# Finite Union of Sets in Additive Function

## Theorem

Let $\AA$ be an algebra of sets.

Let $f: \AA \to \overline {\R}$ be an additive function.

Let $A_1, A_2, \ldots, A_n$ be any finite collection of pairwise disjoint elements of $\AA$.

Then:

- $\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$

That is, for any collection of pairwise disjoint elements of $\AA$, $f$ of their union equals 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 pairwise disjoint elements of $\AA$.

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

- $\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$

$\map P 1$ is the case:

- $\ds \map f {\bigcup_{i \mathop = 1}^1 A_i} = \map f {A_1} = \sum_{i \mathop = 1}^1 \map f {A_i}$

Thus $\map P 1$ is seen to hold.

### Basis for the Induction

$\map P 2$ is the case:

- $\map f {A_1 \cup A_2} = \map f {A_1} + \map f {A_2}$

which comes from the definition of an additive function.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $\ds \map f {\bigcup_{i \mathop = 1}^k A_i} = \sum_{i \mathop = 1}^k \map f {A_i}$

Then we need to show:

- $\ds \map f {\bigcup_{i \mathop = 1}^{k + 1} A_i} = \sum_{i \mathop = 1}^{k + 1} \map f {A_i}$

### Induction Step

This is our induction step:

\(\ds \map f {\bigcup_{i \mathop = 1}^{k + 1} A_i}\) | \(=\) | \(\ds \map f {\bigcup_{i \mathop = 1}^k A_i \cup A_{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f {\bigcup_{i \mathop = 1}^k A_i} + \map f {A_{k + 1} }\) | as $A_{k + 1}$ and $\ds \bigcup_{i \mathop = 1}^k A_i$ are disjoint | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^k \map f {A_i} + \map f {A_{k + 1} }\) | from the induction hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^{k + 1} \map f {A_i}\) |

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

Therefore:

- $\ds \forall n \in \N_{>0}: \paren {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$

$\blacksquare$