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

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


 * $$P(1)$$ 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:
 * $$f \left({\bigcup_{i=1}^k A_i}\right) \le \sum_{i=1}^k f \left({A_i}\right)$$

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

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