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\title{Spaceability and algebrability of the set of non strongly McShane (product) integrable functions} 
\author{Fatemeh Farmanesh and Ali Farokhinia\\{\small Department of Mathematics, Shiraz Branch, Islamic Azad University, Shiraz, Iran}}
\date{}
\begin{document}
\maketitle
\begin{abstract}
Let $X$ be a unitial Banach algebra, $\mathcal{G}$ be the set of functions $f:[0,1]\to X$ that are non strongly McShane integrable and $\mathcal{L}$ be the set of functions $f:[0,1]\to [0,\infty)$ that are non strongly McShane integrable. The aim of this paper is to show that there is an infinite dimensional closed vector space in $\mathcal{G}\cup{0}$ and a $\mathfrak{c}$-generated algebra in $\mathcal{L}$ isomorphic with a free algebra.

\end{abstract}

% ----------------------------------------------------------------
\section{Introductions}

In this paper we investigate large linear and algebraic structures within the set of functions with special integrability properties. The appropriate terminology are lineability, spaceability and algebrability. These terminologies were considered by many authors (see e.g. \cite{AGS,AS,BG,BGP,F1,F2}), whose definitions are as follow:  

%This work is a contribution to search for large linear and algebraic structures within the set of functions which satisfy special properties. These structures are often called lineablity, spaceablity and algebrablity. These terminologies were considered by many authors (see e.g. \cite{1,2,5,6,8}). Definitions follow:

\begin{definition}
Suppose that $\kappa $ is a cardinal number.   
\begin{itemize}
\item Let $\mathcal{L}$ be a vector space and $A$ be a subset of $\mathcal{L}$. We say that $A$ is \textit{$\kappa $-lineable} if $A\cup \{0\}$ contains a $\kappa $-dimensional vector space;
\item Let $\mathcal{L}$ be a Banach space and $A$ be a subset of $\mathcal{L}$. We say that $A$ is \textit{spaceable} if $A\cup \{0\}$ contains an infinite dimensional closed vector space;
\item Let $\mathcal{L}$  be a linear commutative algebra and $A$ be a subset of $\mathcal{L}$. We say that $A$ is \textit{$\kappa $-algebrable} if $A\cup \{0\}$ contains a $\kappa $-generated algebra $B$ (i.e., the minimal system of generators of $B$ has cardinality $\kappa $);
\item Let $\mathcal{L}$  be a linear commutative algebra and $A$ be a subset of $\mathcal{L}$. We say that $A$ is \textit{strongly $\kappa $-algebrable} if $A\cup \{0\}$ contains a $\kappa $-generated algebra $B$ that is isomorphic with a free algebra (denote by $X=\left\{ {{x}_{\alpha }}:\alpha <\kappa \right\}$ the set of generators of this free algebra); 
\item The set $X=\left\{ {{x}_{\alpha }}:\alpha <\kappa \right\}$ is a generating set of some free algebra contained in $A\cup \{0\}$ if and only if the set $\overset{\sim }{\mathop{X}}\,$  
of elements of the form $x_{{{\alpha }_{1}}}^{{{k}_{1}}}x_{{{\alpha }_{2}}}^{{{k}_{2}}}...x_{{{\alpha }_{n}}}^{{{k}_{n}}}$ is linearly independent and all linear combinations of elements from $\overset{\sim }{\mathop{X}}\,$ are in $A\cup \{0\}$; equivalently for any $\kappa\in\mathbb{N},$ any nonzero polynomial $P$ in $\kappa$ variables without a constant term and any distinct ${y_{1},y_{2},...,y_{k}}\in X,$ we have $P({{y}_{1}},...,{{y}_{k}})\ne 0$ and $P({{y}_{1}},...{{y}_{k}})\in A.$ 
\end{itemize}
\end{definition}
So far a variety of sets of functions have turned out to enjoy some of the above properties. For example the set of continuous nowhere differentiable functions on [0,1] is lineable \cite{G}, and also spaceable \cite{FGK}, the set of continuous nowhere H\"older functions is $\mathfrak{c}$-algebrable, the set of differentiable functions on $\mathbb{R}$ which are nowhere monotone is $\mathfrak{c}$-lineable and so on. For a good survey about large algebraic structures see \cite{GPS}.

