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% Research Section from Editor THOMSON
\documentclass{rae}
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%\coverauthor{D. Preiss L. Zaj\' \i \v cek}
%\covertitle{Sigma-Porous Sets in Products of Metric Spaces and Sigma-Directionally Porous Sets in Banach Spaces}
\received{November 6, 1997}
\MathReviews{28A05, 26B05, 46G99}
\keywords{Porous sets, directionally porous set, Fubini theorem }
\firstpagenumber{1}
\markboth{D. Preiss L. Zaj\' \i \v cek}{Sigma Porous Sets in Products of
Metric Spaces}
\author{D. Preiss,\
Department of Mathematics, University College
London, London WC1E~6BT, United Kingdom.
e-mail: {\tt dp@math.ucl.ac.uk}
\and
L. Zaj\' \i\v cek\thanks{Supported by grants GA\v CR 201/97/1161 and
GAUK 190/1996},\
Department of Math. Anal., Charles University, Sokolovsk\' a 83,
186 00 Prague 8, Czech Republic.
e-mail: {\tt zajicek@karlin.mff.cuni.cz}}
\title{SIGMA-POROUS SETS IN PRODUCTS OF METRIC SPACES AND SIGMA-DIRECTIONALLY
POROUS SETS IN BANACH SPACES}
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\begin{document}
\maketitle
\begin{abstract}
We show that no reasonable classical form of
``Fubini type theorems'' can hold for the
$\sigma$-ideal of $\sigma$-porous sets in products of metric spaces
(even in the plane). Then we prove that a ``Fubini type theorem''
in a weak decomposition form remains true also for this $\sigma$-ideal,
and we illustrate how this fact may be applied to the study of
the behavior of measures on small sets in product spaces.
We also prove an analogical decomposition theorem for
$\sigma$-directionally porous sets in Banach spaces; such sets arise
naturally as exceptional sets in some questions concerning
differentiability properties of Lipschitz functions on Banach spaces.
\end{abstract}
\section{Introduction}
The Fubini theorem in $\R^2$ immediately implies the following statements.
\begin{enumerate}
\item[(a)]
If $M \subset \R^2$ is a Lebesgue measurable set and for all $x \in \R$, except
a Lebesgue null set, the section $M_x$ is Lebesgue null, then $M$
is a Lebesgue null set.
\item[(b)]
If $M \subset \R^2$ is of Lebesgue measure zero, then, for all $x \in \R$ except
a (one-dimensional) Lebesgue null set, the section $M_x = \{y \in \R :(x,y)
\in
M \}$ is Lebesgue
null.
\item[(c)]
If $M \subset \R^2$ is a Borel set of Lebesgue measure zero, then
$M$ can be written as
$M=A\cup B$, where $A,B\subset \R^2$ are Borel sets such that, for all $x
\in \R$
the section $A_x = \{y \in \R :(x,y) \in
A \}$ is Lebesgue null
and for all $y \in \R$
the section $B^y = \{x \in \R :(x,y) \in
B \}$ is Lebesgue null.
\end{enumerate}
The question whether Fubini type
theorems; i.e., analogies of the statements (a) and (b),
hold also for sets small in the sense of Baire category
is answered positively by the well-known Kuratowski-Ulam theorem
(cf.\ \cite{O}). The statement (c) seems to be mentioned only rarely, probably
because it is a very easy consequence of (b); we are aware only
of its use in the definition of sets null in the sense of
Aronszajn (see \cite{A}). It is therefore surprising that
for the $\sigma$-ideal of $\sigma$-porous sets
the analogy of (c) appears as the only correct generalization of the
Fubini theorem; and that this seemingly weak generalization has
also interesting applications.
Let us first consider the classical formulations of Fubini type
theorems.
It is well known that the analogy of (a) does not hold for the
$\sigma$-ideal of $\sigma$-porous sets.
In fact, Foran in \cite{F} constructed a continuous function $f: \R \to \R$
which has a non-$\sigma$-porous graph. Thus the ``$\sigma$-porous
analogue'' of (a) is invalid even for some closed sets $M$.
On the other hand, in the (very) special case when
$M$ is an analytic set of the form $M = A \times
B$ the analogue of (a) holds. This follows from \cite{Z4}, where
it is shown that $A \times B$ is non-$\sigma$-porous whenever $A,B$ are
$G_{\delta}$ non-$\sigma$-porous subsets of $\R$ and the fact that each
analytic
non-$\sigma$-porous subset of $\R$ contains a closed non-$\sigma$-porous
set. The
last result was obtained independently by different methods by
J. Pelant and M. Zelen\' y;
the proofs are still unpublished.
In the second section of our article, we give a counterexample showing
that no statement directly analogous to (b) can hold for the
$\sigma$-ideal of $\sigma$-porous sets. We even
show that there exists a set $M \subset \R^2$ which is porous in a
rather strict sense and $\R \setminus M_x$ is Lebesgue null for each $x
\in \R$ except a first category set.
In the third section we prove
our $\sigma$-porous analogy of the weak Fubini theorem (c) saying
that if $X,Y$ are metric spaces and $M \subset X
\times Y$ is a Borel $\sigma$-porous set, then there exists a
decomposition $M = A \cup B$, where $A,B$ are Borel and all sections
$A_x$ and $B^y$ are $\sigma$-porous. In fact, our decomposition
theorem (Theorem \ref{XYc}) shows considerably more. $A$ and $B$
are in certain sense $\sigma$-porous ``in the direction of $Y$ and $X$'',
respectively.
As a consequence, we obtain the following fact (see Proposition
\ref{ac}).
If $\mu$ and $\nu$ are Radon measures
on separable metric spaces $X$ and $Y$, respectively,
which are ``absolutely continuous'' w.r.t.\
$\sigma$-porous sets, then the Radon product measure $\mu \otimes \nu$
is ``absolutely continuous'' w.r.t.\ $\sigma$-porous sets as well.
We also obtain decomposition theorems similar to the one described above
for $\sigma$-directionally
porous sets in separable Banach spaces. It should be noted that the
class of $\sigma$-directionally porous sets arises quite naturally
in some questions concerning
differentiability properties of Lipschitz functions on Banach spaces.
Here we mention only one sample result from \cite{PZ}.
If $f$ is a Lipschitz function
defined on a separable Banach space $X$, then, for all points $a \in X$
except those which belong to a $\sigma$-directionally porous set, the
function $g(v):= \limsup_{h \to 0+} \frac{f(a+hv)-f(a)}{h}$ is convex.
We adopt the following notation.
In a metric space, the open ball with center $x$ and radius $r$
will be denoted by $B(x,r)$. The distance of two sets $A,B$ is denoted
by ${\rm dist \,}(A,B)$ . The closure and the interior of a set $A$ are
denoted
by $\overline A$ and ${\rm int \,} A$, respectively. The Lebesgue measure
on $\R$
is denoted by $\lambda$. The linear span of a subset $M$ of a linear
space is denoted by ${\rm span \,} M$.
