I'm following the book- 'Fourier Analysis' by Javier Duoandikoetxea. In the chapter named- 'The Hardy-Littlewood Maximal Function', the Lebesgue Differentiation theorem was proved as a corollary to the weak $(1,1)$ inequality for the maximal function. The proved theorem states-
If $f\in L^1_{\text{loc}}(\mathbb{R}^n)$ then $$ \lim_{x\rightarrow 0^+}\frac{1}{|B_r|}\int_{B_r}f(x-y)dy=f(x) \text{ a.e}$$
The book says that the corollary can be made sharper, i.e.
If $f\in L^1_{\text{loc}}(\mathbb{R}^n)$ then $$ \lim_{x\rightarrow 0^+}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy=0 \text{ a.e}$$
The given hint states that $$\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy\leq Mf(x) + |f(x)|$$ where $M$ denotes the Hardy Littlewood maximal function, and $|B_r|$ denotes the Lebesgue measure of the ball of radius $r$ centered at the origin. I am unable to understand how to prove the sharper theorem using the hint.
I have been trying to find another function and use the corollary on that function to get the above result but have not succeeded. A proof which I have read earlier approximates $f(x)$ by rationals and uses the countability of rationals along with using the corollary on the function $|f(x-y)-r|$ to obtain the result. However, this does not use the hint. Could someone give another hint on how to proceed using the given hint?
There's nothing to it. Define $$M'f(x)=\sup_{r>0}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy.$$
Since $M'f(x)\le Mf(x)+|f(x)|$ it follows that $$m(\{M'f(x)>\lambda\}) \le\frac c\lambda||f||_1.$$
(Hint for that: $a+b>\lambda$ implies either $a>\lambda/2$ or $b>\lambda/2$.)
And you certainly have $$\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy\to0$$almost everywhere for $f$ is a dense subset of $L^1$; now exactly the proof of the differentiation theorem from the H-L maximal theorem works.
OK, here's the proof. Define $$\omega f(x) = \limsup_{r\to0}\frac{1}{|B_r|}\int_{B_r}|f(x-y)-f(x)|dy.$$We need to show that $\omega f=0$ almost everywhere. So it's enough to show that $$m\{\omega f>\lambda\}=0$$for every $f\in L^1$ and $\lambda>0$.
Note that $$\omega(f+g)\le\omega f+\omega g,$$also $$\omega f\le M'f.$$
Say $f\in L^1$ and $\epsilon>0$. Let $g\in C_c(\Bbb R^n)$ with $||f-g||_1<\epsilon$. Then $\omega g=0$, so $$\omega f\le\omega g+\omega(f-g) =\omega(f-g)\le M'(f-g),$$so $$m\{\omega f>\lambda\}\le\frac c\lambda||f-g||_1<\frac c\lambda\epsilon.$$