Question 39 in Folland's Real Analysis chapter 3
The question "If {$F_j$} is a sequence of nonnegative increasing functions on $[a,b]$ such that $F(x)= \sum_1^\infty F_j(x) < \infty$" for all $x \in [a,b]$, then $F\prime(x)=\sum_1^\infty F\prime_j(x)$ for a.e. $x \in [a,b]$. (It suffices to assume that $F_j \in NBV$. Consider the measures $\mu_{F_j}$.)"
By the theorem 3.23 page 101 in the same book it makes sense to assume that $F_j \in NBV$ for all $j$.
I have very basic questions actually: first of all, how is derivative of F expressed? Are we allowed to write it down as: $F\prime = lim_{r\rightarrow 0} \frac{\mu_F(E_r)}{m(E_r)}$ where $\mu_F$ is the Borel measure , $\mu((a,b])=F(b)-F(a)$, $m$ is the Lebesgue measure, and $E_r = (x,x+h]$.
After that I consider writing $\mu_F$ in terms of $\mu_{F_j}$'s and obtain the equality, but it seems very wrong and there is not any use of the fact that $F_j$'s are nonnegative.
So could anyone please give me some hints and clarifications?
Thank you.
$\endgroup$ 11 Answer
$\begingroup$This is Fubini's theorem on differentiation. The proof goes as follows.
We will use the following lemma:
Lemma: If $G:\mathbb R\to\mathbb R$ is increasing, then $\int_a^bG'(t)dt\leq G(b)-G(a)$.
(proof.)Note that $G$ is differentiable a.e. and the derivative is positive wherever it exists because it is increasing. Let $c\in(a,b)$. By Fatou's lemma, we have $$ \begin{eqnarray} \int_a^{c} G'(t)dt&\leq&\liminf_{n\to\infty}\int_a^c \frac{G(t+n^{-1})-G(t)}{n^{-1}}dt\\ &=&\liminf_{n\to\infty}\left(\int_c^{c+n^{-1}}G(t)dt-\int_a^{a+n^{-1}}G(t)dt \right)\cdot n\\ &\leq&\liminf_{n\to\infty}G(c+n^{-1})-G(a)\\ &\leq& G(b)-G(a). \end{eqnarray} $$Letting $c\to b$ (via a countable sequence and applying MCT) gives the desired inequality.$\square$
Now we prove the claim. Note that $F'(x)$ and $f_j'(x)$ exists for a.e. $x\in[a,b]$ because they are increasing. If $x$ is such point, then for any $n$ we have $F'(x)-\sum_{j=1}^n f_j'(x)\geq 0$ (because $F-\sum_{j=1}^{n}f_j=\sum_{j=n+1}^\infty f_j$ is increasing) and therefore $F'(x)\geq \sum_{j=1}^\infty f_j'(x)$. Thus $F'\geq\sum_{j=1}^\infty f_j'$ a.e. on $[a,b]$.
On the other hand, our lemma implies that$$ \begin{eqnarray} \int_a^b F'(t)dt&=&\int_a^b\sum_{j=1}^nf_j'(t)dt+\int_a^b(F-\sum_{j=1}^nf_j)'(t)dt\\ &\leq&\int_a^b\sum_{j=1}^\infty f_j'(t)dt+(F-\sum_{j=1}^nf_j)(b)-(F-\sum_{j=1}^nf_j)(a), \end{eqnarray} $$so letting $n\to\infty$ yields the inequality $$\int_a^b F'(t)dt\leq\int_a^b\sum_{j=1}^\infty f_j'(t)dt.$$Because $F'\geq\sum_{j=1}^\infty f_j'$ a.e. and $F'\in L^1([a,b],m)$ by the lemma, this inequality implies that $F'=\sum_{j=1}^\infty f_j'$ a.e.
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