Exercise 1.6.13 (Rising sun inequality II)

Question

Show that the left- and right-hand sides in \href{./exercise_1-6-12}{Exercise 1.6.12} are in fact equal for $\lambda > 0$.

Elu Thingol · Accepted Answer

First, additionally assume that $f:[a,b]	o\mathbf{R}$ is a compactly supported function on some compact interval $[a,b]$. As always, set
$$F:[a,b]	o\mathbb{R}, \qquad F(x):= \int_{[a,x]} |f|\ \mathrm{d}m - (x-a)\lambda.$$
Let $U:= \left\{x\in [a,b]: f^{*}(x)>\lambdaight\}$. We then have
\begin{align*}
f^{*}(x)>\lambda 
&\iff \exists h > 0 : \dfrac{1}{h} \int_{[x,x+h]} f\ \mathrm{d}m > \lambda\
&\iff \exists h > 0 : \int_{[a,x+h]} f\ \mathrm{d}m > \int_{[a,x]} f\ \mathrm{d}m + \lambda h\
&\iff \exists h > 0 : \int_{[a,x+h]} f\ \mathrm{d}m - x\lambda> \int_{[a,x]} f\ \mathrm{d}m + (x+h)\lambda\
&\iff \exists h > 0 : F(x+h) > F(x).
\end{align*}
Since $F$ is continuous, we apply the rising sun lemma to $U$ and get a disjoint open partition $U = \bigcup (a_i,b_i)$. Since $U$ is open, we cannot have $a_n = a$ and must have $F(a_i) = F(b_i)$ for any $i\in\mathbf{N}$ (cf. the proof of the rising sun lemma, p.118). Equivalently,
\begin{align*}
F(a_i) = F(b_i)
&\iff \int_{(a,a_i)} f\ \mathrm{d}m - (a_i - a)\lambda = \int_{(a,b_i)} f\ \mathrm{d}m - (b_i - a)\lambda\
&\iff \int_{(a_i,b_i)} f\ \mathrm{d}m = (b_i - a_i)\lambda
\end{align*}

From this, we immediately deduce
\begin{align*}
m \left(\left\{x \in [a,b]: \sup_{h>0} \frac{1}{h} \int_{[x,x+h]} f \ \mathrm{d}m > \lambda ight\} ight)
&= \sum_{i\in\mathbb{N}} (b_i - a_i)\
& = \dfrac{1}{\lambda} \sum_{i\in\mathbb{N}} \int_{(a_i,b_i)} f\ \mathrm{d}m\
&= \dfrac{1}{\lambda}\int_{\{x\in[a,b]:f^{*}(x)>\lambda\}} f\ \mathrm{d}m
\end{align*}

Using the \href{./exercise_1-2-11}{upwards monotonicity} we can generalise this result to any absolutely integrable function $f$, not just compactly supported.

Exercise 1.6.13 (Rising sun inequality II)

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Exercise 1.6.13 (Rising sun inequality II)

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