Real Analysis: Very Difficult Problems with Solutions

Solution:
Put [tex]F(x)=\int_x^b f(t)\,dt[/tex] and [tex]G(x)=\int_x^b g(t)\,dt[/tex]. Since [tex]f[/tex] and [tex]g[/tex] are differentiable, they are continuous, so by the fundamental theorem of calculus [tex]F[/tex] and [tex]G[/tex] are differentiable on [tex][a, b][/tex] and [tex]F'=-f[/tex], [tex]G'=-g[/tex]. Moreover [tex]F(b)=G(b)=0[/tex] and, by the hypothesis, [tex]F(a)=0[/tex].
Step 1. The function [tex]P=FG[/tex] is differentiable and [tex]P(a)=P(b)=0[/tex], so by Rolle's theorem there is [tex]c_1\in(a, b)[/tex] with [tex]P'(c_1)=0[/tex]. Since
[tex]P'=F'G+FG'=-(fG+gF)[/tex]
the function [tex]H=fG+gF[/tex] satisfies [tex]H(c_1)=0[/tex].
Step 2. [tex]H[/tex] is differentiable on [tex][a, b][/tex] as a sum of products of differentiable functions, and [tex]H(b)=f(b)G(b)+g(b)F(b)=0[/tex]. Rolle's theorem on [tex][c_1, b][/tex] gives a point [tex]c\in(c_1, b)\subset(a, b)[/tex] with [tex]H'(c)=0[/tex]. But
[tex]H'=f'G+fG'+g'F+gF'=f'G-fg+g'F-gf=f'G+g'F-2fg[/tex]
so [tex]H'(c)=0[/tex] is exactly the required equality:
[tex]f'(c)\int_c^b g(x)\,dx+g'(c)\int_c^b f(x)\,dx=2f(c)g(c)[/tex]
Note that no condition on [tex]\int_a^b g(x)\,dx[/tex] is needed.

Solution:
Choose [tex]R\gt 0[/tex] with [tex]H\subset[-R, R][/tex] and put [tex]\varphi(t)=m\left((H+t)\setminus H\right)[/tex]. All sets below are obtained from [tex]H[/tex] by translations, differences and unions, hence they are measurable, and Lebesgue measure is translation invariant. Fix [tex]t\ne 0[/tex] and let [tex]k[/tex] be the smallest positive integer with [tex]k|t|\gt 2R[/tex]. Then [tex](k-1)|t|\le 2R[/tex], i.e.
[tex]k\le\frac{2R}{|t|}+1[/tex]
The set [tex]H+kt\subset[-R+kt, R+kt][/tex] is disjoint from [tex][-R, R][/tex] (if [tex]t\gt 0[/tex], then [tex]-R+kt\gt R[/tex]; if [tex]t\lt 0[/tex], then [tex]R+kt\lt -R[/tex]). Hence [tex](H+kt)\setminus H=H+kt[/tex] and
[tex]m\left((H+kt)\setminus H\right)=m(H+kt)=m(H)[/tex]
On the other hand,
[tex](H+kt)\setminus H\subset\bigcup_{i=1}^{k}\big((H+it)\setminus(H+(i-1)t)\big)[/tex]
Indeed, if [tex]x\in H+kt[/tex] and [tex]x\notin H[/tex], let [tex]i[/tex] be the smallest index in [tex]\{0, 1, \ldots, k\}[/tex] with [tex]x\in H+it[/tex]; then [tex]i\ge 1[/tex] and [tex]x\in(H+it)\setminus(H+(i-1)t)[/tex]. Each of these sets is a translate, [tex](H+it)\setminus(H+(i-1)t)=\big((H+t)\setminus H\big)+(i-1)t[/tex], so its measure is [tex]\varphi(t)[/tex]. By the subadditivity of the measure
[tex]m(H)\le k\,\varphi(t)\le\left(\frac{2R}{|t|}+1\right)\varphi(t)\Rightarrow\frac{\varphi(t)}{|t|}\ge\frac{m(H)}{2R+|t|}[/tex]
for every [tex]t\ne 0[/tex]. Letting [tex]t\to 0[/tex], we obtain
[tex]\liminf_{t\to 0}\frac{m\left((H+t)\setminus H\right)}{|t|}\ge\frac{m(H)}{2R}\gt 0[/tex]