0$ (not necessarily an integer). Let $\X$ be a separable metric space endowed with a non-negative ``$n$-dimensional'' Borel measure $\mu$, i.e., a measure satisfying $$ \mu(B(x,r))\le r^n\qquad\text{ for all }x\in\X,\ r>0. $$ Let $L^{p}(\mu)$, $1\le p\le \infty$ be the usual Lebesgue spaces, and let $L^{1,\infty}(\mu)$ be defined by $$ L^{1,\infty}(\mu):=\bigl\{ f:\X\to \C\,:\,\|f\|\ci{L^{1,\infty}(\mu)}:= \sup_{t>0} t\cdot\mu\{x\in\X\,:\,|f(x)|>t\}<+\infty \bigr\}. $$ Note that the ``norm'' $\|f\|\ci{L^{1,\infty}(\mu)}$ is not actually a norm in the sense that it does not satisfy the triangle inequality. Still, we have $$ \|cf\|\ci{L^{1,\infty}(\mu)}=|c|\cdot\|f\|\ci{L^{1,\infty}(\mu)} \quad\text{ and }\quad \|f+g\|\ci{L^{1,\infty}(\mu)}\le 2\bigl(\|f\|\ci{L^{1,\infty}(\mu)}+\|g\|\ci{L^{1,\infty}(\mu)}\bigr) $$ for every $c\in\C$, $f,g\in L^{1,\infty}(\mu)$. (The latter is just the observation that in order to have the sum greater than $t$, one should have at least one term greater than $t/2$). Let $M(\X)$ be the space of all complex-valued Borel measures on $\X$. We will denote by $\|\nu\|$ the total variation of the measure $\nu\in M(\X)$. For $f\in L^p(\mu)$, we will denote by $\supp f$ the essential support of the function $f$, i.e., the smallest closed set $F\subset \X$ for which $f$ vanishes $\mu$-almost everywhere outside $F$. Also, for $\nu\in M(\X)$, we will denote by $\supp \nu$ the smallest closed set $F\subset\X$ for which $\nu$ vanishes on $\X\setminus F$ (i.e., $\nu(E)=0$ for every Borel set $E\subset \X\setminus F$). Since $\X$ is a separable metric space, such smallest closed set always exists. If $\{\B_j\}_{j=1}^\infty$ is some countable base of topology in $\X$, then for $\nu\in M(\X)$, the support $\supp\nu$ is just the complement of the union of those $\B_j$, on which the measure $\nu$ vanishes. For a function $f\in L^p(\mu)$, we obviously have $\supp f=\supp\nu$ where $d\nu=|f|^p\,d\mu$. Let $K:\X\times\X\to \C$ be a classical ``$n$-dimensional'' Calder\'{o}n--Zygmund kernel on $\X$, i.e., for some $A>0$, $\e\in(0,1]$, \begin{enumerate} \item $\qquad{\displaystyle |K(x,y)|\le\frac{A}{\dist(x,y)^n}},$ \item $\qquad{\displaystyle |K(x,y)-K(x',y)|, \ |K(y,x)-K(y,x')|\le A\cdot \frac{\dist(x,x')^\e}{\dist(x,y)^{n+\e}} }$ \newline whenever $x,x',y\in\X$ and $\dist(x,x')\le \frac{1}{2}\dist(x,y)$. \end{enumerate} \begin{Remark} We want to attract the attention of the reader to the fact that, though we call the number $n$ ``dimension'' all the time, it is the dimension of the {\it measure $\mu$} and of the {\it kernel $K(x,y)$}, but {\it by no means} is it the topological (or metric, or whatever else) dimension of the {\it space $\X$}. For instance, for the case of the Cauchy integral operator on the complex plane, $n=1$, not $2$! Actually, the topological dimension of the space $\X$ may be even infinite --- we do not care. \end{Remark} \begin{Df}A bounded linear operator $T$ in $L^2(\mu)$ is called a Calder\'{o}n-Zygmund (integral) operator with the Calder\'{o}n--Zygmund kernel $K$ if for every $f\in L^2(\mu)$, $$ Tf(x)=\int_\X K(x,y)f(y)d\mu(y) $$ for $\mu$-almost every $x\in \X\setminus \supp f$. \end{Df} Obviously, the adjoint operator $T^*$ is also bounded in $L^2(\mu)$ and has the kernel $K^*(x,y)=\overline{K(y,x)}$, which is a Calder\'{o}n--Zygmund kernel as well. Let $\nu\in M(\X)$ and $x\in \X\setminus \supp \nu$. For technical reasons it will be convenient to put {\it by definition} $$ T\nu(x):=\int_\X K(x,y)\,d\nu(y). $$ Note that we {\it do not} attempt here to define the values $T\nu(x)$ for $x\in \supp\nu$. The maximal operator $T^\sharp$ associated with the Calder\'{o}n--Zygmund operator $T$ is defined as follows. For every $r>0$, put $$ T_r f(x):= \int_{\X\setminus B(x,r)}K(x,y)f(y)\,d\mu(y) $$ for $f\in L^p(\mu)$, and $$ T_r\nu(x):= \int_{\X\setminus B(x,r)}K(x,y)\,d\nu(y) $$ for $\nu\in M(\X)$. Define $$ T^\sharp f(x):=\sup_{r>0}|T_r f(x)| $$ for $f\in L^p(\mu)$, and $$ T^\sharp\nu(x):=\sup_{r>0}|T_r \nu(x)| $$ for $\nu\in M(\X)$. Now we are able to formulate the main result of this paper: \begin{thm} For any (bounded in $L^2(\mu)$\,) Calder\'{o}n-Zygmund operator $T$, the following statements hold: \label{t01} \begin{enumerate} \item {\em\bf $L^p$-action:} For every $p\in(1,+\infty)$, the operator $T$ is bounded in $L^p(\mu)$ in the sense that for every $f\in L^p(\mu)\cap L^2(\mu)$, $$ \|Tf\|\ci{L^p(\mu)}\le C\|f\|\ci{L^p(\mu)} $$ with some constant $C>0$ not depending on $f$. \item {\em \bf Weak type 1-1 estimate:} The operator $T$ is bounded from $L^1(\mu)$ to $L^{1,\infty}(\mu)$ in the sense that for every $f\in L^1(\mu)\cap L^2(\mu)$, $$ \|Tf\|\ci{L^{1,\infty}(\mu)}\le C\|f\|\ci{L^1(\mu)} $$ with some constant $C>0$ not depending on $f$. \item {\em \bf Action of the maximal operator in $L^p(\mu)$:} For every $p\in(1,+\infty)$, the operator $T^\sharp$ is bounded in $L^p(\mu)$ in the sense that for every $f\in L^p(\mu)$, $$ \|T^\sharp f\|\ci{L^p(\mu)}\le C\|f\|\ci{L^p(\mu)} $$ with some constant $C>0$ not depending on $f$. \item {\em \bf Weak type 1-1 estimate for the maximal operator:} The operator $T^\sharp$ is bounded from $M(\X)$ to $L^{1,\infty}(\mu)$ in the sense that for every $\nu\in M(\X)$, $$ \|T^\sharp \nu \|\ci{L^{1,\infty}(\mu)}\le C\|\nu\| $$ with some constant $C>0$ not depending on $\nu$. \end{enumerate} \end{thm} \begin{Remark} \label{r0.1} The above theorem will remain true if we replace the a priori assumption $\|T\|\ci{L^2(\mu)\to L^2(\mu)}<+\infty$ in the definition of the Calderon-Zygmund operator $T$ by the assumption that $T$ is bounded in some other $L^{p\cci{0}}(\mu)$ with $1

0}\frac{1}{\mu(B(x,r))}\int_{B(x,r)}|f|\,d\mu.
$$
Note that if $x\in\supp\mu$, then $\mu(B(x,r))>0$ for every $r>0$
(otherwise one could drop a small open ball centered at $x$
from the support of $\mu$), so the definition makes sense $\mu$-almost
everywhere.