The present paper deals with strongly McShane (product) integrable functions (see definitions 1.4 and 1.5) from the point of view of large algebraic structures. In fact we show that the set of non strongly McShane (product) integrable functions is spaceabl as well as strongly $\mathfrak{c}$-algebrable.

We need the following notation.

A \textit{tagged partition} of an interval $[a,b]$ is a collection of point-interval pairs $D=({{\xi }_{i}},[{{t}_{i-1}},{{t}_{i}}])_{i=1}^{m}$, where $a={{t}_{0}}\le {{t}_{1}}\le ...\le {{t}_{m}}=b$ and ${{\xi }_{i}}\in [{{t}_{i-1}},{{t}_{i}}]$ for every $i\in \left\{ 1,2,...,m \right\}$.  

%Anothere notion that we will be used in this paper is product integrability. The notion of product integral has been introduced by Vito Volterra, about the end of the 19th century (see \cite{19,20}); 
%Let a tagged partition of an interval $[a,b]$ is a collection of point-interval pairs $D=({{\xi }_{i}},[{{t}_{i-1}},{{t}_{i}}])_{i=1}^{m}$, where $a={{t}_{0}}\le {{t}_{1}}\le ...\le {{t}_{m}}=b$ and ${{\xi }_{i}}\in [{{t}_{i-1}},{{t}_{i}}]$ for every $i\in \left\{ 1,2,...,m \right\}$. We refer to ${{t}_{0}},{{t}_{1}},...,{{t}_{m}}$ as the division points of $D$, while ${{\xi }_{1}},{{\xi }_{2}},...,{{\xi }_{m}}$ are the tags of $D$. Now consider a matrix function $A:[a,b]\to {{\mathbb{R}}^{n\times n}}$ with entries $\left\{ {{a}_{ij}} \right\}_{i,j=1}^{n}$. For every tagged partiton $D:a={{t}_{0}}\le {{\xi }_{1}}\le {{t}_{1}}\le ...\le {{t}_{m-1}}\le {{\xi }_{m}}\le {{t}_{m}}=b$ of interval $[a,b]$ with division points ${{t}_{i}}$ and tags ${{\xi }_{i}}$, denote
%\begin{equation*}
%\Delta\ {{t}_{i}}={{t}_{i}}-{{t}_{i-1}} ~~,i=1,2,...,m ,~~ \upsilon (D)=\underset{1\le i\le m}{\mathop{\max \Delta {{t}_{i}}}},
%\end{equation*}
%and put 
%\begin{align*}
%P(A,D)&=\prod\limits_{i=m}^{1}{(I+A({{\xi }_{i}}})\Delta {{t}_{i}})\\
%&=(I+A({{\xi }_{m}})\Delta {{t}_{m}})(I+A({{\xi }_{m-1}})\Delta {{t}_{m-1}})....(I+A({{\xi }_{1}})\Delta {{t}_{1}}).
%\end{align*}
%\begin{definition}
%Consider a function $A:[a,b]\to {{\mathbb{R}}^{n\times n}}.$ If $\underset{\upsilon (D)\to 0}{\mathop{\lim P(A,D)}}\,$ exists, it is called the (left) product integral of the function $A$ on interval $[a,b]$ and denoted by 
%$\prod\limits_{a}^{b}{}(I+A(t)dt).$
% \end{definition}
%Volterra showed that the indefinit product integral $Y(t)=\prod\limits_{a}^{t}{(I+A(s)ds)}$ is the solution of the linear system of differential equation of the form
%\begin{align*}
%Y'(t)&=A(t)Y(t) , t \in [a,b] \\
%Y(a)&=I,
%\end{align*}
%where $I$ is the identity matrix, $A:[a,b]\to {{\mathbb{R}}^{n\times n}}$ is a given continuous function and $Y:[a,b]\to {{\mathbb{R}}^{n\times n}}$ is the unknown function. He also observed that this procedure still works if $A$ is a Riemann integrable function and in this case we have:
%\begin{align*}
%Y(t)=I+\int\limits_{a}^{t}{A(s)Y(S)ds},~~~t\in[a,b],
%\end{align*}
%where the integral on the right-hand side is the Riemann integral, and hense $Y'(t)=A(t)Y(t)$ a.e., on $[a,b].$ 
%After that the concept of product integrale was extended to Lebesgue, Bochner, Kurzweil/Henstock, strong Kurzweil/Henstock, McShane and strong McShane integrable functions. We reffer the interested reader to \cite{13,15,16,17,18} for a wider range of results in product integrability.   
%\begin{remark}
If we replace ${{\xi }_{i}}\in [{{t}_{i-1}},{{t}_{i}}]$ by ${{\xi }_{i}}\in [a,b]$, then the collection $D$ is called a \textit{free tagged} partition. Given a function $\delta :[a,b]\to {{\mathbb{R}}^{+}}$ (called a gauge on $[a,b]$), a free tagged partition is called $\textit{$\delta$-fine}$ if
\begin{align*}
[{{t}_{i-1}},{{t}_{i}}]\subset ({{\xi }_{i}}-\delta ({{\xi }_{i}}),{{\xi }_{i}}+\delta ({{\xi }_{i}})),~~~~i=\{1,2,...,m\}.
\end{align*}
%\end{remark}
In the rest of this paper, we assume that $X$ is a real and unital Banach algebra with infinite dimension, and for all $f:[a,b]\to X$, we define 
 ${{\left\| f \right\|}_{X}}(x):={{\left\| f(x) \right\|}_{X}}$, for each $x\in [a,b]$.
\begin{definition}[{\cite[Theorem 1.3.11]{S2}}]
A function $A:[a,b]\to X$ is called \textit{Bochner integrable} if there is a sequence of simple functions $A_{n}:[a,b]\to X$, $n\in\mathbb{N}$ such that
\begin{align*}
 \underset{n\to \infty }{\mathop{\lim }}\,{{A}_{n}}(x) 
=A(x) \text {~~a.e.,~~ on~~} [a,b],
\end{align*}
 and
\begin{align*}   
\underset{n\to \infty }{\mathop{\lim }}\,\int\limits_{a}^{b}{{{\left\| {{A}_{n}}-A \right\|}_{X}}}=0.
\end{align*}

