Finite Union of Sets in Additive Function

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

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

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


$\displaystyle f \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$

That is, for any collection of pairwise disjoint elements of $\mathcal A$, $f$ of their union equals the sum of $f$ of the individual elements.


Proof by induction:

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

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle f \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$

$P(1)$ is true, as this just says:

$f \left({A_1}\right) = f \left({A_1}\right)$

Basis for the Induction

$P(2)$ is the case:

$f \left({A_1 \cup A_2}\right) = f \left({A_1}\right) + f \left({A_2}\right)$

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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle f \left({\bigcup_{i \mathop = 1}^k A_i}\right) = \sum_{i \mathop = 1}^k f \left({A_i}\right)$

Then we need to show:

$\displaystyle f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right) = \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)$

Induction Step

This is our induction step:

\(\displaystyle f \left({\bigcup_{i \mathop = 1}^{k+1} A_i}\right)\) \(=\) \(\displaystyle f \left({\bigcup_{i \mathop = 1}^k A_i \cup A_{k+1} }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({\bigcup_{i \mathop = 1}^k A_i}\right) + f \left({A_{k+1} }\right)\) $\quad$ as $A_{k+1}$ and $\displaystyle \bigcup_{i \mathop = 1}^k A_i$ are disjoint $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^k f \left({A_i}\right) + f \left({A_{k+1} }\right)\) $\quad$ from the induction hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{k+1} f \left({A_i}\right)\) $\quad$ $\quad$

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


$\displaystyle \forall n \in \N_{>0}: \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$