If the measure $\mu$ satisfies the doubling property, or if $\X$ has nice
geometric structure (similar to that of $\R^N$), the Hardy-Littlewood maximal
function operator is well-known to be bounded in all $L^p(\mu)$ with
$1< p\le +\infty$ and from $L^1(\mu)$ to $L^{1,\infty}(\mu)$. But,
unfortunately, for
arbitrary separable metric space $\X$ and measure $\mu$, the best one can say
is that $M$ is bounded in $L^\infty(\mu)$ (which is just
the obvious observation that the
integral does not exceed the essential supremum of the integrand times the
measure of the domain of integration). Fortunately, to save the game
is not too hard: all one needs is to replace the measure of the ball
$B(x,r)$ in the denominator by the measure of the {\it three times larger}
ball, i.e., to define
$$
\wt Mf(x):=
\sup_{r>0}
\frac{1}{\mu(B(x,3r))}\int_{B(x,r)}|f|\,d\mu.
$$
Note that always $\wt Mf(x)\le Mf(x)$ and,
if the measure $\mu$ satisfies the doubling condition,
$Mf(x)\le C\cdot \wt Mf(x)$ for some constant $C>0$ (the square of the
constant in the doubling condition).
\begin{lm}
\label{l3.1}
The modified maximal function operator
$\wt M$ is bounded in $L^p(\mu)$ for each $p\in(1,+\infty]$
and acts from $L^1(\mu)$ to $L^{1,\infty}(\mu)$.
\end{lm}
\begin{proof}%[Proof of Lemma \ref{l3.1}]
The boundedness in $L^\infty(\mu)$ is obvious.
To prove the weak type $1-1$ estimate, we will use the celebrated%
% Vitali covering theorem.
\begin{Vitali}%[Vitali covering theorem]
Fix some $R>0$. Let $E\subset \X$ be any set and let \linebreak
$\{B(x,r_x)\}\ci{x\in E}$
be a family of balls of radii $0 0}\frac{\nu(B(x,r))}{\mu(B(x,3r))}
$$
belongs to $L^{1,\infty}(\mu)$ and satisfies the estimate
$$
\|\wt M\nu\|\ci{L^{1,\infty}(\mu)}\le \nu(\X).
$$
\end{Remark}
\bigskip
We shall also need the modification of the maximal function $\wt Mf$,
in which the averaging of $|f|$ over balls is done with some power
$\beta\ne 1$. Namely, for each $\beta>0$, put
$$
\wt M_\beta f(x):=
\bigl[
\wt M(|f|^\beta)(x)
\bigr]^{\frac1\beta}=
\sup_{r>0}\Bigl[\frac{1}{\mu(B(x,3r))}\int_{B(x,r)}
|f|^\beta\,d\mu \Bigr]^{\frac1\beta}.
$$
Note that the greater $\beta$ is, the greater $\wt M_\beta f(x)$ is
(the Holder inequality). Note also that $\wt M_\beta$ is bounded in
$L^p(\mu)$ for every $p\in(\beta,+\infty]$ (to say that $\wt M_\beta$
is bounded in $L^p(\mu)$ is exactly the same as to say that $\wt M$
is bounded in $L^{{}^{\scriptstyle p}\!/\!{\scriptscriptstyle \beta}}(\mu)$).
We shall however need one less trivial (though no less standard) observation:
\begin{lm}
\label{l3.2}
For any $\beta\in(0,1)$, the maximal operator $\wt M_\beta$ is bounded
in $L^{1,\infty}(\mu)$, i.e.,
$$
\|\wt M_\beta f\|\ci{L^{1,\infty}(\mu)}\le C
\|f\|\ci{L^{1,\infty}(\mu)}
$$
with some constant $C>0$ not depending on $f$.
\end{lm}
\begin{proof}
Let $f\in L^{1,\infty}(\mu)$. Write $f=f_t+f^t$ where
$$
f_t(x)=
\left\{
\begin{array}{ll}
f(x),\quad &\text{if }|f(x)|\le t;
\\
0,& \text{if }|f(x)|> t;
\end{array}
\right.
\qquad \text{ and } \qquad
f^t(x)=
\left\{
\begin{array}{ll}
0,&\quad \text{if }|f(x)|\le t;
\\
f(x),&\quad \text{if }|f(x)|> t.
\end{array}
\right.
$$
Since
$\|\wt M_\beta f_t\|\ci{L^\infty(\mu)}\le \|f_t\|\ci{L^\infty(\mu)}\le t$
and
$[\wt M_\beta f]^\beta\le [\wt M_\beta f_t]^\beta+[\wt M_\beta f^t]^\beta$
(additivity of integral), we have
%%
\begin{multline*}
\mu\{x\in\X\,:\,\wt M_\beta f>2^{\frac 1\beta}t\}\le
\mu\{x\in\X\,:\,\wt M_\beta f^t> t\}
\\
=
\mu\{x\in\X\,:\,\wt M|f^t|^\beta> t^\beta\}\le
t^{-\beta}\int_\X |f^t|^\beta\,d\mu
\end{multline*}
%%
according to the weak type $1-1$ estimate for $\wt M$.
On the other hand, we have
%%
\begin{multline*}
\int_\X |f^t|^\beta\,d\mu= t^\beta \mu\{|f|>t\}+
\int_t^{+\infty}\beta s^{\beta-1}\mu\{|f|>s\}ds
\\
\le
t^\beta\frac1t\|f\|\ci{L^{1,\infty}(\mu)}+
\|f\|\ci{L^{1,\infty}(\mu)}\int_t^{+\infty}\beta s^{\beta-2}ds
=\frac{1}{1-\beta}\frac{1}{t}t^{\beta}
\|f\|\ci{L^{1,\infty}(\mu)}.
\end{multline*}
%%
So, finally we get
$$
\mu\{x\in\X\,:\,\wt M_\beta f>2^{\frac 1\beta}t\}\le
\frac{1}{1-\beta}\frac{1}{t}
\|f\|\ci{L^{1,\infty}(\mu)},
$$
i.e.,
$$
\|\wt M_\beta f\|\ci{L^{1,\infty}}\le
\frac{2^
%{\frac{1}{\beta}}}
{1/\beta}}
{1-\beta}
\,\|f\|\ci{L^{1,\infty}},
$$
proving the lemma.
\end{proof}
\subsection{Comparison lemma}
\label{s3.1}
\begin{lm}
\label{l3.3}
Let $U:(0,+\infty)\to [0,+\infty)$ be a continuous
non-negative decreasing function.
Let $\nu$ be any non-negative Borel measure on $\X$.
