Primitive of Pointwise Sum of Functions/Proof 1

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:

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:

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

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