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Let <math>h(t)=\sum_{k=1}^{what the hell is this}f_{k}^{'}(t)</math> | Let <math>h(t)=\sum_{k=1}^{what the hell is this}f_{k}^{'}(t)</math> | ||
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| + | Lebesgue Monotone Convergence Theorem gives us <math>g(x)=\int_{0}^{x}h(t)dt</math> | ||
Revision as of 09:31, 10 July 2008
Since all the $ f_{n} $ are AC, there exists $ f_{n}^{'} $ such that $ f_{n}(x)=f_{n}(x)-f_{n}(0)=\int_{0}^{x}f_{n}^{'}(t)dt $ and $ f_{n}^{'} $ are nonnegative almost everywhere.
Let $ g_{n}(x)= \sum_{k=1}^{n}f_{k}(x)=\sum_{1}^{n}\int_{0}^{x}f_{k}^{'}(t)dt=\int_{0}^{x}(\sum_{k=0}^{n}f_{k}^{'}(t))dt $
Let $ h(t)=\sum_{k=1}^{what the hell is this}f_{k}^{'}(t) $
Lebesgue Monotone Convergence Theorem gives us $ g(x)=\int_{0}^{x}h(t)dt $