Then for every $x\in\X$ and $R>0$,
$$
\int_{\X\setminus B(x,R)}U(\dist(x,y))\,d\nu(y)\le
3^{n}\wt M \nu(x)
\Bigl[R^n U(R)+n\int_{R}^{+\infty}t^{n-1}U(t)\,dt\Bigr].
$$
\end{lm}
\begin{proof}
Consider first the case when $U$ is a ``step-function'', i.e.,
$U(t)=\chi\ci{(0,T]}$ for some $T>0$ (as usual, by $\chi\ci E$ we denote the
characteristic function of the set $E$).
%
If $T\le R$, the inequality is obvious because the left hand part is $0$.
For $T>R$, it is equivalent to the estimate
$$
\nu(B(x,T)\setminus B(x,R))\le 3^n\wt M\nu (x)\cdot T^n,
$$
which easily follows from
the definition of $\wt M\nu(x)$ and the inequality
$\mu(B(x,3T))\le 3^n\cdot T^n$.
Now to obtain the lemma,
it is enough to recall that every non-negative continuous decreasing
function $U(t)$ can be represented as the limit of an increasing sequence
of linear combinations of step-functions with non-negative coefficients.
\end{proof}
\subsection{H\"{o}rmander inequality}
\label{s3.2}
We shall need one more standard observation about Calder\'{o}n--Zygmund kernels.
\begin{lm}
\label{l3.4}
Let $\eta\in M(\X)$, $\eta(\X)=0$, and
$\supp\eta\subset B(x,\rho)$ for some $\rho>0$.
Then for every non-negative Borel measure $\nu$ on $\X$, we have
$$
\int_{\X\setminus B(x,2\rho)}|T\eta|d\nu\le A_1\,\wt M\nu(x)\,\|\eta\|
%\var(\eta)
$$
where $A_1>0$ depends only on the dimension $n$ and the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$.
In particular,
$$
\int_{\X\setminus B(x,2\rho)}|T\eta|\cdot |f|d\mu\le A_1\,\wt Mf(x)\,
\|\eta\|
%\var(\eta)
$$
for every Borel measurable function $f$ on $\X$, and
$$
\int_{\X\setminus B(x,2\rho)}|T\eta|d\mu\le A_1\,
\|\eta\|
%\var(\eta).
$$
\end{lm}
\begin{proof}
For any $y\in \X\setminus B(x,2\rho)$, we have
%%
\begin{multline*}
|T\eta(y)|=\Bigl|\int_{B(x,\rho)}K(y,x')d\eta(x')\Bigr|=
\Bigl|\int_{B(x,\rho)}[K(y,x')-K(y,x)]d\eta(x')\Bigr|\le
\\
\le
\|\eta\|\sup_{x'\in B(x,\rho)}|K(y,x')-K(y,x)|\le
\|\eta\|\frac{A\rho^\e}{\dist(x,y)^{n+\e}}.
\end{multline*}
%%
It remains only to notice that, according to the Comparison Lemma
(Lemma \ref{l3.3})
applied to $R=2\rho$ and $U(t)=\dfrac{\rho^\e}{t^{n+\e}}$,
%%
\begin{multline*}
\int_{\X\setminus
B(x,2\rho)}\frac{\rho^\e\,d\nu(y)}{\dist(x,y)^{1+\e}}
\\
\le
%
3^n \wt M\nu(x) \Bigl[(2\rho)^n \frac{\rho^\e}{(2\rho)^{n+\e}}+
n \int_{2\rho}^{+\infty} t^{n-1}\frac{\rho^\e}{t^{n+\e}}\Bigr]dt
=3^n 2^{-\e}(1+\tfrac n\e)\wt M\nu(x).
\end{multline*}
%%
\end{proof}
\section{The Guy David lemma.}
\label{s4}
The following lemma is implicitly contained in \cite{D}.
\begin{lm}
\label{l4.1}
For any Borel set $F\in \X$ of finite measure
and for any point $x\in\supp\mu$,
$$
T^\sharp\chi\ci F(x)\le 2\cdot 3^n\,\wt MT\chi\ci F(x)+A_2
$$
where $A_2>0$ depends only on the dimension $n$, the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$, and the norm
$\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
%
%(the condition
%$x\in\supp\mu$ is imposed only to have the maximal function
%$\wt MT\chi\ci F(x)$ well-defined at the point $x$).
%
\end{lm}
\begin{proof}
Let $
%%%% x\in \supp\mu,
r>0$. Consider the sequence of balls
$B(x,r_j)$ where $r_j:=3^j r$,
and the corresponding sequence of measures $\mu_j:=\mu(B(x,r_j))$
($j=0,1,\dots)$.
Note that
we cannot have $\mu_j>2\cdot 3^n\mu_{j-1}$ for every $j\ge 1$.
Indeed, otherwise we would have for every $j=1,2,\dots$,
$$
\mu(B(x,r))=\mu_0\le [2\cdot 3^n]^{-j}\mu_j\le
[2\cdot 3^n]^{-j} r_j^n=2^{-j}r^n.
$$
Since the right hand part tends to $0$ as $j\to+\infty$,
we could conclude from here that $\mu(B(x,r))=0$, which is impossible.
Therefore there exists
the smallest positive integer $k$ for which $\mu_k\le 2\cdot 3^n\mu_{k-1}$.
Put $R:=r_{k-1}=3^{k-1}r$. Observe that
%%
\begin{multline*}
|T_r\chi\ci F(x)-T\ci{3R}\chi\ci F(x)|\le
\int_{B(x,3R)\setminus B(x,r)}|K(x,y)|\,d\mu(y)
\\
=
\sum_{j=1}^k\int_{B(x, r_j)\setminus B(x,r_{j-1})}|K(x,y)|\,d\mu(y)
=:
\sum_{j=1}^k\mathcal I_j.
\end{multline*}
%%
Now recall that $|K(x,y)|\le\frac{A}{\dist(x,y)^n}$ and therefore
$\mathcal I_j\le A\frac{\mu_j}{r_{j-1}^n}$ for every $j=1,\dots,k$.
Note that $\mu_j\le [2\cdot 3^n]^{(j+1-k)}\mu_{k-1}$ and
$r_{j-1}=3^{j-k}r_{k-1}$ for $j=1,\dots, k$.
Hence
$$
\sum_{j=1}^k\mathcal I_j\le
A
\sum_{j=1}^k \frac{\mu_j}{r_{j-1}^n}\le
A\cdot 2\cdot 3^n\,\frac{\mu_{k-1}}{r_{k-1}^n}\,
\sum_{j=1}^k 2^{j-k}\le 4\cdot 3^n A
$$
(for $\mu_{k-1}=\mu(B(x,r_{k-1}))\le r_{k-1}^n$).
And that is basically the main part of the reasoning,
because now it is enough to pick
any standard proof based on the doubling condition to get the
desired
estimate for $T\ci{3R}\chi\ci F(x)$ (recall that $\mu(B(x,3R))\le
2\cdot 3^n\mu(B(x,R))\,!)$.
One of such standard ways is to compare
$T\ci{3R}\chi\ci F(x)$ to the average
$$
V\ci R(x):=\frac{1}{\mu(B(x,R))}\int_{B(x,R)}T\chi\ci Fd\mu
$$
(the quantity, which is clearly bounded by
$\frac{\mu(B(x,3R))}{\mu(B(x,R))}\,
\wt MT\chi\ci F(x)\le 3\cdot 2^n\wt MT\chi\ci F(x)$).
