Linear Combination of Integrals/Definite/Proof 2

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Theorem

$\displaystyle \int_a^b \left({\lambda f \left({t}\right) + \mu g \left({t}\right)}\right) \rd t = \lambda \int_a^b f \left({t}\right) \rd t + \mu \int_a^b g \left({t}\right) \rd t$


Proof

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


$\displaystyle \int_a^b \lambda s \left({x}\right) + \mu t \left({x}\right) \rd x = \lambda \int_a^b s \left({x}\right) \rd x + \mu \int_a^b t \left({x}\right) \rd x$

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 $\left[{a \,.\,.\, b}\right]$, 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.

\(\displaystyle \int_a^b \left({\lambda f \left({t}\right) + \mu g \left({t}\right)}\right) \rd t\) \(=\) \(\displaystyle \sup \left\{ {\sum_{\nu \mathop = 1}^n m_\nu^{\left({\lambda f + \mu g}\right)} \left({x_\nu - x_{\nu - 1} }\right): \forall \nu \in \left[{1 \,.\,.\, n}\right] x_\nu > x_{\nu - 1} }\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \sup \left\{ {\sum_{\nu \mathop = 1}^n m_\nu^{\left({f}\right)} \left({x_\nu - x_{\nu - 1} }\right): \forall \nu \in \left[{1 \,.\,.\, n}\right] x_\nu > x_{\nu - 1} }\right\} + \mu \sup \left\{ {\sum_{\nu \mathop = 1}^n m_\nu^{\left({g}\right)} \left({x_\nu - x_{\nu - 1} }\right): \forall \nu \in \left[{1 \,.\,.\, n}\right] x_\nu > x_{\nu - 1} }\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \int_a^b f \left({x}\right) \rd x + \mu \int_a^b g \left({x}\right) \rd x\)

$\blacksquare$


Sources