Is the following theorem true? If so, does my proof prove it?
Theorem:
Let $\mathcal{A} \subset 2^{\Omega}$ be a $\sigma$-algebra and let $\mu$ be a measure on $\mathcal{A}$. If $\mu$ is
- $\sigma$-finite, that is, $\exists\{A_{i}\}_{i\geq{1}}\subset\mathcal{A}$ s.t. $\Omega = \bigcup\limits_{i\geq{1}}A_{i}$ and $\mu(A_{i})<\infty$
- additive, that is, $\{A_{i}\}_{i=1}^{n}\subset\mathcal{A} \implies \mu\biggl(\biguplus\limits_{i=1}^{n}A_{i}\biggr) = \sum\limits_{i=1}^{n}\mu(A_{i})$
then $\mu$ is $\sigma$-subadditive, that is, if $\{A_{i}\}_{i\geq{1}}\subset\mathcal{A}$ is a countable covering of $A\in\mathcal{A}$ then $\mu(A)\leq\sum\limits_{i\geq{1}}\mu(A_{i})$.
Note: $\biguplus$ represents the disjoint union.
Proof:
From the hypothesis there exists $\{A_{i}\}_{i\geq{1}}\subset\mathcal{A}$ s.t. $\Omega=\bigcup\limits_{i\geq{1}}A_{i}$. Let $B_{1} = A_{1}$ and $B_{i}=A_{i}\setminus\biggl(\bigcup\limits_{1\leq{j}<i}A_{j}\biggr)$
$\implies A_{i}=A_{i}\bigcap\biggl( \bigcup\limits_{1\leq{j}<i}A_{j} \biggr) \bigcup B_{i} \implies B_{i}\subset A_{i}$
$\implies \mu(B_{i})\leq \mu(A_{i})$
therefore, $\mu\biggl(\biguplus\limits_{i\geq{1}}B_i\biggr)=\sum\limits_{i\geq{1}}\mu(B_i)\leq\sum\limits_{j\geq{1}}\mu(A_{j})$. Since $\mu$ is defined over $\mathcal{A}$ a similar construction can be made for any $A\in\mathcal{A}$; hence, $\mu$ is $\sigma$-subadditive on $\mathcal{A}$ if $\mu$ is $\sigma$-finite and additive.