We have (here and below $\d_x$ is the unit point mass at the point
$x\in \X$):
%%
\begin{multline*}
T\ci{3R}\chi\ci F(x)-V\ci{R}(x)=
\\
\int_{F\setminus B(x,3R)} T^*[\d_x-\tfrac
{1}{\mu(B(x,R))}\chi\ci{B(x,R)}d\mu]d\mu-
\frac{1}{\mu(B(x,R))}\int_{\X}\chi\ci{B(x,R)}\cdot T\chi\ci{F\cap
B(x,3R)}d\mu.\kern-8pt
\end{multline*}
%%
The first term does not exceed $2A_1$ according to Lemma 3.4
(applied to the adjoint operator $T^*$ instead of $T$),
while the second can be estimated by
%%
\begin{multline*}
\frac{1}{\mu(B(x,R))}
\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}\cdot
\|\chi\ci{B(x,R)}\|\ci{L^2(\mu)}
\cdot \|\chi\ci{F\cap
B(x,3R)}\|\ci{L^2(\mu)}\le
\\
%\le
\frac{1}{\mu(B(x,R))}
\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}\sqrt{\mu(B(x,R))}\sqrt{\mu(B(x,3R))}
\le \sqrt{2\cdot 3^n}\,\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}.
\end{multline*}
%%
Combining all the above inequalities, we see that one can take
$A_2=4\cdot 3^n\,A+2A_1+\sqrt{2\cdot 3^n}\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
\end{proof}
The lemma we just outlined is crucial for our proof of the weak type $1-1$
estimate, but, unfortunately, not sufficient alone. In the next section
we will present a construction that, however simple and natural, seems
to have been completely overlooked
(at any rate we do not know about any other paper in which it is used).
\section{An alternative to the Calder\'{o}n--Zygmund decomposition}
\label{s5}
Let $\nu\in M(\X)$ be a finite
linear combination of unit point masses with positive coefficients,
i.e.,
$$
\nu=\sum_{i=1}^N \a_i\d_{x_i}.
$$
\begin{thm}
\label{t5.1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%$$
%\mu\{x\in\X\,:\,|T\nu(x)|>t\}\le \frac{A_4\|\nu\|}{t}
%$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$
\|T\nu\|\ci{L^{1,\infty}(\mu)}\le A_4\|\nu\|
$$
with some
$A_4>0$ depending only on the dimension $n$, the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$, and the norm
$\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
\end{thm}
Here there is no problem with the definition of $T\nu$: it is just the finite
sum \linebreak $\sum_{i=1}^N \a_i K(x,x_i)$, which makes sense everywhere except
finitely many points.
\begin{proof}
Without loss of generality, we may assume
that $\|\nu\|=\sum_i\a_i=1$ (this is just a matter of normalization).
Thus we have to prove that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%for every $t>0$,
%$$
%\mu\{|T\nu|>t\}\le \frac {A_4}t.
%$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\|T\nu\|\ci{L^{1,\infty}(\mu)}\le A_4$.
Fix some $t>0$ and
suppose first that $\mu(\X)>\frac1t$. Let
$B(x_1,\rho_1)$ be the smallest (closed) ball such that
$\mu(B(x_1,\rho_1))\ge\dfrac{\a_1}{t}$ (since the function
$\rho\to \mu(B(x_1,\rho))$ is increasing and continuous from the right,
tends to $0$ as $\rho\to 0$, and is greater than $\dfrac1t\ge
\dfrac{\a_1}{t}$ for
sufficiently large $\rho>0$,
such $\rho_1$ exists and is strictly positive).
Note that for the corresponding {\it open} ball $B'(x_1,\rho_1):=
\{y\in\X\,:\,\dist(x_1,y)<\rho_1\}$, we
have $\mu(B'(x_1,\rho_1))=\lim_{\rho\to \rho_1-0}\mu(B(x_1,\rho))\le
\dfrac{\a_1}{t}$.
Since the measure $\mu$ is $\sigma$-finite and
non-atomic, one can choose a Borel set
$E_1$ satisfying
$$
B'(x_1,\rho_1)\subset E_1\subset B(x_1,\rho_1)\qquad\text{ and }\qquad
\mu(E_1)=\frac{\a_1}{t}.
$$
Let
$B(x_2,\rho_2)$ be the smallest ball such that $\mu(B(x_2,\rho_2)\setminus
E_1)\ge\dfrac{\a_2}{t}$ (since $\mu(\X)>\frac1t$, the measure of the remaining
part $\X\setminus E_1$ is still greater than
$\dfrac{1-\a_1}{t}\ge\dfrac{\a_2}{t}$).
Again for the corresponding open ball $B'(x_2,\rho_2)$,
we have $\mu(B'(x_2,\rho_2)\setminus
E_1)\le\dfrac{\a_2}{t}$, and therefore there exists a Borel set $E_2$
satisfying
$$
B'(x_2,\rho_2)\setminus E_1
\subset E_2
\subset B(x_1,\rho_1)\setminus E_1
\qquad\text{ and }\qquad
\mu(E_2)=\frac{\a_2}{t}.
$$
In general, for $i=3,4,\dots,N$, let
$B(x_i,\rho_i)$ be the smallest ball such that
$$
\mu\Bigl(B(x_i,\rho_i)\setminus
\bigcup_{\ell=1}^{i-1}E_\ell
\Bigr)\ge\frac{\a_i}{t},
$$
and let $E_i$ be a Borel set satisfying
$$
B'(x_i,\rho_i)\setminus
\bigcup_{\ell=1}^{i-1}E_\ell
\subset
E_i
\subset
B(x_i,\rho_i)\setminus
\bigcup_{\ell=1}^{i-1}E_\ell
\quad\text{ and }\quad
\mu(E_i)=\frac{\a_i}{t}.
$$
Put $E:=\bigcup_i E_i$.
Clearly
$$
\bigcup_i B'(x_i,\rho_i)
\subset
E
\subset
\bigcup_i B(x_i,\rho_i)
\qquad\text{ and }\qquad
\mu(E)=\frac1t.
$$
Now let us compare $T\nu$ to $t\,\sum_i\chi\ci{\X\setminus
B(x_i,2\rho_i)}
\cdot T\chi\ci{E_i}=:t\sigma$ outside $E$.
We have
$$
T\nu-t\s=\sum_i\f_i
$$
where
$$
\f_i=\a_i T\d_{x_i}- t\,\chi\ci{\X\setminus B(x_i,2\rho_i)}\cdot
T\chi\ci{E_i}.
$$
Note now that
$$
\int_{\X\setminus E}|\f_i|d\mu\le \int_{\X\setminus
B(x_i,2\rho_i)}\bigl|T[\a_i
\d_{x_i}-t\chi\ci{E_i}d\mu]\bigr|d\mu + \int_{B(x_i,2\rho_i)\setminus
B'(x_i,\rho_i)}
\a_i|T\d_{x_i}|d\mu.
$$
But, according to Lemma \ref{l3.4}, the first integral does not exceed
$$
A_1
%\var(
\|
\a_i
\d_{x_i}-t\,\chi\ci{E_i}d\mu
\|
%)
=2A_1 \a_i,
$$
while $|T\d_{x_i}|\le A \rho_i^{-n}$ outside $B'(x_i,\rho_i)$ and therefore
the second integral is not greater than
$\a_i A \rho_i^{-n}\mu(B(x_i,2\rho_i))\le 2^n A \a_i$.
