Finite Union of Sets in Subadditive Function

Theorem
Let $\mathcal A$ be an algebra of sets.

Let $f: \mathcal A \to \overline {\R}$ be a subadditive function.

Let $A_1, A_2, \ldots, A_n$ be any finite collection of elements of $\mathcal A$.

Then:
 * $\displaystyle f \left({\bigcup_{i=1}^n A_i}\right) \le \sum_{i=1}^n f \left({A_i}\right)$

That is, for any collection of elements of $\mathcal A$, $f$ of their union is less than or equal to the sum of $f$ of the individual elements.

Proof
Proof by induction:

In the below, we assume that $A_1, A_2, \ldots$ are all elements of $\mathcal A$.

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle f \left({\bigcup_{i=1}^n A_i}\right) \le \sum_{i=1}^n f \left({A_i}\right)$

$P(1)$ is trivially true, as this just says $f \left({A_1}\right) \le f \left({A_1}\right)$.

Basis for the Induction
$P(2)$ is the case $f \left({A_1 \cup A_2}\right) \le f \left({A_1}\right) + f \left({A_2}\right)$, which comes from the definition of a subadditive function.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle f \left({\bigcup_{i=1}^k A_i}\right) \le \sum_{i=1}^k f \left({A_i}\right)$

Then we need to show:
 * $\displaystyle f \left({\bigcup_{i=1}^{k+1} A_i}\right) \le \sum_{i=1}^{k+1} f \left({A_i}\right)$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N^*: \left({\bigcup_{i=1}^n A_i}\right) \le \sum_{i=1}^n f \left({A_i}\right)$