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When $f_n$ if non-negative and increasing on $(0,\ \infty)$

$$\lim_{x\to \infty}\sum_{n=1}^{\infty}f_n(x)<\infty$$

Prove that $$\sum_{n=1}^{\infty}f_n'(x)<\infty$$ on $(0,\ 1)$ a.e $[m]$.

Is there the question means $f$ is differentiable? If so I will try mean value theorem.

If not, I am totally stuck at the beginning, since $f$ is not mentioned absolutely continuous or f' belong to $L^1(m)$, I have no idea how to connect $f'$ and $f$ here.

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monotonic functions are differentiable almost everywhere. –  Alex R. Nov 29 '12 at 3:00
    
What is $f_n$? $\ $ –  Martin Argerami Nov 29 '12 at 3:22
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2 Answers

This result, when stated in full generality, is known as Fubini's Theorem on the Termwise-Differentiation of Series with Monotone Terms.

A proof that uses the theory of Lebesgue integration may be found in the following document: www.math.sc.edu/~howard/Notes/fubini.ps.gz.

To see an elementary proof that does not rely on integration theory at all, please consult the classic text Functional Analysis by Frigyes Riesz and Béla Nagy. For your convenience, I shall provide a link to the relevant page of the text:

http://books.google.com/books?id=jlQnThDV41UC&pg=PA11&lpg=PA11&dq=Fubini#v=onepage&q=Fubini&f=false.

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Hint: there is a positive measure $\mu_n$ on $(0,\infty)$ such that $\mu_n((0, x)) = f_n(x) - f_n(0)$. Decompose $\mu_n$ into its singular and absolutely continuous components.

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