Now we recall definitions of some porosity notions and present
basic relevant comments.
\begin{definition}\label{1}
Let $(X, \rho)$ be a metric space, $M \subset X$ and $a \in X$. Then we
say that
\begin{enumerate}
\item
$M$ is porous at $a$ if there exists $c>0$ such that for each $\varepsilon >0$
there exists $b \in X$ and $r > c \rho(a,b)$ such that $\rho(a,b) <
\varepsilon$
and
$M \cap B(b,r) = \emptyset$.
\item
If moreover $X$ is a normed linear space and also a set $V \subset X$
is given, then we say that $M$ is porous at $a$ in direction $V$ if
the $b \in X$ from (i) verifying the porosity of $M$ at $a$ can be
always found in the form $b = a + tv$, where $t \geq 0$ (or,
equivalently, $t>0$) and $v \in V$.
If $V = \{v\}$, then we say that $M$ is porous at $a$ in direction $v$.
We say that $M$ is directionally porous at $a$ if there exists $v \in X$
such that $M$ is porous at $a$ in direction $v$.
\item
If in the above definitions a fixed $c > 0$ can be used, we speak about
$c$-porosity, $c$-porosity in direction $V$, $c$-porosity in direction
$v$ and $c$-directional porosity.
If $M \subset X$ is $c$-porous at $a \in X$ for each $0 < c < 1$,
then we say that $M$ is strongly porous at $a$.
\item
We say that $M$ is porous ($c$-porous, porous in direction $V$,...)
if $M$ is porous ($c$-porous, porous in direction $V$,...) at each of
its points.
\item
We say that $M$ is $\sigma$-porous ($\sigma$-c-porous, $\sigma$-porous
in direction $V$,...) if it is a countable union of porous sets
($c$-porous sets, sets porous in direction $V$,...).
\end{enumerate}
\end{definition}
Note that $M$ is porous at $a$ in direction $0$ iff $ a \notin
\overline{M}$; in this case $M$ is $c$-porous at $a$ in all directions
for any $c>0$.
Clearly each directionally porous ($\sigma$-directionally porous) set is
porous ($\sigma$-porous) and it is an easy well-known fact that these
concepts coincide in finite-dimensional spaces (cf.\ Lemma~\ref {zakl},
(iv)).
The notion of $\sigma$-porosity was introduced by Dolzhenko \cite{D}
and since then, has been used and investigated by many authors; in some
applications other variants of porosity notions are natural (cf.\
the survey article \cite{Z3}).
Each $\sigma$-porous set is clearly a first category set and the
Lebesgue density theorem easily implies that each $\sigma$-porous subset
of $\R^n$ is of Lebesgue measure zero. Unfortunately, a $\sigma$-porous
subset of an infinite-dimensional separable Banach space need not be
null in any ``natural measure sense''. In fact, in \cite{PT} an example
of an $F_{\sigma}$, $\sigma$-porous subset $S$ of a separable Hilbert space
$H$ is constructed in such a way that the complement $ C:=H \setminus S$
intersects any line in a set of null one-dimensional Lebesgue measure
(on this line); consequently $C$ is null in the Aronszajn sense
and therefore it is also both
null for each
non-degenerate Gaussian measure on $H$ (see \cite{Ph}) and
of Haar
measure zero in Christensen's sense (see \cite{Ch} for the definition).
A substantial strengthening of this example has been
recently found in \cite{MM}.
On the other hand, each Borel $\sigma$-directionally porous subset of a
separable Banach space $X$ is null in the Aronszajn sense. This fact can
be easily deduced from Aronszajn's theorem, which says that a Lipschitz
function on $X$ is G\^ ateaux differentiable at all points except
those belonging to a set
which is Aronszajn null, and
from the easy observation (cf.\ \cite{Z2}, p.~299) that a
set $E \subset X$ is directionally porous at a point $a \in E$ iff
the distance function $d(x):={\rm dist \,}(x,E)$ is not G\^ ateaux
differentiable at
$a$. Nevertheless, it is interesting to try
to deduce this result without any direct or indirect use
of Aronszajn's Theorem. We therefore point out that an application of
our decomposition theorem for
$\sigma$-directionally porous sets (Theorem \ref{dec})
supplies such an alternative argument (see
Remark \ref{borel}).
\section {Fubini Type Theorems for $\sigma$-Porous Sets:
Counterexamples}
\begin{definition}\label{h}
Let $h:[0,\infty) \to [0,\infty)$ be a continuous function for which
$h(0)=0$ and $h(x) > x$ for each $x>0$.
We shall say that $M \subset \R^2$ is $h$-right porous at a point
$c=(a,b) \in \R^2$ in direction of the $x$-axis (simply, $M$ is
$(h,+)$-porous at $c$) if for each $\varepsilon > 0$ there exist
$t,r>0$ such that $t < \varepsilon$, $ ((a+t-r,a+t+r) \times (b-r,b+r))
\cap M = \emptyset$ and $h(r) > t$.
The notion of a set which is $(h,-)$-porous at $c$ is defined in the
symmetrical way.
We say $M \subset \R^2$ is $(h,bil)$-porous at $c$ if $M$ is both
$(h,+)$-porous and $(h,-)$-porous at $c$.
The notions of a $(h,bil)$-porous set and a $\sigma$-$(h,bil)$-porous
set are defined in the obvious way.
\end{definition}
\begin{remark}\label{hstrongly}
Let $h(x)=x + x^2$ and $M \subset \R^2$ be $(h,+)$-porous at $c$ (or
$(h,-)$-porous at $c$). Then clearly $M$ is also strongly porous at $c$.
\end{remark}
\begin{definition}\label{W}
Let $h$ be as in Definition~\ref{h}, $G \subset \R^2$ and $\delta >0$.
Then we define $W^+(G,h,\delta)$ as the set of all points $c=(a,b) \in
\R^2$ such that there exist $00$ such that
$(a+t-r,a+t+r) \times (b-r,b+r) \subset G$ and $h(r) > t$.
The set $W^-(G,h,\delta)$ is defined in the obvious symmetrical way.
\end{definition}
\begin{remark}\label{hW}
Obviously $M$ is $(h,+)$-porous at $x$ iff $c \in
\bigcap_{k=1}^{\infty} W^+(\R^2 \setminus M,h,\frac{1}{k})$. The
corresponding result is true for
$(h,-)$-porosity.
\end{remark}
\begin{lemma}\label{alpha}
Let $h$ be as in Definition~\ref{h}, let $-\infty 0$ be
given.
Then there exists a number $t < T < s$ such that
\begin{eqnarray*}
\lefteqn{(T,2s-T) \times (v+\eta, w - \eta)}
\phantom{T,2s-T}\\
&\subset& W^-((t,T) \times
(v,w),h,\delta) \cap W^+((2s-T,2s-t) \times (v,w),h,\delta).
\end{eqnarray*}
\end{lemma}
\pf
Put $\varepsilon = \min (\frac{\delta}{3}, \eta, \frac{s-t}{2})$.
Choose $T$ such that $s-\varepsilon < T < s $ and
$h(\varepsilon - (s-T)) > \varepsilon + (s-T)$; this is clearly possible.
Now suppose that $a \in (T,2s-T)$
and $b \in (v+\eta, w-\eta)$ are given.