Finally we conclude that
$$
\int_{\X\setminus E}|T\nu-t\s|d\mu\le
(2A_1+2^nA)\sum_i\a_i=2A_1+2^nA,
$$
and thereby $|T\nu-t\s|\le (2A_1+2^nA)t$
everywhere on $\X\setminus E$, except, maybe, a set of
measure $\frac{1}{t}$. To accomplish the proof of the theorem,
we will show that for sufficiently large $A_3$,
$$
\mu\{|\s|>A_3\}\le \frac2t.
$$
Then, combining all the above estimates, we shall get
$$
\mu\bigl\{x\in\X\,:\,|T\nu(x)|>(A_3+2A_1+2^nA)t\bigr\}\le\frac4t.
$$
Since the same inequality is obviously true in the
case when $\mu(\X)\le\frac1t$,
one may take $A_4=4(A_3+2A_1+2^nA)$.
We will apply the standard Stein-Weiss duality trick.
Assume that the inverse inequality
$\mu\{|\s|>A_3\}> \frac2t$
holds. Then either
$
\mu\{\s>A_3\}> \frac1t,
$
or
$
\mu\{\s<-A_3\}> \frac1t.
$
Assume for definiteness that the first case takes place
and choose some set $F\subset \X$ of measure exactly
$\frac1t$
such that $\s>A_3$ everywhere on $F$.
Then, clearly,
$$
\int_\X\s\chi\ci F d\mu>\frac{A_3}{t}.
$$
On the other hand, this integral can be computed as
$$
\sum_i \int_{\X} [T\chi\ci{E_i}]\cdot\chi\ci{F\setminus
B(x_i,2\rho_i)}\,d\mu
=\sum_i \int_{\X} \chi\ci{E_i}\cdot [T^*\chi\ci{F\setminus
B(x_i,2\rho_i)}]\,d\mu.
$$
Note that for every point $x\in E_i\subset B(x_i,\rho_i)$,
%%
\begin{multline*}
|T^*\chi\ci{F\setminus B(x_i,2\rho_i)}(x)-T^*\chi\ci{F\setminus
B(x,\rho_i)}(x)|
\le\int_{B(x_i,2\rho_i)\setminus B(x,\rho_i)}|K(y,x)|\,d\mu(y)
\le
\\
\le A\rho_i^{-n}\mu(B(x_i,2\rho_i))\le 2^nA
\end{multline*}
%%
and thereby for every $x\in E_i\cap\supp\mu$,
$$
|T^*\chi\ci{F\setminus B(x_i,2\rho_i)}(x)|\le (T^*)^\sharp \chi\ci
F(x)+2^nA\le
2\cdot 3^n\,\wt MT^*\chi\ci F(x)+A_2+2^nA
$$
according to the Guy David lemma (Lemma \ref{l4.1}).
Hence
$$
\int_{\X} \s\chi\ci F d\mu\le (A_2+2^nA)\mu(E)+2\cdot 3^n\int_{\X}\chi\ci
E\cdot
\wt MT^*\chi\ci F
d\mu.
$$
But the first term equals $\dfrac{A_2+2^nA}{t}$ while the second one does
not exceed
$$
2\cdot 3^n\,
\|\chi\ci E\|\ci{L^2(\mu)}
\|\wt MT^*\chi\ci F\|\ci{L^2(\mu)}\le
\frac{2\cdot 3^n}{t}\|\wt M\|\ci{L^2(\mu)\to
L^2(\mu)}
\|T^*\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}.
$$
Recalling that
$
\|T^*\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}
=\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}
$, we see that one can take
$$
A_3=A_2+2^nA+2\cdot 3^n\,\|\wt M\|\ci{L^2(\mu)\to
L^2(\mu)}\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}
$$
to get a contradiction. Since the norm $\|\wt M\|\ci{L^2(\mu)\to
L^2(\mu)}$ is bounded by some absolute constant (the constant in the
Marcinkiewicz interpolation theorem), we are done.
\end{proof}
\section{From finite linear
combinations of point masses to $L^1(\mu)$-functions}
\label{s6}
Note first of all that Theorem 5.1 remains valid (with twice larger constant)
for finite linear combinations of point masses with {\it arbitrary} real
coefficients. Indeed, every such measure $\nu$ can be represented as
$\nu_+-\nu_-$ where $\nu_\pm$ are
finite linear combinations of point masses with {\it positive}
coefficients and $\|\nu\|=\|\nu_+\|+\|\nu_-\|$. Hence
$$
\|T\nu\|\ci{L^{1,\infty}(\mu)}\le 2\bigl(
\|T\nu_+\|\ci{L^{1,\infty}(\mu)}+
\|T\nu_-\|\ci{L^{1,\infty}(\mu)}
\bigr)\le
2A_4 (\|\nu_+\|+\|\nu_-\|)= 2A_4\|\nu\|.
$$
Now we are ready to prove
\begin{thm}
\label{t6.1}
Let $f\in L^1(\mu)\cap L^2(\mu)$. Then
$$
\|Tf\|\ci{L^{1,\infty}(\mu)}\le A_5\|f\|\ci{L^1(\mu)}
$$
with some
$A_5>0$ depending only on the dimension $n$, the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$, and the norm
$\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
\end{thm}
\begin{proof}
Let $C_0(\X)$ be the space of bounded continuous functions on $\X$
with bounded support (a function is said to have bounded support
if it vanishes outside some (large) ball of finite radius).
Clearly, $C_0(\X)\subset L^1(\mu)\cap L^2(\mu)$, and it
is a standard fact from
measure theory that $C_0(\X)$ is dense in $L^1(\mu)\cap L^2(\mu)$
with respect to the norm
\linebreak
$\|\cdot\|\ci{L^1(\mu)}+\|\cdot\|\ci{L^2(\mu)}$.
Therefore it is enough to prove the desired inequality for $f\in C_0(\X)$.
Fix $t>0$ and put $G:=\{x\in \X\,:\,|f(x)|>t\}$,
$f^t:=f\cdot\chi\ci G$ and $f_t=f\cdot\chi\ci{\X\setminus G}$.
We have
$
T f= T f^t+T f_t.
$
Now observe, as usual, that
$$
\int_\X |f_t|^2\,d\mu\le
t\int_\X |f_t|\,d\mu
\le t\|f\|\ci{L^1(\mu)}.
$$
Therefore $\int_\X|Tf_t|^2\,d\mu\le \|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}^2
t\|f\|\ci{L^1(\mu)}$, and
$$
\mu\bigl\{x\in\X\,:\, |Tf_t(x)|>t\cdot\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}\bigr\}\le
\frac{\|f\|\cci{L^1(\mu)}}{t}.
$$
Note now that $G$ is an {\it open} set (this is the only place where we
use the continuity of $f$) and that $\mu(G)\le \frac1t \|f\|\ci{L^1(\mu)}$.
Recall that every open set $G$ in a separable metric space allows a ``Whitney
decomposition'', i.e., it can be represented as a union of countably many
pairwise disjoint Borel sets $G_i$ ($i=1,2,\dots$) satisfying
$$
\diam G_i\le \tfrac12 \dist(G_i,\X\setminus G).
$$
Put $f_i:=f\cdot\chi\ci{G_i}$. Then $f^t=\sum_{i=1}^\infty f_i$ where the
series converges at least in $L^2(\mu)$.
Let $f^{(N)}$ be the $N$-th partial sum of this series.
Define
$$
\a_i:=\int_{\X}f_i\,d\mu=\int_{G_i}f\,d\mu.
$$
Obviously, $\sum_{i=1}^\infty|\a_i|\le \|f\|\ci{L^1(\mu)}$.
