-(A_1\cup A_2),A_4-(A_1\cup A_2\cup A_3),\cdots$, each of which is in $R$ (because unions and differences of sets in $R$ are again in $R$), then apply countable additivity, then use monotonicity on the fact that $A_n-(A_1\cup\cdots\cup A_{n-1})\subseteq A_n$.
Is that right? I just want to make sure I didn't make any mistake in my logic.
Best Answer
Your proof is essentially right, though you might want to also mention that your new sequence $A_1, A_2-A_1,\dots$ has the same union as the original sequence.