\end{definition}
\begin{definition}[{\cite{S3}}]
A function $A:[a,b]\to X$ is called \textit{Henstock-Kurzweil integrable} if there exists a vector ${{S}_{f}}\in X$ with the following property: For each $\epsilon>0$ there is a gauge $\delta :[a,b]\to {{\mathbb{R}}^{+}}$ such that 
\begin{align}
{{\left\| \sum\limits_{i=1}^{m}{f({{\xi }_{i}})({{t}_{i}}-{{t}_{i-1}})-{{S}_{f}}} \right\|}_{X}}<\varepsilon,
\end{align}
for every $\delta$-fine tagged partition of $[a,b]$. In this case, ${{S}_{f}}\in X$ is called the \textit{Henstock-Kurzweil integral} of $f$ over $[a,b],$ and is denoted by $\int\limits_{a}^{b}{f(t)dt}$. If (1) holds for all $\delta$-fine free tagged partitions of $[a,b]$, then $f$ is called the \textit{McShane integrable} over $[a,b]$. 
\end{definition}
%Note that if the gauge $\delta$ is assumed to be constant on $[a,b]$, then the above integral is a Riemann integral.
%J. Jarn\'ik and J. Kurzweil in \cite{JK} defined a Kurzweil product integral as Perron product integral and pointed out the important properties in the finite dimensional space $X={{\mathbb{R}}^{n\times n}}$ wich play a key role in their proof that $W'(t)=A(t)W(t)$ a.e., on $ [a,b] $. Howevere, the original proof of these properties is no longer applicable in infinite dimensional space, (for more details see \cite[Lemma 2.2 and Theorem 2.4]{JK} ). But one can overcome this problem by working with the \textit{strong Henstock-Kurzweil} and \textit{strong McShane} integral instead. These facts provide a motivation for introducing the strong Kurzweil and McShane product integrals (see \cite{S4}). Let us recall some definitions.
\begin{definition}[{\cite{S3}}]
A function $A:[a,b]\to X$ is called \textit{strongly Henstock-Kurzweil integrable} if there is a function $B:[a,b]\to X$ such that for each $\epsilon>0$ there is a gauge $\delta :[a,b]\to {{\mathbb{R}}^{+}}$ such that
\begin{align} 
\sum\limits_{i=1}^{m}{{{\left\| A({{\xi }_{i}})({{t}_{i}}-{{t}_{i-1}})-(B({{t}_{i}})-B({{t}_{i-1}})) \right\|}_{X}}}<\varepsilon,
\end{align}
for every $\delta$-fine tagged partition of $[a,b]$. If (2) holds for all $\delta$-fine free tagged partitions of $[a,b]$, then $f$ is called \textit{strongly McShane integrable} over $[a,b]$.
\end{definition}
\begin{definition}[{\cite[definition 3.4]{S3}}]
A function $A:[a,b]\to X$ is called \textit{strongly Kurzweil product integrable} if there is a function $W:[a,b]\to X$ such that $W(t)^{-1}$ exists for all $t\in[a,b]$, both $W$ and $W^{-1}$ are bounded, and for every $\epsilon>0$, there is a gauge $\delta :[a,b]\to {{\mathbb{R}}^{+}}$ such that
\begin{align}
\sum\limits_{i=1}^{m}{{{\left\| I+A({{\xi }_{i}})({{t}_{i}}-{{t}_{i-1}})-(W({{t}_{i}})W({{t}_{i-1}})^{-1}) \right\|_X}}}<\varepsilon,
\end{align}
for every $\delta$-fine tagged partition of $[a,b]$. In this case, we define the \textit{strongly McShane product integral} as 
$\prod\limits_{a}^{b}{{}}(I+A(t)dt)=W(b)^{-1}W(a).$