Put $p = a-(s-\varepsilon)$ and $r = \varepsilon -(s-T)$. Then clearly
\begin{eqnarray*}
&&0 < p < \varepsilon +(s-T) < 2 \varepsilon < \delta, \\
&& h(r) > \varepsilon +
(s-T) > p, \\\noalign{\noindent\mbox{and}}
&&V(a-p-r,a-p+r) \times (b-r,b+r) \subset (t,T) \times (v,w).
\end{eqnarray*}
Consequently $(a,b) \in W^- ((t,T) \times (v,w),h,\delta)$. A quite
symmetrical argument gives $(a,b) \in W^+((2s-T,2s-t)\times
(v,w),h,\delta)$.
\qed
\begin{lemma}\label{omega}
Let $h$ be a function as in Definition~\ref{h}, let
$\varepsilon,\delta,\omega >0$ and let $I=(a,b)$,
$K=(c,d)$ be bounded intervals. Then there exists an open interval $J
\subset I$,
an open set $G \subset
\R^2$ and a Borel set $V \subset \R^2$ such that
\begin{enumerate}
\item
$\lambda (G_x) \leq \varepsilon $ \ for each \ $x \in \R$,
\item
$\lambda (V_x) \leq \omega$ \ for each \ $x \in \R$ \ and
\item
$(J \times K) \setminus V \subset W^-(G,h,\delta) \cap
W^+(G,h,\delta)$.
\end{enumerate}
\end{lemma}
\pf
We may and will suppose $\omega < d-c$.
Put $s = \frac{a+b}{2}$. Further choose a natural
number $n$ for which $\frac{d-c}{n} < \varepsilon$ and put $\eta =
\frac{\omega}{2n}$. Now we will use Lemma~\ref{alpha} $n$-times; each
time with the same $h, \eta,\delta$ and $s$ but each time with different
$t,v$ and $w$.
In the first step we apply Lemma~\ref{alpha} to $t = t_1 = a, v=c$ and $w
= c + \frac{d-c}{n}$; we obtain the corresponding $ t_1 < T = T_1 ~~0$.
Then there exists a Borel set $M \subset \R \times K$ such that
\begin{enumerate}
\item
$M$ is $(h,bil)$-porous.
\item
There exists a residual set $A \subset \R$ such that $ \lambda (M_x) >
\lambda(K) - \xi$ for each $x \in A$.
\end{enumerate}
\end{proposition}
\pf
Let $I_1,I_2,\dots$ be a sequence of all intervals with rational
endpoints. For each natural number $n$ we apply Lemma~\ref{omega}
to
$$I = I_n, K, \delta = \frac{1}{n}, \varepsilon = \omega =
\frac{\xi}{2^{n+1}};$$
we obtain corresponding $J=J_n, G = G_n $ and $V = V_n$.
Put
$$ A = \limsup_{n \to \infty} J_n = \bigcap_{k=1}^{\infty}
\bigcup_{n=k}^{\infty} J_n \ \mbox{and}\ M = (A \times K) \setminus
(\bigcup_{n=1}^{\infty} V_n \cup \bigcup_{n=1}^{\infty} G_n).$$
Obviously $A$ is a residual subset of $\R$ and $M$ is a Borel subset of
$\R^2$. Since
$$ \lambda(M_x) > \lambda (K) - (\sum_{n=1}^{\infty}
\lambda((V_n)_x) + \sum_{n=1}^{\infty} \lambda((G_n)_x) \geq \lambda(K)
- \xi$$
for each $x \in A$, we have proved (ii).
If now $c \in M$ and a natural number $l$ are given, then we find $n >
l$ such that $c \in (J_n \times K) \setminus V_n$. Thus by our
construction (cf.\ Lemma~\ref{omega},(iii)) we have
$$ c \in W^-(G_n,h, \frac{1}{l}) \cap W^+(G_n,h,\frac{1}{l}) \subset
W^-(\R^2 \setminus M,h,\frac{1}{l}) \cap W^+(\R^2 \setminus M,h,
\frac{1}{l}).$$
From Remark~\ref{hW} we obtain also (i).
\qed
\begin{theorem}\label{sigma}
Let $h$ be a function satisfying the conditions of Definition~\ref{h}. Then
there exists
a Borel
$\sigma-(h,bil)$-porous set $M \subset \R^2$ such that the set $\{ x
\in \R:
\lambda(\R \setminus M_x) = 0 \}$ is residual.
\end{theorem}
\pf
Let $K_1,K_2,\dots $ be a sequence of all intervals with rational
endpoints. By Proposition~\ref{hpor} for each $n$ there are a residual
set $A_n \subset \R$ and a Borel $(h,bil)$-porous
set $M_n \subset \R \times K_n$ such that
$ \lambda ((M_n)_x) > \frac{1}{2} \lambda(K_n)$ for every $x \in A_n$.
If we put $ M = \bigcup_{n=1}^{\infty} M_n$, the Lebesgue density
theorem easily implies that $ \lambda(\R \setminus M_x) = 0$ for each
$x$ from the residual set $A:= \bigcap_{n=1}^{\infty} A_n$.
\qed
\smallskip
By Remark~\ref{hstrongly} we immediately obtain the following corollary.
\begin{proposition}\label{strongly}
There exists a $\sigma$-strongly porous set $M \subset \R^2$ such that
the set $\{ x
\in \R:
\lambda(\R \setminus M_x) = 0 \}$ is residual.
\end{proposition}
The following remark concerns ``very porosity'' (cf.\ \cite{Z3})
which is stronger than porosity and is incomparable with strong porosity.
\begin{remark}\label{very}
It is easy to prove that
$ \{ x \in \R : \lambda(M_x) > 0 \}$ is a first category set whenever
$M \subset \R^2$ is a subset of an $F_{\sigma}$ set of Lebesgue measure
zero (in particular, if $M$ is $\sigma$-very porous). Thus the analogue
of Proposition \ref{strongly} for $\sigma$-very porous sets does not
hold.
\end{remark}
\section{Directional Porosity in Products of Metric Spaces}
In the following, if $(X,\rho) $ and $(Y,\eta)$ are metric spaces,
then we shall denote by $(\rho \times \eta)_m $ and $(\rho \times
\eta)_s$ the maximum and the sum metric on $X \times Y$, respectively.
\begin{definition}\label{c}
Let $X,Y$ be sets and let $X \times Y$ be equipped with a metric
$\omega$ and let $c > 0$. Then we say that a set $A \subset X \times Y$
is $X$-directionally $c$-porous at a point $a = (a_1,a_2) \in X \times
Y$ if, for each $\varepsilon > 0$,
\begin{equation}\label{defc}
\begin{split}
&\mbox{there exist}\ \ b \in X \ \mbox{and}\ r>0 \ \mbox{such that}\
\omega(a,(b,a_2)) < \varepsilon, \\
& r > c \omega(a,(b,a_2))
\ \mbox{and} \ B((b,a_2),r) \cap A = \emptyset.
\end{split}
\end{equation}
The notion of a set $Y$-directionally $c$-porous at $a$ is defined in
the obvious way.
\end{definition}
For proofs we shall need the following technical notion.
\begin{definition}\label{cxcy}
Let $(X,\rho), (Y, \eta)$ be metric spaces and let $c_x,c_y > 0$.
