$\displaystyle=I(f)-m_{n}\left(I\left(\frac{g_{3}}{m_{n}}\right)+\ldots+I\left(\frac{f_{n}}{m_{n}}\right)\right)$
$\displaystyle=\ldots$
$\displaystyle=I(f)-m_{n}I\left(\frac{g_{n}}{m_{n}}\right)$
$\displaystyle=m_{n}m\left(I\left(\frac{f}{m_{n}m}\right)-I\left(\frac{g_{n}}{m_{n}m}\right)\right)$
$\displaystyle=m_{n}mI(h_{n})<\epsilon$

Because $(I(f)-\sum_{k\leq m}I(f_{k}))_{m}$ is a decreasing sequence of
positive numbers, we can conclude that $I(f)=\sum_{n\in\mathbb{N}}I(f_{n})$. ∎

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