If (3) holds for all $\delta$-fine free tagged partitions of $[a,b]$, then $A$ is called \textit{strongly McShane product integrable} over $[a,b],$ and is defined as $\prod\limits_{a}^{b}{{}}(I+A(t)dt)=W(b)^{-1}W(a).$ 
\end{definition}
% ----------------------------------------------------------------
\section{Spaceability and algebrability of sets of non strongly McShane (product) integrable functions}
Let $\mathcal{G}$ be the set of functions $f:[0,1]\to X$ that are non strongly McShane (product) integrable, and $\mathcal{L}$ be the set of functions $f:[0,1]\to [0,\infty)$ that are non strongly McShane (product) integrable. Our aim is to show that the set $\mathcal{G}$ is spaceable and the set $\mathcal{L}$ is strongly $\mathfrak{c}$-algebrable. In order to do this, let us recall some theorems that will be needed, and use the idea from \cite{GPKP1, GPKP2} to produce the assertions.
%\end{defn}
\begin{theorem}[{\cite[Theorem 5.1.4]{S2}}]
A function $f:[a,b]\to X$ is Bochner integrable if and only if $f$ is strongly McShane integrable.
\end{theorem}
\begin{theorem}[{\cite[Theorem 4.14]{S3}}]
For every function $f:[a,b]\to X$, $A$ is strongly McShane product integrable if and only if $f$ is Bochner integrable.
   
