Prove that $\mu$ is $\sigma-$subadditive.

Question

Let $X$ be a set. $\mathcal{A}$ an algebra over $X,$ and $\mu : \mathcal{A} \rightarrow [0, \infty]$ a $\sigma-$additive pre-measure. Prove that $\mu$ is $\sigma-$subadditive.

First of all, I beleive that, as $\mu$ is $\sigma-$additive, for all finite sequences of disjoint measurable sets $A_1, A_2, ..., A_n$ we have $$\mu(\bigcup_{i \in \mathbb{N}}A_i) = \sum\mu(A_i) ,$$ and I should try to prove that for all finite sequences of measurable sets $B_1, B_2,.., B_n$, not necessariliy disjoint, $$\mu(\bigcup_{i \in \mathbb{N}}B_i) \leq \sum\mu(B_i) $$ holds.

Let's suppose not. e.g. $$\mu(\bigcup_{i \in \mathbb{N}}B_i) > \sum\mu(B_i) ,$$ for all sequences. This would contradict the hypothesis, which says that a disjoint sequence of sets exists, for which the equality holds.

I beleive that my proof might be wrong, because my argument is too easy, and does not require something else than the $\sigma -$additivity.

Answer 1

The issue here is that $\mu$ is just a pre-measure, not a measure. More specifically, the fact that $\mathcal{A}$ is an algebra means that finite unions of members of $\mathcal{A}$ are in $\mathcal{A}$, the same need not be true for infinite unions.

What you need to show is that:

Let $A_n\in \mathcal{A}$ for every $n\in\mathbb{N}$, and suppose additionally that $A:=\bigcup_{n=0}^\infty A_n\in \mathcal{A}$. Then $\mu(A)\le \sum_{n=0}^\infty \mu(A_n)$.

It is key that $A\in\mathcal{A}$, for otherwise $\mu(A)$ need not be defined.

As for the proof, take $B_n:=A_n\setminus\bigcup_{k<n}A_k$, and note that the $B_n$ are pairwise disjoint and that $A=\bigcup_{n=0}^\infty B_n$. Then,

$$\mu(A)=\sum_{n=0}^\infty \mu(B_n)\le \sum_{n=0}^\infty \mu(A_n)$$ since $B_n\subset A_n$.

Answer 2

I think your definitions of $\sigma$-additivity and $\sigma$-subadditivity are incorrect. You should replace “there exists” with “for all”.