# Primitive of Pointwise Sum of Functions

## Theorem

Let $f_1, f_2, \ldots, f_n$ be real functions which are integrable.

Then:

$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$

## Proof

Proof by induction:

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

$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$

$\map P 1$ is true, as this just says:

$\ds \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$

### Basis for the Induction

$\map P 2$ is the case:

 $\ds \int \map {\paren {f_1 \pm f_2} } x \rd x$ $=$ $\ds \int \paren {\map {f_1} x \pm \map {f_2} x} \rd x$ Definition of Pointwise Addition of Real-Valued Functions $\ds$ $=$ $\ds \int \map {f_1} x \rd x \pm \map {f_2} x \rd x$ Linear Combination of Primitives

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:

$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$

Then we need to show:

$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$

### Induction Step

This is our induction step:

 $\ds$  $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x$ $\ds$ $=$ $\ds \int \paren {\map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x + \map {f_{k + 1} } x} \rd x$ Definition of Pointwise Addition of Real-Valued Functions $\ds$ $=$ $\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x + \int \map {f_{k + 1} } x \rd x$ Basis for the Induction $\ds$ $=$ $\ds \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x + \int \map {f_{k + 1} } x \rd x$ Induction Hypothesis

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

Therefore:

$\forall n \in \N: \ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$

$\blacksquare$