Suppose $f$ is a Lebesgue measurable function on $\mathbb R$, and $\forall x\in \mathbb R$, define the Hardy-Littlewood maximal function $f^*(x)=\sup_{t>0}$$1\over{2t}$$\int_{x-t}^{x+t} |f|$
My question is how to show $|f(x)|\le f^*(x)$ for almost every $x$?
For the case $f$ is integrable, this can be proved using Lebesgue's Differentiation Theorem. But I do not know how to extend this argument to general measurable functions.
Best Answer
Case 1: $f^*(x) = \infty$ $\forall x$. The inequality holds.
Case 2: $f^*(x) < \infty$ for some $ x$.
We have $\frac{1}{2t}\int_{x-t}^{x+t} |f|<\infty$ and thus $\int_{x-t}^{x+t} |f|<\infty$ $\forall t>0$. For any compact $K \subset \mathbb R$, there exists $t$ such that $K\subset (x-t,x+t)$, hence $$\int_K |f| \le\int_{x-t}^{x+t} |f|<\infty$$This shows $f$ is locally integrable. By Lebesgue's Differentiation Theorem, $$|f(x)|=\lim_{t\to 0}\frac{1}{2t}\int_{x-t}^{x+t}|f|\le\sup_{t>0}\frac{1}{2t}\int_{x-t}^{x+t}|f|=f^*(x)$$ for almost every $x\in \mathbb R$.