# hardy-littlewood maximal function

I am following Stein's real analysis book and he defines the maximal function of an integrable function $f$ by $\displaystyle f^*(x) = \text{sup}_{x\in B} \frac{1}{m(B)}\int_B|f(y)|dy$ where the sup is taken over all balls containing $x$.

Now the following is proven in the book: $m(\{x\in \mathbb{R}^d:f^*(x)>\alpha\}) \leq \frac{A}{\alpha}||f||_{L^1(\mathbb{R}^d)}$ where $A=3^d$ is a constant.

It can be easily shown that $E_\alpha = \{x\in \mathbb{R}^d:f^*(x)>\alpha\}$ open set.

The book proves it by considering any compact subset $K$ of $E_\alpha$ and showing that $m(K) \leq \frac{A}{\alpha}||f||_{L^1(\mathbb{R}^d)}$. It then concludes that because the inequality holds for any compact subset $K$ then it also holds for $E_\alpha$.

Why is it that we can generalize a property for compact subsets to the open set containing it? I am not getting this part. Is there a theorem for this?

• $m(A)=\sup\{m(K),\; K\subset A, \; K \mbox{compact} \}$. – Pocho la pantera Oct 3 '13 at 22:07

Every open subset $U$ of $\mathbb{R}^d$ is the union of an increasing sequence of compact subsets,
$$U = \bigcup_{n=0}^\infty K_n,$$
where the compact subsets can be chosen so that $K_n \subset \overset{\circ}{K}_{n+1}$. Then, here, we have by the monotone convergence theorem
$$m(E_\alpha) = \int_{\mathbb{R}^d} \chi_{E_\alpha} \, dm = \int_{\mathbb{R}^d} \lim_{n\to\infty}\chi_{K_n}\, dm = \lim_{n\to\infty} \int_{\mathbb{R}^d} \chi_{K_n}\, dm \leqslant \frac{A}{\alpha} \lVert f\rVert_1.$$