Real Analysis: Difficult Problems with Solutions

Solution:
For [tex]n\ge 1[/tex] put [tex]U_n=\bigcup_{k\ge n}E_k[/tex]. Each [tex]U_n[/tex] is measurable and contains [tex]E_n[/tex], so [tex]m(U_n)\ge s[/tex]. The sets decrease, [tex]U_1\supset U_2\supset\ldots[/tex], and [tex]m(U_1)\le 1\lt\infty[/tex], so by the continuity of the measure from above the set [tex]L=\bigcap_{n=1}^{\infty}U_n[/tex] satisfies
[tex]m(L)=\lim_{n\to\infty}m(U_n)\ge s\gt 0[/tex]
A point [tex]x[/tex] belongs to [tex]L[/tex] exactly when for every [tex]n[/tex] there is [tex]k\ge n[/tex] with [tex]x\in E_k[/tex], i.e. when [tex]x[/tex] belongs to infinitely many of the sets [tex]E_k[/tex]. Since [tex]m(L)\gt 0[/tex], the set [tex]L[/tex] is not empty; take [tex]x\in L[/tex]. Then the set of indices [tex]\{k: x\in E_k\}[/tex] is infinite; list its elements in increasing order as [tex]n_1\lt n_2\lt n_3\lt\ldots[/tex]. By construction [tex]x\in E_{n_j}[/tex] for every [tex]j[/tex], hence
[tex]x\in\bigcap_{j=1}^{\infty}E_{n_j}\Rightarrow\bigcap_{j=1}^{\infty}E_{n_j}\ne\varnothing[/tex]
Remark: the intersection may have measure zero. For example, if [tex]E_j[/tex] is the set of the numbers [tex]x\in(0, 1)[/tex] whose [tex]j[/tex]-th binary digit is [tex]0[/tex], then [tex]m(E_j)=\frac{1}{2}[/tex], but any [tex]r[/tex] of these sets intersect in a set of measure [tex]2^{-r}[/tex], so every infinite intersection has measure [tex]0[/tex].

Solution:
If [tex]B=\varnothing[/tex], take [tex]A=\varnothing[/tex]. Let [tex]B\ne\varnothing[/tex]. A nonempty compact subset of [tex]\mathbb{R}[/tex] is closed and bounded, so it has a least and a greatest element:
[tex]p=\min B,\qquad q=\max B[/tex]
Since [tex]B\subset f(\mathbb{R}^2)[/tex], there are points [tex]u,v\in\mathbb{R}^2[/tex] with [tex]f(u)=p[/tex] and [tex]f(v)=q[/tex]. Let [tex]r=\max\{|u|, |v|\}[/tex] and let [tex]K=\{x\in\mathbb{R}^2: |x|\le r\}[/tex] be the closed disc of radius [tex]r[/tex]. It is compact (closed and bounded, by the Heine-Borel theorem) and connected (it is convex). Hence its continuous image [tex]f(K)[/tex] is a compact connected subset of [tex]\mathbb{R}[/tex], i.e. a closed interval. This interval contains [tex]p=f(u)[/tex] and [tex]q=f(v)[/tex], therefore
[tex]B\subset[p, q]\subset f(K)[/tex]
Now put [tex]A=K\cap f^{-1}(B)[/tex]. The set [tex]B[/tex] is closed and [tex]f[/tex] is continuous, so [tex]f^{-1}(B)[/tex] is closed; thus [tex]A[/tex] is a closed subset of the compact set [tex]K[/tex] and is therefore compact. Clearly [tex]f(A)\subset B[/tex]. Conversely, for every [tex]y\in B\subset f(K)[/tex] there is [tex]x\in K[/tex] with [tex]f(x)=y[/tex], and then [tex]x\in A[/tex]. Hence
[tex]f(A)=B[/tex]

Solution:
Upper bound. Let [tex]f\in F[/tex] and [tex]x\in(0, 1)[/tex]. By Lagrange's mean value theorem on [tex][0, x][/tex] and on [tex][x, 1][/tex] there are points [tex]\xi[/tex] and [tex]\eta[/tex] such that
[tex]f(x)=f'(\xi)\,x\le ax,\qquad b-f(x)=f'(\eta)(1-x)\ge -a(1-x)[/tex]
Hence [tex]f(x)\le h(x)[/tex] on [tex][0, 1][/tex], where
[tex]h(x)=\min\{ax, a+b-ax\}[/tex]
The two lines meet at [tex]x_0=\frac{a+b}{2a}\in[0, 1][/tex] (because [tex]|b|\le a[/tex]), at height [tex]y_0=ax_0=\frac{a+b}{2}[/tex]. Therefore
[tex]\int_0^1 f(x)\,dx\le\int_0^{x_0}ax\,dx+\int_{x_0}^{1}(a+b-ax)\,dx=\frac{a^2+2ab-b^2}{4a}[/tex]
Sharpness. If [tex]|b|=a[/tex], then [tex]h(x)=bx[/tex], and this function itself belongs to [tex]F[/tex]. If [tex]|b|\lt a[/tex], then [tex]0\lt x_0\lt 1[/tex] and [tex]h[/tex] has a corner at [tex]x_0[/tex], which we round off. For [tex]0\lt\delta\lt\min\{x_0, 1-x_0\}[/tex] let [tex]f_\delta=h[/tex] for [tex]|x-x_0|\ge\delta[/tex] and
[tex]f_\delta(x)=y_0-\frac{a\delta}{2}-\frac{a(x-x_0)^2}{2\delta},\qquad |x-x_0|\lt\delta[/tex]
At [tex]x_0-\delta[/tex] and [tex]x_0+\delta[/tex] the pieces have equal values ([tex]y_0-a\delta[/tex]) and equal derivatives ([tex]a[/tex] and [tex]-a[/tex], respectively), so [tex]f_\delta[/tex] is differentiable, [tex]|f_\delta'|\le a[/tex] with equality outside [tex](x_0-\delta, x_0+\delta)[/tex], [tex]f_\delta(0)=0[/tex] and [tex]f_\delta(1)=b[/tex]; thus [tex]f_\delta\in F[/tex]. With [tex]t=x-x_0[/tex] we have [tex]h-f_\delta=\frac{a(|t|-\delta)^2}{2\delta}[/tex] for [tex]|t|\lt\delta[/tex], so
[tex]\int_0^1 h(x)\,dx-\int_0^1 f_\delta(x)\,dx=2\int_0^{\delta}\frac{a(\delta-t)^2}{2\delta}\,dt=\frac{a\delta^2}{3}\to 0\quad(\delta\to 0)[/tex]
Hence the supremum equals [tex]\frac{a^2+2ab-b^2}{4a}[/tex]. (For [tex]|b|\lt a[/tex] it is not attained: equality would force [tex]f=h[/tex], which is not differentiable at [tex]x_0[/tex].)