Then we say that a set $A \subset X \times Y$ is $X$-directionally
$(c_x,c_y)$-porous at a point $a = (a_1,a_2)\in X \times Y$ if, for each
$\varepsilon > 0$,
\begin{equation}\label{defcxcy}
\begin{split}
&\mbox{there exist} \ b \in X \ \mbox{and} \ r,s >0 \ \mbox{such that} \
\rho(a_1,b) < \varepsilon, \\
& r > c_x \rho(a_1,b), s > c_y \rho(a_1,b) \ \mbox{and}\ (B(b,r)
\times B(a_2,s)) \cap A = \emptyset.
\end{split}
\end{equation}
The notion of $Y$-directional $(c_x,c_y)$-porosity is defined in an
obvious way.
\end{definition}
The corresponding notions of $X$-directionally (or $Y$-directionally)
$c$-porous (or $\sigma$-$c$-porous, or $(c_1,c_2)$-porous, or
$\sigma$-$(c_1,c_2)$-porous) sets are defined in the obvious way.
\begin{remark}\label{ccxcy}
It is easy to see that the notion of $X$-directional
$(c_x,c_y)$-porosity is a special case of $X$-directional $c$-porosity.
In fact, it is easy to see that $A$ is $X$-directionally
$(c_x,c_y)$-porous in $X \times Y$(where $X,Y$ are equipped with metrics
$\rho, \eta$, respectively) iff $A$ is $X$-directionally $c_x$-porous in
$(X \times Y, \omega)$, where $\omega = (\rho \times \frac{c_y}{c_x} \eta
)_m$.
\end{remark}
To prove the existence of ``small'' Borel covers of ``small'' sets we
shall need the following technical notions.
\begin{definition}\label{R}
Let $(X \times Y, \omega)$ be a metric space, $A \subset X \times Y$ and
let $c, \varepsilon > 0$ be given. Then we define $R(A,c,\varepsilon)$ as the
set of all points $a=(a_1,a_2) \in X \times Y$ for which (\ref{defc})
holds.
\end{definition}
\begin{definition}\label{Q}
Let $(X,\rho)$ and $(Y,\eta)$ be metric spaces, $A \subset X \times Y$ and let
$c_x,c_y,\varepsilon >0$ be given. Then we define $Q(A,c_x,c_y,\varepsilon)$
as the set of all points $a=(a_1,a_2) \in X \times Y$ for which
(\ref{defcxcy}) holds.
\end{definition}
\begin{lemma}\label{obal2}
Let $(X,\rho), (Y,\eta)$ be metric spaces and let $X \times Y$ be equipped
with a metric $\omega$ which is (topologically)
equivalent to $(\rho \times \eta)_m$.
Let $A \subset X \times Y$ and $c,c_x,c_y,\varepsilon >0$ be given. Then
each of the following holds.
\begin{enumerate}
\item
$R(A,c,\varepsilon)$ and $Q(A,c_x,c_y,\varepsilon)$ are open sets in $(X\times
Y, \omega).$
\item
$A$ is $X$-directionally $c$-porous at a point $a = (a_1,a_2) \in X
\times Y$ iff
$$ a \in \bigcap_{\varepsilon > 0} R(A,c, \varepsilon) =
\bigcap_{n=1}^{\infty} R(A,c, \frac{1}{n}).$$
\item
$A$ is $X$-directionally $(c_x,c_y)$-porous at a point $a = (a_1,a_2) \in X
\times Y$ iff
$$ a \in \bigcap_{\varepsilon > 0} Q(A,c_x,c_y, \varepsilon) =
\bigcap_{n=1}^{\infty} Q(A,c_x,c_y, \frac{1}{n}).$$
\item
If $A$ is $X$-directionally $c$-porous ($(c_x,c_y)$-porous), then there
exists
a $G_{\delta}$ set $\tilde{A} \supset A$ ($A^* \supset A)$ which is
$X$-directionally $c$-porous (($c_x,c_y)$-porous).
\item
If $A$ is $\sigma$-$X$-directionally $c$-porous ($(c_x,c_y)$-porous), then there
exists
a $G_{\delta\sigma}$ set $\tilde{A} \supset A$ ($A^* \supset A)$ which is
$\sigma$-$X$-directionally $c$-porous (($c_x,c_y)$-porous).
\end{enumerate}
\end{lemma}
\pf
The statements (i), (ii) and (iii) are obvious. If $A$ is
$X$-directionally $c$-porous, we put $\tilde A = \overline A \cap
\bigcap_{n=1}^{\infty} R(A,c,1/n)$. By (i), $\tilde A$ is a $G_{\delta}$
set. By (ii) and by the obvious fact, that if $A$ is $X$-directionally
$c$-porous at a point $a$, then any subset of $\overline A$ is
$X$-directionally $c$-porous at $a$, we obtain that $A \subset \tilde A$
and $\tilde A$ is $X$-directionally $c$-porous. This proves the
first part of (iv). The second part is an immediate consequence of the
first one and Remark \ref{ccxcy}. The assertion (v) follows easily by
(iv).
\qed
\begin{lemma}\label{smetric}
Let $(X,\rho), (Y,\eta)$ be metric spaces , let $X \times Y$ be
equipped with the sum metric $\omega = (\rho \times \eta)_s$, and let
$0 < \alpha < \frac{1}{2}$ be given. Then each of the following holds.
\begin{enumerate}
\item
If a set $A \subset X \times Y$ is
$(1-\alpha)$-porous at a point $a = (a_1,a_2) \in X \times Y$, then
either $A$ is $X$-directionally $(1-2\alpha)$-porous at $a$ or $A$ is
$Y$-directionally $(1-2\alpha)$-porous at $a$.
\item
If $A$ is $(1-\alpha)$-porous, then we can write $A = A_1 \cup A_2$,
where $A_1$ is $X$-directionally $(1-2\alpha)$-porous and $A_2$ is
$Y$-directionally $(1-2\alpha)$-porous.
\end{enumerate}
\end{lemma}
\pf
Suppose that the assertion of (i) is not true. Then there exists
$\varepsilon > 0$ such that
\begin{equation}\label{h1}
r_1 \leq (1-2\alpha) \rho(a_1,x_1)\ \mbox{if}\ B((x_1,a_2),r_1)
\cap A = \emptyset \ \mbox{and} \ \rho(a_1,x_1) < \varepsilon
\end{equation}
and
\begin{equation}\label{h2}
r_2 \leq (1-2\alpha) \eta(a_2,x_2)\ \mbox{if}\ B((a_1,x_2),r_2)
\cap A = \emptyset \ \mbox{and} \ \eta(a_2,x_2) < \varepsilon.
\end{equation}
By our assumption we can find a point $x=(x_1,x_2)$ and $r > 0$ such
that $\omega(a,x) < \varepsilon, B(x,r) \cap A = \emptyset$ and
\begin{equation}\label{h3}
r > (1-\alpha) \omega(a,x) = (1-\alpha) (\rho(a_1,x_1)+ \eta(a_2,x_2)).
