Would anyone be able to provide a proof of the following lemma:
Where $| \cdot |$ is the Lebesgue measure and $M_L$ and $M_R$ are the left and right maximal functions defined on $\mathbb{R}$ respectively as $$M_Lf(x) = \sup_{r>0} \frac{1}{r} \int_{x-r}^r |f(y)|dy$$ $$M_Rf(x) = \sup_{r>0} \frac{1}{r} \int_{x}^{x+r} |f(y)|dy$$ I am interested because I would like to determine if there is a relationship between $\lambda|C_\lambda \cup D_\lambda |$ and $\int_{C\lambda \cup D_\lambda} f dt$ (as well as the same idea for the intersection). Thanks.
Proof. Suppose $E\neq\emptyset$. If $x\in E$, then by definition there exists a $h>0$ such that $\epsilon:=G(x+h)-G(x)>0$. As $G$ is uniformly continuous on the compact interval $[-2(x+h),2(x+h)]$, there exists $\delta>0$ such that \begin{align*} \left|y-y'\right|<\delta\Rightarrow \left|G(y)-G(y')\right|<\epsilon/3 \end{align*} From the triangle inequalities, it follows that \begin{align*} G(y+h)-G(y)&=[G(y+h)-G(x+h)]+[G(x+h)-G(x)]+[G(x)-G(y)]\\ &>G(x+h)-G(x)-\epsilon/3-\epsilon/3=\epsilon/3 \end{align*}
Being an open subset of $\mathbb{R}$, we can be written as the countable union $\bigcup (a_{k},b_{k})$ of disjoint open intervals $(a_{k},b_{k})$, where $-\infty\leq a_{k}<b_{k}\leq\infty$. I claim that $G(b_{k})-G(a_{k})=0$. Since $a_{k}\notin E$, it follows $G(b_{k})\leq G(a_{k})$. Suppose $G(b_{k})<G(a_{k})$. Then by the intermediate value theorem (IVT), there exists $c\in (a_{k},b_{k})$ such that \begin{align*} G(c)=\dfrac{G(a_{k})+G(b_{k})}{2} \end{align*} Moreover, we can choose $c=\sup\left\{a_{k}<c'<b_{k}: G(c')=(G(a_{k})+G(b_{k}))/2\right\}$ so that there does not exists $c'>c$ with this property. Since $c\in E$, there exists $h>0$ such that $G(c+h)>G(c)$. As $G(c)>G(b_{k})$ and $b_{k}\notin E$, we have that $c+h\in (c,b_{k})$. By another application of IVT, there exists $d\in (c+h,b_{k})$ such that $G(d)=(G(a_{k})+G(b_{k}))/2$. But this contradicts our choice of $c$. This completes the proof. $\Box$
As the argument is completely analogous, we only prove the case for $D_{\lambda}$. If $D_{\lambda}=\emptyset$, then both sides are zero; so assume that $D_{\lambda}\neq\emptyset$. Additionally, if $\left|D_{\lambda}\right|=\infty$, then both sides are infinite; so assume $\left|D_{\lambda}\right|<\infty$. First, observe that for $\lambda>0$ fixed, \begin{align*} x\in D_{\lambda}&\Leftrightarrow \exists h=h_{x}>0 \text{ s.t. } \int_{x}^{x+h}\left|f\right|\mathrm{d}y>h\lambda\\ &\Leftrightarrow \int_{-\infty}^{x+h}\left|f\right|\mathrm{d}y-(h+x)\lambda>\int_{-\infty}^{x}\left|f\right|\mathrm{d}y-x\lambda \end{align*} If we define a function $F(x)$ by \begin{align*} F(x):=\int_{-\infty}^{x}\left|f\right|\mathrm{d}y, \qquad\forall x\in\mathbb{R} \end{align*} then the last inequality may be written as $F(x+h)>F(x)$.
The function $F$ is continuous on $\mathbb{R}$. Whence by Riesz's sunrise lemma, $D_{\lambda}=\bigcup_{k}(a_{k},b_{k})$, where the intervals are disjoint, and $F(b_{k})-F(a_{k})=0$; equivalently, \begin{align*} \int_{a_{k}}^{b_{k}}\left|f\right|\mathrm{d}y=\int_{0}^{b_{k}}\left|f\right|\mathrm{d}y-\int_{0}^{a_{k}}\left|f\right|\mathrm{d}y=(b_{k}-a_{k})\lambda=\lambda\left|(a_{k},b_{k})\right| \end{align*} Summing over all the $k$, we have by $\sigma$-additivity of Lebesgue measure that \begin{align*} \int_{D_{\lambda}}\left|f\right|\mathrm{d}y=\lambda\sum_{k}\left|(a_{k},b_{k})\right|=\lambda\left|D_{\lambda}\right| \end{align*}
$\Box$
I can't really comment the relationship between $\int_{C_{\lambda}\cup D_{\lambda}}\left|f\right|\mathrm{d}x$ and $\lambda\left|C_{\lambda}\cup D_{\lambda}\right|$--analogously for the intersection--as I am not sure what you're looking for. If I come up with anything, I'll edit my answer.