Choose one point $x_i$ in every set $G_i$ and
put $\nu\ci N=\sum_{i=1}^N \a_i\d_{x_i}$.
Consider the difference $Tf^{(N)}-T\nu\ci N$ outside $G$.
We have
$$
\int_{\X\setminus G}\bigl|Tf^{(N)}-T\nu\ci N\bigr|\,d\mu{\le}
\sum_{i=1}^N \int_{\X\setminus
G}\bigl|T[f_id\mu-\a_i\d_{x_i}]\bigl|\,d\mu{\le}
2 A_1 \sum_{i=1}^{N}|\a_i|\le 2 A_1 \|f\|\ci{L^1(\mu)}
$$
according to Lemma \ref{l3.4}.
Thus $|Tf^{(N)}-T\nu\ci N|\le 2 A_1 t$
everywhere outside $G$ save, maybe, some exceptional set of
measure at most $\frac1t\|f\|\ci{L^1(\mu)}$.
As we have seen above,
$$
\mu\{x\in\X\,:\,|T\nu\ci N(x)|> 2 A_4 t\}\le \frac1t\|\nu\ci N\|
\le\frac1t\|f\|\ci{L^1(\mu)}.
$$
Hence
$$
\mu\bigl\{x\in\X\setminus G\,:\,|Tf^{(N)}(x)|> 2(A_1+A_4) t
\bigr\}\le
\frac2t\|f\|\ci{L^1(\mu)},
$$
and
$$
\mu\bigl\{x\in\X\,:\,|Tf^{(N)}(x)|> 2(A_1+A_4) t
\bigr\}\le
\frac3t\|f\|\ci{L^1(\mu)}.
$$
Since $f^{(N)}\to f^t$ in $L^2(\mu)$ as $N\to +\infty$, we have
$Tf^{(N)}\to Tf^t$ in $L^2(\mu)$ as $N\to +\infty$, which is more than enough
to pass to the limit and to conclude that
$$
\mu\bigl\{x\in\X\,:\,|Tf^{t}(x)|> 2(A_1+A_4) t
\bigr\}\le
\frac3t\|f\|\ci{L^1(\mu)}.
$$
Thus, we can take $A_5=4\bigl[\,\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}+2(A_1+A_4)
\,\bigr]$.
As usual, by the Marcinkiewicz interpolation theorem, we obtain
that the operator $T$ is bounded in $L^p(\mu)$
for every $1 1$ and $x\in \supp\mu$,
$$
T^\sharp f(x)\le 4\cdot 9^n \wt M T f(x) + B(\beta)\,\wt M\ci\beta f(x)
$$
where the constant $B(\beta)>0$ depends on the parameter $\beta>1$, the
dimension $n$, the constants
$\e$ and $A$ in the definition of the Calder\'{o}n--Zygmund kernel $K$,
and the norm $\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$ only.
\end{thm}
\begin{proof}
It is just a minor modification of the proof of the Guy David lemma.
Let again $r>0$. Put $r_j=3^j r$ and $\mu_j=\mu(B(x,r_j))$ as before,
but let now $k$ be the smallest positive integer
for which $\mu_{k+1}\le 4\cdot 9^n\mu_{k-1}$ (i.e., we look now {\it two
steps forward\/} when checking for the doubling).
Note that such an integer $k$ exists, because otherwise
for every {\it even} $j$,
$$
\mu(B(x,r))\le 2^{-j}3^{-nj}\mu(B(x,r_j))\le
2^{-j}r^n,
$$
and thereby $\mu(B(x,r))=0$, which is impossible.
Put $R=r_{k-1}=3^{k-1}r$ exactly as before.
We have
%%
\begin{multline*}
|T_r f(x)-T\ci{3R} f(x)|\le
\int_{B(x,3R)\setminus B(x,r)}|K(x,y)|\cdot|f(y)|\,d\mu(y)
\\
=\sum_{j=1}^k
\int_{B(x,r_j)\setminus B(x,r_{j-1})}|K(x,y)|\cdot|f(y)|\,d\mu(y)
=:\sum_{j=1}^k \mathcal I_j.
\end{multline*}
%%
Note that
$$
\mathcal I_j\le Ar_{j-1}^{-n}\int_{B(x,r_j)}|f|\,d\mu
\le
A r_{j-1}^{-n}\mu_{j+1}
\wt Mf(x).
$$
Observe now that $r_{j-1}=3^{j-1-k}r_{k}$ and
$\mu_{j+1}\le [4\cdot 9^n]^{\frac{j+2-k}{2}}\mu_{k}$ for $1\le j\le k$
(it is enough to check this inequality for $j=k, k-1$ and $k-2$).
Hence
$$
\sum_{j=1}^k \mathcal I_j
\le 4\cdot 27^n\, A\, \wt M f(x)\,
\frac{\mu_{k}}{r_k^n}\,
\sum_{j=1}^k 2^{j-k}\le
8\cdot 27^n\,
A\,\wt Mf(x)
\le 8\cdot 27^n\,
A\,\wt M\ci\beta f(x).
$$
So, again, we need only to estimate $T\ci{3R} f(x)$.
As before, consider the average
$$
V\ci R(x):= \frac{1}{\mu(B(x,R))}\int_{B(x,R)}T f\, d\mu,
$$
which is bounded by $\frac{\mu(B(x,3R))}{\mu(B(x,R))}\wt MTf(x)
\le 4\cdot 9^n\wt MTf(x)$ according to our choice of $k$,
and write
%%
\begin{multline*}
T\ci{3R} f(x)-V\ci{R}(x)=
\\
\int_{\!\X\setminus B(x,3R)}\! T^*[\d_x-\tfrac
{1}{\mu(B(x,R))}\chi\ci{B(x,R)}d\mu]\, f\,d\mu-
\frac{1}{\mu(B(x,R))}\!\int_\X\chi\ci{B(x,R)}\!\cdot
T[ f\chi\ci{
B(x,3R)}]d\mu.\kern-12pt
\end{multline*}
%%
Using Lemma \ref{l3.4}, we can now estimate
the absolute value of the minuend by \linebreak
$2A_1 \wt Mf(x)\le 2A_1 \wt M\ci\beta f(x)$.
As to the subtrahend, at this stage we know that $T$ is bounded
in $L^\beta(\mu)$,
and therefore the absolute value of the subtrahend does not exceed
$$
\frac{1}{\mu(B(x,R))}\|T\|\ci{L^\beta(\mu)\to L^\beta(\mu)}
\|\chi\ci{B(x,R)}\|\ci{L^{\beta'}(\mu)}\cdot \| f\chi\ci{
B(x,3R)}\|\ci{L^\beta(\mu)}
$$
where ${\beta'}:= \frac{\beta}{\beta-1}$ is the conjugate exponent to
$\beta$.
