Linear Combination of Integrals/Definite/Proof 2

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Theorem

$\displaystyle \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$


Proof

It is clear that for step functions $s$ and $t$:


$\displaystyle \int_a^b \lambda \map s x + \mu \map t x \rd x = \lambda \int_a^b \map s x \rd x + \mu \int_a^b \map t x \rd x$

Under any partition, the lower sums and upper sums of $f$ and $g$ are step functions, so the above formula relates the lower and upper sums of $f$ and $g$ to the lower and upper sums of the linear combinations of $f$ and $g$.

Because this identity is preserved for all possible partitions of $\closedint a b$, it is preserved for the supremum and infimum of all possible lower and upper sums, so the linear combinations of $f$ and $g$ are integrable.

\(\displaystyle \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t\) \(=\) \(\displaystyle \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren {\lambda f + \mu g} } } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren f} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} } + \mu \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren g} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \int_a^b \map f x \rd x + \mu \int_a^b \map g x \rd x\)

$\blacksquare$


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