\end{equation}
Put $r_1 = r - \eta(a_2,x_2)$ and $r_2 = r - \rho(a_1,x_1)$.
Clearly we have (if $r_1>0$) $B((x_1,a_2),r_1) \subset B(x,r) \subset
(X \times Y) \setminus A$ and $ \rho(a_1,x_1) < \varepsilon$. Thus (\ref{h1})
implies
$$ r - \eta(a_2,x_2) \leq (1 - 2\alpha) \rho(a_1,x_1).$$
By a symmetrical argument, (\ref{h2}) implies
$$r - \rho(a_1,x_1) \leq (1 - 2\alpha) \eta(a_2,x_2).$$
Adding the last two inequalities, we obtain
$$ r \leq (1-\alpha) (\rho(a_1,x_1) + \eta(a_2,x_2)),$$
which contradicts (\ref{h3}). Thus (i) is proved.
To prove (ii), it is clearly sufficient to define $A_1$ ($A_2$) as the
set of points in $A$ at which $A$ is $X$-directionally
($Y$-directionally) $(1-2\alpha)$-porous.
\qed
Using the above lemma and Theorem 4.5 of \cite{Z1}
according to which, for any $00$ such that $c < 1 - 2 \varepsilon$. By
Theorem 4.5 of \cite{Z1}, $A$ is $\sigma$-$(1-\varepsilon)$-porous set.
Using this fact and
Lemma~\ref{smetric},(ii), we obtain our assertion.
\qed
\begin{remark}\label{decbor}
\begin{enumerate}
\item
If $A$ is Borel, then the sets $A_1,A_2$ in Theorem \ref{XYc} can be
chosen to be Borel. In fact, we can write $A = (A \cap \tilde{A_1})
\cup (A \cap \tilde{A_2})$, where $\tilde{A_1},\tilde{A_2}$ are
$G_{\delta \sigma}$ covers from Lemma \ref{obal2},(v).
\item
Of course, $A_1$ ($A_2$) has the property that all sections $(A_1)^y$
($(A_2)_x$) are $\sigma$-porous subsets of $X$ ($Y$).
\end{enumerate}
\end{remark}
If $P$ is a metric space and $\mu$ is a measure on $P$, we say that
$\mu$ is absolutely continuous w.r.t.\ $\sigma$-porous sets if $\mu(M)=0$
for each $\sigma$-porous set $M \subset P$. Note that every Radon
measure on $\R$ which is absolutely continuous w.r.t.\ Lebesgue
measure is clearly absolutely continuous w.r.t.\ $\sigma$-porous sets
but the opposite implication does not hold (see \cite{T}).
Remark \ref{decbor} easily implies the following proposition which
seems to be of some independent interest and which suggests that Theorem
\ref{XYc} is in fact a substitute for the (invalid) ``Fubini type''
theorem for $\sigma$-porous sets.
\begin{proposition}\label{ac}
Let $\mu$ and $\nu$ be Radon measures
on separable metric spaces $X$ and $Y$, respectively,
which are absolutely continuous w.r.t.\
$\sigma$-porous sets. Then the Radon product measure $\mu \otimes \nu$
is absolutely continuous w.r.t.\ $\sigma$-porous sets as well.
\end{proposition}
If we use Theorem~\ref{XYc} in $\R^2$ equipped with the Euclidean
norm, we easily obtain the following result, in which we put
$e_1=(1,0)$ and $e_2=(0,1)$.
\begin{proposition}\label{euclidean}
Let $A \subset \R^2$ be a $\sigma$-porous set and let $0 < c<
\frac{1}{\sqrt{2}}$. Then
$$ A = A_1^+ \cup A_1^- \cup A_2^+ \cup A_2^-,$$
where $A_i^+ (A_i^-)$ is a set which is $\sigma$-$c$-porous in direction
$e_i (-e_i), \ i=1,2$.
\end{proposition}
\begin{remark}\label{cmax}
If we consider $A = \{(x,x): x \in \R \}$, we easily observe (using the
Baire category theorem) that the assertion of
Proposition~\ref{euclidean} does not hold for $c= \frac{1}{\sqrt{2}}$.
\end{remark}
\begin{remark}\label{coneni}
Lemma~\ref{smetric} and subsequently also Theorem~\ref{XYc} can be easily
generalized to the case of the product of several metric spaces. Further
we can easily generalize Proposition~\ref{euclidean} to $\R^n$ (with the
assumption $0 < c < \frac{1}{\sqrt{n}}$).
\end{remark}
In the proof of Theorem \ref{XYc} we applied the method of
``enlargement of the porosity index''
established in Proposition 4.1 from \cite{Z1} (which implies
Theorem 4.5 from
\cite{Z1}) to the
sum metric in the product of two metric spaces. In the following lemma and
its consequence Proposition \ref{beta}
we now show how this method may be generalized
in the special situation of product spaces and directional porosity.
These results will be used in the following section (see
(\ref{aplzv})--(\ref{posl})).
\begin{lemma}\label{zvetsovani}
Let $(X,\rho)$ and $(Y,\eta)$ be metric spaces, $1>c_x>0$, $c_y>0$ and let $A
\subset X \times Y$ be an $X$-directionally $(c_x,c_y)$-porous set. Then
$A$ is $\sigma$-$X$-directionally $(\sqrt{c_x},\frac{c_y}{2})$-porous
set.
\end{lemma}
\pf
Put $A_0 = A \cap \bigcap_{n=1}^{\infty}
Q(A,\sqrt{c_x},\frac{c_y}{2},\frac{1}{n})$. By Lemma~\ref{obal2},(iii)
we have that $A_0$ is an $X$-directionally
$(\sqrt{c_x},\frac{c_y}{2})$-porous set. Since $A \setminus A_0 =
\bigcup_{n=1}^{\infty} \ (A \setminus
Q(A,\sqrt{c_x},\frac{c_y}{2},\frac{1}{n}))$, it is sufficient to prove
that each of the sets $A_n := A \setminus
Q(A,\sqrt{c_x},\frac{c_y}{2},\frac{1}{n})$ is $X$-directionally
$(\sqrt{c_x},\frac{c_y}{2})$-porous. To this end choose an arbitrary
natural number $n,\ 0 < \varepsilon < \frac{1}{n}$ and $(x,y) \in A_n$. It is
sufficient to prove that
\begin{equation}\label{xyQ}
(x,y) \in Q(A_n,\sqrt{c_x},\frac{c_y}{2},\varepsilon).
\end{equation}
Since $(x,y) \in A$, we know that $(x,y) \in Q(A,c_x,c_y,\varepsilon)$ and
therefore we can choose $b \in X$ and $r,s > 0$ such that $ \rho(x,b) <
\varepsilon, r > c_x \rho(x,b), s > c_y \rho_(x,b)$ and $ (B(b,r) \times
B(y,s)) \cap A = \emptyset$. Since $r > c_x \rho(x,b)$, we can choose
$r^* > 0$ such that $\sqrt{c_x} \rho(x,b) < r^* < \rho(x,b)$ and $r >
\sqrt{c_x} r^*$.
To prove (\ref{xyQ}), it is clearly sufficient to prove that $A_n \cap
(B(b,r^*)\times B(y,\frac{s}{2})) = \emptyset$.