Clearly
$$
\|\chi\ci{B(x,R)}\|\ci{L^{\beta'}(\mu)}=\bigl\{\mu(B(x,R))
\bigr\}^{1/{\beta'}}.
$$
The point is that now, according to our choice of $k$, we
have
$\mu(B(x,9R))\le 4\cdot 9^n\mu(B(x,R))$,
and therefore
$$
\| f\chi\ci{
B(x,3R)}\|\ci{L^\beta(\mu)}\le
\wt M\ci{\beta} f(x) \bigl\{\mu(B(x,9R))
\bigr\}^{1/\beta} \le
\wt M\ci{\beta} f(x) \bigl\{4\cdot 9^n\mu(B(x,R))
\bigr\}^{1/\beta}.
$$
This allows us to conclude finally that the subtrahend is bounded by
$$
[4\cdot 9^n]^{1/\beta}\|T\|\ci{L^\beta(\mu)\to L^\beta(\mu)}\wt M\ci{\beta}
f(x),
$$
proving the theorem with $B(\beta)=8\cdot 27^n A+2A_1+
[4\cdot 9^n]^{1/\beta}\|T\|\ci{L^\beta(\mu)\to L^\beta(\mu)}$.
\end{proof}
\section{Weak type 1-1 estimate for the maximal operator $T^\sharp$}
\label{s8}
Now, to complete the ``classical $L^p$-theory'', it remains to prove
that the maximal operator $T^\sharp$ is bounded from $M(\X)$ to
$L^{1,\infty}(\mu)$, i.e., that for every signed measure $\nu\in
M(\X)$,
$$
\|T^\sharp\nu\|\ci{L^{1,\infty}(\mu)}\le C\|\nu\|
$$
with some
constant $C>0$, not depending on $\nu$.
We will start again with ``elementary'' measures $\nu\in M(\X)$,
i.e., with the measures of the kind $\nu=\sum_{i=1}^N \a_i\d_{x_i}$
where $x_i\in \X$, $\a_i>0$ ($i=1,\dots,N$).
\begin{thm}
\label{t8.1}
Let $\beta\in(0,1)$.
For every elementary measure $\nu\in M(\X)$ and for every $x\in\supp\mu$,
$$
[T^\sharp\nu(x)]^\beta \le 4\cdot 9^n[\wt M_\beta T\nu(x)]^{\beta}+
B(\beta)\,[\wt M\nu(x)]^{\beta}
$$
with some
$B(\beta)>0$ depending only on the parameter $\beta<1$,
dimension $n$, the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$,
and the norm
$\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
\end{thm}
Note again that $T\nu$ is well-defined everywhere except finitely many
points, so the first term on the right does make sense.
\begin{cor}
\label{c8.2}
For every elementary measure $\nu\in M(\X)$,
$$
\|T^\sharp\nu\|\ci{L^{1,\infty}(\mu)}\le
A_6\|\nu\|
$$
with
$A_6>0$ depending only on the
dimension $n$, the constants $A$ and $\e$
in the definition of the Calder\'{o}n--Zygmund kernel $K$, and the norm
$\|T\|\ci{\!L^2(\mu){\to} L^2(\mu)\!}$.
\end{cor}
\begin{proof}[Proof of Theorem 8.1]
%
Take some $r>0$. Put $r_j=3^j r$ and $\mu_j=\mu(B(x,r_j))$ as usual,
and let again (like in Section 7) $k$ be the smallest positive integer
for which $\mu_{k+1}\le 4\cdot 9^n\mu_{k-1}$.
Put $R=r_{k-1}=3^{k-1}r$.
The same reasoning as in the proof of Theorem \ref{t7.1} yields
$$
|T_r\nu(x)-T\ci{3R}\nu(x)|
\le 8\cdot 27^n\,
A\,\wt M\nu(x)
$$
Now represent the measure $\nu$ as $\nu_1+\nu_2$, where
$$
\nu_1:=\sum_{i:\,x_i\in B(x,3R)}\a_i\d_{x_i}
\quad\text{ and }\quad
\nu_2:=\sum_{i:\,x_i\notin B(x,3R)}\a_i\d_{x_i}.
$$
For any $x'\in B(x,R)$, we have
%%
\begin{multline*}
|T\ci{3R}\nu(x)-T\nu_2(x')|= |T\nu_2(x)-T\nu_2(x')|
%=\Bigl|\int_\X T\nu_2\,d[\d_x-\d_{x'}] \Bigr|
=\Bigl|\int_\X T^*[\d_x-\d_{x'}]\,d\nu_2 \Bigr|
\\
\le
\int_\X |T^*[\d_x-\d_{x'}]|\,d\nu_2
=
\int_{\X\setminus B(x,3R)} |T^*[\d_x-\d_{x'}]|\,d\nu\le
2A_1\wt M\nu(x)
\end{multline*}
%%
(see Lemma \ref{l3.4}).
Hence
$$
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\ci{3R}\nu(x)-T\nu_2(x')|^\beta\,d\mu(x')
\le
\bigl[2A_1\wt M\nu(x)\bigr]^\beta.
$$
On the other hand,
%%
\begin{multline*}
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\nu_2(x')-T\nu(x')|^\beta\,d\mu(x')
\\
=
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\nu_1(x')|^\beta\,d\mu(x')
\\
=
\frac{1}{\mu(B(x,R))}
\int_0^{+\infty}\beta s^{\beta-1}\mu\{x'\in B(x,R)\,:\,|T\nu_1(x')|>s\}\,ds.
\end{multline*}
%%
Note now that for every $s>0$,
%%
\begin{multline*}
\mu\{x'\in B(x,R)\,:\,|T\nu_1(x')|>s\}
\le
\min\Bigl(\mu(B(x,R)),\frac{A_4\,\|\nu_1\|}{s}\Bigr)
\\
\le
\mu(B(x,R))
\min\bigl(1,\tfrac{\mu(B(x,9R))}{\mu(B(x,R))}\,
\tfrac {A_4\,\wt M\nu(x)}{s}\bigr)
\le
\mu(B(x,R))
\min\bigl(1,\tfrac{4\cdot 9^n\,A_4\,\wt M\nu(x)}{s}\bigr).
\end{multline*}
%%
Therefore
%%
\begin{multline*}
\frac{1}{\mu(B(x,R))}
\int_0^{+\infty}\beta s^{\beta-1}\mu\{x'\in B(x,R)\,:\,|T\nu_1(x')|>s\}\,ds
\\
\le
\int_0^{+\infty}\beta s^{\beta-1}
\min\Bigl(1,\frac{4\cdot 9^n\,A_4\,\wt M\nu(x)}{s}\Bigr)\,ds
\\
=
\bigl[4\cdot 9^n\,A_4\,\wt M\nu(x)\bigr]^{\beta}
\int_0^{+\infty}\beta s^{\beta-1}\min(1,\tfrac1s)\,ds
=\tfrac1{1-\beta}
\bigl[4\cdot 9^n\,A_4\,\wt M\nu(x)\bigr]^{\beta}.
\end{multline*}
%%
Using the elementary inequality $|a+b|^\beta\le |a|^{\beta}+|b|^{\beta}$
($a,b\in\R;\,\beta\in(0,1)$\,), we obtain
$$
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\ci{3R}\nu(x)-T\nu(x')|^\beta\,d\mu(x')
{\le}
\bigl([2A_1]^\beta
+\tfrac1{1-\beta}
[4\cdot 9^n\,A_4]^\beta\bigr)\,[\wt M\nu(x)]^{\beta}.
$$
Using it twice more, we finally get
%%
\begin{multline*}
|T\ci{r}\nu(x)|^{\beta}
\le
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\nu|^\beta\,d\mu
\\
+
\bigl([8\cdot 27^n A]^\beta+[2A_1]^\beta
+\tfrac1{1-\beta}
[4\cdot 9^n\,A_4]^\beta\bigr)\,[\wt M\nu(x)]^{\beta}.\!\!
