Let μ be a
Borel non-negative measure on
RN, finite on compact subsets and let f be a \mu-integrable function. Define the
maximal function f^* by setting for every x (using the convention \infty \times 0 = 0) :f^*(x) = \sup_{r > 0} \Bigl( \mu(B(x, r))^{-1} \int_{B(x, r)} |f(y)| \, d\mu(y) \Bigr). This maximal function is lower
semicontinuous, hence
measurable. The following maximal inequality is satisfied for every λ > 0: :\lambda \, \mu \bigl( \{ x : f^*(x) > \lambda \} \bigr) \le b_N \, \int |f| \, d\mu. ;Proof. The set
Eλ of the points
x such that f^*(x) > \lambda clearly admits a Besicovitch cover
Fλ by balls
B such that :\int \mathbf{1}_B \, |f| \ d\mu = \int_{B} |f(y)| \, d\mu(y) > \lambda \, \mu(B). For every bounded Borel subset
E´ of
Eλ, one can find a subcollection
G extracted from
Fλ that covers
E´ and such that
SG ≤
bN, hence :\begin{align} \lambda \, \mu(E') &\le \lambda \, \sum_{B \in \mathbf{G}} \mu(B)\\ &\le \sum_{B \in \mathbf{G}} \int \mathbf{1}_B \, |f| \, d\mu = \int S_{\mathbf {G}} \, |f| \, d\mu \le b_N \, \int |f| \, d\mu, \end{align} which implies the inequality above. When dealing with the
Lebesgue measure on
RN, it is more customary to use the easier (and older)
Vitali covering lemma in order to derive the previous maximal inequality (with a different constant). ==See also==