Assuming that this is not the case, we choose an $(x_1,y_1) \in
A_n \cap (B(b,r^*) \times B(y,\frac{s}{2}))$.
Clearly,
$$ B(b,r) \times B(y_1,\frac{s}{2}) \subset B(b,r) \times B(y,s) \subset
(X \times Y) \setminus A.$$
Since
\begin{eqnarray*}
&& \rho(x_1,b) < r^* < \rho(x,b) < \varepsilon < \frac{1}{n} , \\
&& r > \sqrt{c_x} r^* > \sqrt{c_x} \rho(x_1,b) \\
\noalign{\noindent\mbox{and}}
&&\frac{s}{2} > \frac{c_y}{2} \rho(x,b) > \frac{c_y}{2} r^* >
\frac{c_y}{2} \rho(x_1,b),
\end{eqnarray*}
we conclude that $(x_1,y_1) \in
Q(A,\sqrt{c_x},\frac{c_y}{2},\frac{1}{n})$,
which contradicts the fact that $(x_1,y_1) \in A_n$.
\qed
\begin{proposition}\label{beta}
Let $(X,\rho), (Y,\eta)$ be metric spaces, let $c_x,c_y > 0, 0 < \alpha
< 1$, and let $A \subset X \times Y$ be a $\sigma$-$X$-directionally
$(c_x,c_y)$-porous set. Then there exists $\beta > 0$ such that $A$ is
$\sigma$-$X$-directionally $(\alpha,\beta)$-porous.
\end{proposition}
\pf
We may suppose $c_x < 1$. Choose a natural number $n$ so large that $
(c_x)^{2^{-n}} > \alpha$ and put $\beta = 2^{-n} c_y$. Then, applying
Lemma~\ref{zvetsovani} $n$-times, we obtain that $A$ is
$\sigma$-$X$-directionally $(\alpha,\beta)$-porous.
\qed
\section {Sigma-Directionally Porous Sets in Banach Spaces}
We start with the following easy properties of directional porosity.
\begin{lemma}\label{zakl}
Let $X$ be a normed linear space, $M \subset X, V \subset X, W \subset
X , a \in X$ and $c > 0$. Then the following assertions hold.
\begin{enumerate}
\item
If $0 \notin V$, then $M$ is porous ($c$-porous) at $a$ in direction $V$
iff $M$ is porous ($c$-porous) at $a$ in direction $V_n$, where
$V_n= \{ \frac{v}{\|v\|} : v \in V \}$. $M$ is porous ($c$-porous) at
$a$ in direction $v$ iff it is porous ($c$-porous) at $a$ in direction
$\{tv: t \geq 0\}$.
\item
If $V = V_1 \cup V_2 \cup ...\cup V_n$ and $M$ is $c$-porous at $a$ in
direction $V$, then there exists $1\leq i \leq n$ such that $M$ is
$c$-porous at $a$ in direction $V_i$. If $M$ is porous ($c$-porous)
in direction $V$, then we can write $M = \bigcup_{i=1}^n M_i$, where
$M_i$ is porous ($c$-porous) in direction $V_i$.
\item
Let $V,W \subset \{ x \in X: \|x\|=1\}, 0< \omega< c$ and let ${\rm dist \,}
(v,W) < c-\omega$ for each $v \in V$. Let $M$ be $c$-porous at $a$ in
direction $V$. Then
$M$ is $\omega$-porous at $a$ in direction $W$.
\item
If $V$ is a compact subset of $\{x \in X: \|x\|=1\}$, then $M$ is
porous at $a$ in direction $V$ iff there exists $v \in V$ such that $M$
is porous at $a$ in direction $v$. In particular, if $X$ is a
finite-dimensional space, then $M$ is porous at $a$ iff $M$ is
directionally porous at $a$.
\end{enumerate}
\end{lemma}
\pf
The statement (i) and the first part of (ii) are obvious. To prove the
second part of (ii) it is sufficient to define $M_i$ as the set of
points in $M$ at which $M$ is porous ($c$-porous) in direction $V_i$.
To prove (iii), choose an arbitrary $\varepsilon > 0$. Since $M$ is
$c$-porous at $a$ in direction $V$, there exist $t \geq 0, v \in V$ and
$r>0$ such that $B(a+tv,r) \cap M = \emptyset,\ \|tv\|=t <
\varepsilon$ and $r > c \|tv\|=ct$. Find $w \in W$ such that
$\|v-w\|< c-\omega$ and put $b^*=a+tw,\ r^*=r-t\|v-w\|$.
Then $\|b^*-a\|=\|tw\|=t < \varepsilon, \ \omega \|tw\| = \omega t =ct -
(c-\omega)t 0$ and
$\varepsilon > 0$. Then we denote by $P(A,Z,c,\varepsilon)$ the set of all
points $a \in X$ for which there exist $z \in Z,\ t \geq 0$ and $r >0$
such that $B(a+tz,r) \cap A = \emptyset, \|tz\| < \varepsilon$ and $r >
c\|tz\|$.
\end{definition}
\begin{lemma}\label{obal}
Let $X$ be a normed linear space, $A \subset X, Z \subset X,a\in X, c>0$ and
$\varepsilon >0$. Then the following assertions are true.
\begin{enumerate}
\item
$P(A,Z,c,\varepsilon)$ is an open set.
\item
$A$ is $c$-porous at $a$ in direction $Z$ iff
$$\textstyle a \in \bigcap_{\varepsilon
>0} P(A,Z,c,\varepsilon) = \bigcap_{n=1}^{\infty} P(A,Z,c,\frac{1}{n}).$$
\item
$A$ is porous at $a$ in direction $Z$ iff
$$\textstyle a \in \bigcup_{k=1}^{\infty}
\bigcap_{n=1}^{\infty} P(A,Z,\frac{1}{k},\frac{1}{n}).$$
\item
If $A$ is $c$-porous in direction $Z$, then there exists a
$G_{\delta}$-set $\tilde{A} \supset A$ which is $c$-porous in direction $Z$.
\item
If $A$ is porous ($\sigma$-porous) in direction $Z$, then there exists a
$G_{\delta\sigma}$-set $A^* \supset A$ which is porous
($\sigma$-porous) in direction $Z$.
\item
If $A$ is porous in direction $Z$
(directionally porous), then we can write $A =
\bigcup_{k=1}^{\infty} A_k$, where each $A_k$ is $\frac{1}{k}$-porous in
direction $Z$ ($\frac{1}{k}$-directionally porous).
\end{enumerate}
\end{lemma}
\pf
The statements (i), (ii) and (iii) are obvious.
If $A$ is $c$-porous in
direction $Z$, we put $ \tilde{A} = \overline{A} \cap
\bigcap_{n=1}^{\infty}
P(A,Z,c,\frac{1}{n})$.