\end{multline*}
%%
To prove the theorem, it remains only to note that
$$
\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|T\nu|^\beta\,d\mu
\le
\frac{\mu(B(x,3R))}{\mu(B(x,R))}[\wt M\ci\beta T\nu]^\beta
\le
4\cdot 9^n[\wt M\ci\beta T\nu]^\beta.
$$
\end{proof}
To prove Corollary \ref{c8.2}, it is enough to recall that $\wt M_\beta$
is bounded in $L^{1,\infty}(\mu)$ for any $\beta\in(0,1)$,
and that
$\|T\nu\|\ci{L^{1,\infty}(\mu)}\le A_4\|\nu\|$ and
$\|\wt M\nu\|\ci{L^{1,\infty}(\mu)}\le\|\nu\|$.
%\bye
\section{The weak type $1-1$ estimate for arbitrary measures $\nu\in M(X)$}
\label{s9}
\begin{thm}
\label{t9.1}
For any finite non-negative measure $\nu\in M(\X)$, one has
$$
\|T^\sharp\nu\|\ci{L^{1,\infty}(\mu)}\le A_6\|\nu\|,
$$
where $A_6$ is the same constant as in the corollary 8.2.
\end{thm}
Theorem 9.1 essentially says that elementary measures are ``weakly
dense'' in the set of all finite non-negative measures. Though by no
means surprising, it is not
completely obvious
(or, maybe, it is, but we just do not see how),
because we work with a space that is not locally
compact and with a kernel that is not everywhere continuous.
That is why we decided to include a formal proof.
\begin{cor}
\label{c9.2}
For every $\nu\in M(\X)$,
$$
\|T^\sharp \nu \|\ci{L^{1,\infty}(\mu)}\le 2A_6\|\nu\|.
$$
\end{cor}
\begin{proof}[Proof of Theorem 9.1]
%Without loss of generality, we may assume that $\|\nu\|=\nu(\X)=1$.
Fix $t>0$. Our aim is to show that
$$
\mu\{x\in\X\,:\,T^\sharp\nu(x)>t\}\le \frac{A_6\|\nu\|}{t}.
$$
Take $R>0$ and consider the truncated maximal operator
$$
T\ci R^\sharp\nu(x):=\sup_{r>R}|T_r\nu(x)|.
$$
Since $T\ci R^\sharp\nu\nearrow T^\sharp\nu$ pointwise on $\X$ as $R\to 0$,
it is enough to check that
$$
\mu\{x\in\X\,:\,T\ci R^\sharp\nu(x)>t\}\le \frac{A_6\|\nu\|}{t}
$$
for every $R>0$.
For every $N\in\mathbb{N}$, consider the random elementary measure
$$
\nu\ci N:=\frac{\|\nu\|}N\sum_{i=1}^N \d_{x_i}
$$
where the random points $x_i\in \X$ are independent and
$\mathcal P\{x_i\in E\}=\frac{\nu(E)}{\|\nu\|}$
for every Borel set $E\subset \X$.
(Here and below we denote by $\mathcal P\{X\}$ the probability of the event $X$,
by $\mathcal E\xi$ the mathematical expectation of a random variable $\xi$, and
by $\mathcal D\xi:=\mathcal E|\xi-\mathcal E\xi|^2=\mathcal E|\xi|^2-|\mathcal E\xi|^2$
the dispersion of the random variable $\xi$).
Note that for every fixed $x\in \X$ and $r>R$,
$$
\mathcal E\, T_r\d_{x_i}(x)= T_r(\mathcal E\,\d_{x_i})(x)= \frac{1}{\|\nu\|}T_r\nu(x)
$$
and
$$
\mathcal D\, T_r\d_{x_i}(x)\le
\mathcal E\, |T_r\d_{x_i}(x)|^2\le \frac{A^2}{r^{2n}}\le \frac{A^2}{R^{2n}}.
$$
Hence
$$
\mathcal E\, T_r\nu\ci N(x)= T_r\nu(x)
\qquad \text{ and }\qquad
\mathcal D\, T_r\nu\ci N(x)\le
\frac1N\,\frac{A^2\|\nu\|^2}{R^{2n}}.
$$
Fix a very small number $\gamma>0$ and note that for every point $x\in\X $
satisfying $|T_r\nu(x)|>t$, we have
%%
\begin{multline*}
\mathcal P\{|T_r\nu\ci N(x)|\le (1-\gamma)t\}
\le
\mathcal P\{|T_r\nu\ci N(x)-T_r\nu(x)|> \gamma t\}
\\
\le
\frac{\mathcal D\,T_r\nu\cci N(x)}{\gamma^2 t^2}\le
\frac1N\,\frac{A^2\|\nu\|^2}{R^{2n}\gamma^2 t^2}\le\gamma,
\end{multline*}
%%
provided that $N\in\mathbb{N}$ is large enough.
From here we incur that for every point $x\in\X $
satisfying $|T^\sharp\ci R\nu(x)|>t$, we have
$$
\mathcal P\{|T^\sharp\nu\ci N(x)|\le (1-\gamma)t\}
\le\gamma.
$$
Let now $E$ be any Borel set {\it of finite measure} such that
$T^\sharp\ci R\nu(x)>t$ for every $x\in E$.
We have
$$
\mathcal E\,\mu\{x\in E\,:\,|T^\sharp\nu\ci N(x)|\le (1-\gamma)t\}=
\int_E
\mathcal P\{|T^\sharp\nu\ci N(x)|\le (1-\gamma)t\}
\,d\mu(x)
\le\gamma\mu(E).
$$
Thus there exists at least one choice of points $x_i$ ($i=1,\dots,N$)
for which $\mu\{x\in E\,:\,
|T^\sharp\nu\ci N(x)|\le (1-\gamma)t\}\le \gamma\mu(E)$ and therefore
$$
\mu\{x\in E\,:\,
|T^\sharp\nu\ci N(x)|> (1-\gamma)t\}\ge (1-\gamma)\mu(E).
$$
According to the weak type $1-1$ estimate for elementary measures, this
implies
$$
\mu(E)\le\frac{A_6\|\nu\ci N\|}{(1-\gamma)^2 t}=
\frac{A_6\|\nu\|}{(1-\gamma)^2 t}.
$$
Since $\gamma>0$ was arbitrary, we get $\mu(E)\le\frac{A_6\|\nu\|}{t}$.
At last, since $\mu$ is $\sigma$-finite and $E$ was an
arbitrary subset of finite measure
of the set of the points $x\in\X$ for which $T\ci
R^\sharp\nu(x)>t$, we conclude that
$$
\mu\{x\in\X\,:\,T\ci R^\sharp\nu(x)>t\}\le \frac{A_6\|\nu\|}{t},
$$
proving the theorem.
\end{proof}
To prove Corollary \ref{c9.2}, it is enough to recall that every signed
measure $\nu\in M(\X)$ can be represented as $\nu_+-\nu_-$, where $\nu_\pm$
are {\it finite non-negative} measures and $\|\nu_+\|+\|\nu_-\|=\|\nu\|$.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% %
% %
% \bigskip %
% %
% Address: %
% Department of Mathematics, %
% Michigan State University, %
% East Lansing, Michigan 48824, USA. %
% %
% \bigskip %
% %
% Current address of A.Volberg: %
% Department of Mathematics, %
% Univ. of Hawaii, %
% 2565 The Mall, %
% Honolulu, HI. 96822. %
% %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}