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.
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.
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.