If $A$ is porous in direction $Z$, then we put
$$ A^*: = \overline{A} \cap \bigcup_{k=1}^{\infty}
\bigcap_{n=1}^{\infty} P(A,Z,\frac{1}{k},\frac{1}{n}). $$
By (i) $\tilde{A}$ is a $G_{\delta}$ set and
$A^*$ is a $G_{\delta \sigma}$-set. By (ii), (iii) and by the
obvious fact,
that if $A$ is porous ($c$-porous)
at a point $a$ in direction $Z$, then any subset of
$\overline{A}$ is porous ($c$-porous) at $a$
in direction $Z$, we obtain that $\tilde{A} \supset A$, $A^* \supset A$,
$\tilde{A}$
is $c$-porous in direction $Z$ and $A^*$ is porous in direction $Z$.
This proves (iv) and the statement of (v)
concerning porosity; it then easily implies the second statement of (v)
concerning $\sigma$-porosity. To prove (vi), it is clearly sufficient to
define $A_k$ as the set of all points in $A$ at which $A$ is
$\frac{1}{k}$-porous in direction $Z$ ($\frac{1}{k}$-directionally
porous).
\qed
For the brevity of formulations, we now define an auxiliary notion of
admissible pair of sets.
\begin{definition}\label{admissible}
Let $X$ be a normed linear space and let $E,F \subset X$ be given. We
say that $E,F$ is an admissible pair, if there exists $\eta >0$ such
that
$$ \|e+f\| \geq \eta \max(\|e\|,\|f\|) \ \ \mbox{whenever}\ \ e\in
E,f\in F.$$
\end{definition}
\begin{remark}\label{topcom}
Obviously, if there exist two topologically complementary closed spaces
$V_1,V_2 \subset X$ such that $E \subset V_1$ and $F \subset V_2$, then
$E,F$ is an admissible pair.
\end{remark}
\begin{lemma}\label{EF}
Let $X$ be a normed linear space, $E,F \subset X$ be an admissible pair,
$c > 0$ and let $A \subset X$ be $c$-porous in direction $E+F$. Then $A$
can be written in the form $A = A_1 \cup A_2$, where $A_1$ is porous in
direction $E$ and $A_2$ is $\sigma$-porous in direction $F$.
\end{lemma}
\pf
Let $\eta > 0$ be a number corresponding to $E,F$ from
Definition~\ref{admissible} and put $A_1 = A \cap \bigcap_{m=1}^{\infty}
P(A,E,\frac{c \eta}{2},\frac{1}{m}).$ By Lemma~\ref{obal},(ii) we obtain
that $A_1$ is $\frac{c \eta}{2}$-porous in direction $E$. Thus it is
sufficient to prove that
$$ A\setminus A_1 = \bigcup_{m=1}^{\infty} (A \setminus P(A,E,\frac{c
\eta}{2},\frac{1}{m})) \ \ \mbox{is} \ \sigma\mbox{-porous in
direction}\ F.$$
To this end we shall show that each set $D_m := A \setminus
P(A,E,\frac{c \eta}{2},\frac{1}{m})$ is $\frac{c \eta}{2}$-porous in
direction $F$. To prove this, choose arbitrary $x \in D_m$ and $0 <
\varepsilon < \frac{1}{m}$. We need to prove that
\begin{equation}\label{xinp}
x \in P(D_m,F,\frac{c \eta}{2}, \varepsilon).
\end{equation}
Since $x \in A$, we know that $A$ is $c$-porous at $x$ in direction
$E+F$ and consequently we can find $e \in E, f \in F, t > 0$ and $r >
0$ such that
$$B(x+t(e+f),r) \cap A = \emptyset, \| t(e+f)\| < \varepsilon
\eta\text{ and }r > c \|t (e+f)\|.$$
By the definition of $\eta$,
\begin{equation}\label{max}
\max(\|te\|,\|tf\|) \leq \frac{1}{\eta} \|te +tf\| < \varepsilon <
\frac{1}{m}.
\end{equation}
To prove (\ref{xinp}), it is sufficient to show
that $B(x+tf,\frac{r}{2}) \cap D_m =
\emptyset$, since (\ref{max}) implies $\|tf\| < \varepsilon$ and also
$ \frac{r}{2} > \frac{c}{2} \|t(e+f)\| \geq \frac{c \eta}{2} \|tf\|.$
Thus suppose to the contrary that there is a $y \in
B(x+tf,\frac{r}{2})\cap D_m$.
Clearly $B(y+te,\frac{r}{2}) \subset B(x+t(e+f),r)$ and therefore
$B(y+te,\frac{r}{2}) \cap A = \emptyset$. Since (\ref{max}) implies $
\|te\| < \frac{1}{m}$ and
$$ \frac{r}{2} > \frac{c}{2} \|t(e+f)\| \geq \frac{c \eta}{2} \|te\|, $$
we obtain that $ y \in P(A,E, \frac{c \eta}{2}, \frac{1}{m})$, which
contradicts the fact that $y \in D_m$.\qed
\smallskip
Lemma~\ref{EF} and Lemma~\ref{obal},(vi) clearly imply the following
proposition.
\begin{proposition}\label{PropEF}
Let $X$ be a normed linear space and let $E,F \subset X$ be an
admissible pair. Then each $A \subset X$ which is $\sigma$-porous in
direction $E+F$ can be written in the form $A = A_1 \cup A_2$, where
$A_1$ is $\sigma$-porous in direction $E$ and $A_2$ is $\sigma$-porous
in direction $F$.
\end{proposition}
\begin{proposition}\label{kondim}
Let $X$ be a normed linear space and let
its finite dimensional subspace $V$ be spanned by
$v_1, \dots ,v_n$. Let $A \subset X$ be $\sigma$-porous in
direction $V$. Then we can write $A = \bigcup_{k=1}^{n} (A_k^+ \cup
A_k^-)$, where $A_k^+,A_k^-$ are sets $\sigma$-porous in directions
$v_k,-v_k$, respectively.
\end{proposition}
\pf
We shall proceed by induction. For $n=1$ we infer the statement from
Lemma~\ref{zakl}, (i) and (ii).
Assume now that $n>1$ and the statement holds with $n$ replaced by $n-1$.
Suppose that $V$, $v_1,\dots,v_n$ , and $A$ are as in the assumptions of the
proposition, and that, without loss of generality,
$v_1,\dots,v_n$ are linearly independent. Since $V=E+F$, where $E =
{\rm span \,}\{v_1,\dots,v_{n-1}\}$ and $F={\rm span \,} \{v_n\}$,
and the pair $E,F$ is clearly
admissible, Proposition \ref{PropEF} and the induction assumption
easily imply the assertion.
\qed
Now we are ready to deduce the announced decomposition results for
directionally porous sets.
\begin{proposition}\label {1*}
Let $V= \overline{{\rm span \,}\{v_1,v_2,\dots\}}$ be a subspace of a
normed linear space $X$. Let $A \subset X$
be a set such that for each point $a \in A$ there exists $v_a \in V$
such that $A$ is porous at $a$ in direction $v_a$. Then
we can write $A = \bigcup_{n=1}^{\infty} (A_n^+ \cup A_n^-)$, where
$A_n^+,A_n^-$ are sets $\sigma$-porous in directions $v_n,-v_n,$
respectively.
\end{proposition}
\pf
We may and will suppose $\|v_a\| = 1$.
Let $A_{k,n}$ be the set of all points $a \in A$, at which $A$ is
$\frac{1}{k}$-porous in direction $v_a$ and
$$ {\rm dist \,}(v_a, {\rm span \,} \{v_1,\dots,v_n\} \cap \{v\in X:
\|v\|=1\}) <
\frac{1}{2k}.$$
It is easy to see that $A = \bigcup_{k,n=1}^{\infty} A_{k,n}$.
Lemma~\ref{zakl}, (iii)
(applied to $V:= \{v_a\}$,\ $W:= {\rm span \,} \{v_1,\dots,v_n\}
\cap \{v\in X: \|v\|=1\},\ c:= \frac{1}{k}, \ \omega:= \frac{1}{2k})$
easily implies that $A_{k,n}$ is porous in direction
${\rm span \,} \{v_1,\dots,v_n\}$.
Thus Proposition~\ref{kondim} immediately gives our
assertion.
\qed
The following theorem captures the most interesting special case
of Proposition \ref{1*}.
\begin{theorem}\label{dec}
Let $X$ be a separable normed linear space and let $(v_n)_1^{\infty}$ be
a complete sequence in $X$ (i.e., $\overline{{\rm span \,}\{v_1,\dots \}}= X$).
Let $A \subset X$ be a $\sigma$-directionally porous set.
Then
we can write $A = \bigcup_{n=1}^{\infty} (A_n^+ \cup A_n^-)$, where
$A_n^+,A_n^-$ are sets $\sigma$-porous in directions $v_n,-v_n,$
respectively.
\end{theorem}
\begin{remark}\label{borel}
We can demand in Proposition \ref{1*} and Theorem~\ref{dec}
that the sets $A_n^+ ,A_n^-$ are Borel if $A$ is Borel. In fact,
on account of Lemma~\ref{obal},(v) we see that they can be replaced
by sets $(A_n^+)^* \cap A$ and $(A_n^-)^* \cap A$. This observation
immediately implies that each Borel $\sigma$-directionally porous subset
of a separable Banach space is null in the Aronszajn sense - the fact
which was deduced in Introduction from Aronszajn's differentiability
theorem.
\end{remark}
\begin{remark}\label{aplzv}
Our method of proof of the decomposition results,
based on Lemma \ref{EF}, works also in cases when
we want to distinguish between the ``one-sided'' and the ``bilateral''
directional porosity in Banach spaces; in such cases we may apply
Proposition \ref{PropEF}.
In certain other circumstances,
it is possible to use Proposition \ref{beta} instead of
Lemma \ref{EF}. For example, Proposition~\ref{kondim}
may be easily obtained by using
Proposition~\ref{beta} in $(X_1 \times \dots \times
X_n) \times Y$, where $X_1 \times \dots \times X_n$ is equipped with the sum
metric, and a generalization of Lemma~\ref{smetric}.
This provides an alternative approach to the proof of Theorem \ref{dec}.
\end{remark}
We will now strengthen Theorem \ref{dec} by showing that the sets $A_n^+,
A_n^-$
can be chosen to be even ``more $\sigma$-porous'' in the sense
of the following definition. This application of
Proposition \ref{beta} is different from that
indicated in the preceding remark, since our arguments use Proposition
\ref{beta} together
with Theorem \ref{dec}.
\begin{definition}\label{H}
Let $X$ be a normed linear space and let $a,v \in X$.
If $c,\gamma > 0$ and $A \subset X$ are given, then we say that $A$ is
$[c,\gamma]$-porous at $a$ in direction $v$ if for each $\varepsilon > 0$
there exist $t\geq 0$ and $r,s > 0$ such that $\|tv\| < \varepsilon$,
$r > ct, s
> \gamma \|tv\|$ and
\begin{equation}\label{Ha}
B(a+(t+\tau)v,s) \cap A = \emptyset \ \mbox{for each} \
-r \leq \tau \leq r.
\end{equation}
We say that $A$ is $[c,\gamma]$-porous in direction $v$ if $A$
is $[c,\gamma]$-porous at each of its points.
\end{definition}
\begin{theorem}\label{theorem1**}
Let $X$ be a separable normed linear space, $0 0$, an index $n$ and $v^k \in X$ such that $v^k \in
\{v_n,-v_n\}$ and the set $C_k$ is $[c,\gamma_k]$-porous in direction
$v^k$.
\end{theorem}
\pf
We may assume that $v_n \neq 0$ for each $n$. On account of
Theorem~\ref{dec}
and Lemma~\ref{obal},(vi) we may suppose that $A$ is $c$-porous in
direction $v$, where $c > 0$ and $v=v_n$ or $v=-v_n$ for some $n$.
Let $P$ be a norm one projection of $X$ onto $V:={\rm span \,} \{ v\}$,
and let $W$ be the kernel of $P$.
Put $\alpha = \frac{c+1}{2}=c + \frac{1-c}{2}$.
For any $a = a_1 +a_2$,
where $a_1 \in V$ and $a_2 \in W$, and for any $r>0$ the ball
$B(a,r)$ contains the sum of balls $B_V(a_1, r/2)$ and $B_W(a_2, r/2)$,
where the symbols $B_V$ and $B_W$ are used for balls in $V$ and $W$,
respectively. Thus we infer that, in the natural identification of
$X=V\oplus W$ with $V\times W$,
the set $A$ is $V$-directionally $(c/2,c/2)$-porous,
and we conclude from Proposition~\ref{beta} that there is $\beta>0$ such that
$A = \bigcup_{n=1}^{\infty} A_n$ for
some $V$-directionally $(\alpha,\beta)$-porous sets $A_n$.
Fix an index $n$ and choose arbitrary $a \in A_n$ and $\varepsilon > 0$.
Let $a = a_1 +a_2$, where $a_1 \in V$ and $a_2 \in W$. Since
$A_n$ is $V$-directionally $(\alpha,\beta)$-porous, we can find
$b \in V$, $r^* > \alpha \|b-a_1\|$ and $s^* > \beta \|b-a_1\|$
such that $\|b-a_1\| < \varepsilon$
and $ (B_V(b,r^*) \times B_W(a_2,s^*)) \cap A_n
= \emptyset$.
Now let $t \ne 0$ be such that $b-a_1 = tv$.
Further put
$$ \gamma= \min (\frac{1-c}{8},\frac{\beta}{4}), \ r=
(\frac{3c+1}{4})|t| \ \mbox{and}\ s= 2 \gamma \|tv\|.$$
Then clearly \ $\| tv\| < \varepsilon$, $r > c|t|$ and $s > \gamma \|tv\|$;
so, to finish the proof, it suffices to show that
$B(a+(t+\tau)v,s) \subset B_V(b,r^*) + B_W(a_2,s^*)$
whenever $-r\le \tau\le r$. But this is clear, because
for any $-r\le \tau\le r$
and $z\in B(a+(t+\tau)v,s)$ we have that $z=P(z)+(z-P(z))$,
where $P(z)\in V$, $z-P(z)\in W$,
$P(z)\in B_V(b,r^*)$ since
$$\|P(z)-b\|\le \|\tau v\|+\|z-(a+(t+\tau)v)\|
< (\frac{3c+1}{4}+2\gamma )\|tv\|\le\alpha \|b-a_1\|~~