\end{theorem}
\begin{theorem}[{\cite[Theorem 1.4.3]{S2}}]
A measurable function $f:[a,b]\to X$ is Bochner integrable if and only if ${{\left\| f \right\|}_{X}}:[a,b]\to [0, \infty)$ is Bochner integrable.
\end{theorem}
%Note that for ${{\left\| f \right\|}_{X}}:[a,b]\to [0,\infty )$, we define ${{\left\| f \right\|}_{X}}(x):={{\left\| f(x) \right\|}_{X}}$, for each $x\in [a,b]$.


The next proposition is a usefull technique of proving strong McShane integrability of functions. Assume that $\{{{z}_{k}}\}_{k\in \mathbb{N}}^{{}}$ is a sequence in $X$ and define the function $f:[0,1]\to X$ as follows:
\begin{align}
f:=\sum\limits_{k=1}^{\infty }{\chi_{(\frac{1}{2^k},\frac{1}{2^{k-1}}]}}{z_k},~~ f(0)=0.
\end{align}
\begin{proposition}[{\cite[Proposition 5.4.1]{S2}}]
If ${{\left\| \frac{1}{2^k}{z_k}\right\|}_X}<B$ for all $k\in \mathbb{N}$ and $B>1,$ then the function $f:[0,1]\to X$ given by (4) is strongly McShane integrable if and only if the series $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z}_{k}}$ is absolutely convergent.
\end{proposition}

We also need the following.
%\begin{remark}


Assume that  $\{{{x}_{n}}\}$ is a sequence in the Banach spase $X$. Then $\{{{x}_{n}}\}$ is said to be a \textit{basic sequence} whenevere for each vector $x\in \overline{\Span}\{{{x}_{n}}:n\in \mathbb{N}\}$, we can find a unique sequence $\{{{a}_{n}}\}$ of scalars such that $\sum\limits_{n=1}^{\infty }{{{a}_{n}}{{x}_{n}}}=x$. The last equality means that $\left\| \sum\limits_{n=1}^{N}{{{a}_{n}}{{x}_{n}}-x} \right\|_X\to 0$ as $N\to \infty .$ In other words, $\{{{x}_{n}}\}$ is the \textit{Schauder basis} of the subspace $\overline{\Span}\{{{x}_{n}}:n\in \mathbb{N}\}$.
%\end{remark}
\begin{lemma}[{\cite[Lemma 2.1]{BO}}]
Let $E$ be a Banach space and $\{{{x}_{n}}\}\subseteq E\backslash \{0\}$. The following properties are equivalent:
\begin{itemize}
\item[1.]$\{{{x}_{n}}\}$ is a  basic sequence.
\item[2.]There is a constant $ 0<C<\infty$ such that for every pair $r,s\in \mathbb{N} $ with $s\ge r$ and every finite sequence of scalars ${{a}_{1}},...,{{a}_{s}}$, one has $\left\| \sum\limits_{n=1}^{r}{{{a}_{n}}{{x}_{n}}} \right\|\le C\left\| \sum\limits_{n=1}^{s}{{{a}_{n}}{{x}_{n}}} \right\|$.
\end{itemize}

\end{lemma}
%Now for each $k\in\mathbb{N}$, put $B_k=(\frac{1}{2^k},\frac{1}{2^{k-1}}]$. Then for each $j\ne k$, $m(B_i\cap B_k)=0$,  and $m({\cup _k}{B_k})=1$, where $m$ denotes Lebesgue measure.
 Assume that $\{{{z}_{k}}\}_{k\in \mathbb{N}}^{{}}$ is a sequence in $X$ and $f:[0,1]\to X$ is the function given by (4) with the following property: If ${{\left\| \frac{1}{2^k}{z_k}\right\|}_X}<B$ for all $k\in \mathbb{N}$ and $B>1,$ then the series $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z}_{k}}$ is non absolutely convergent. 

Thus due to Theorem 2.4 the function $f$ is non strongly McShane integrable and hence by Theorem 2.1 and Theorem 2.2, $f$ is non strongly McShane (product) integrable. 
%( Note that by Theorem 2.3, without loose of generality we can assume that $f:[0,1]\to \mathbb{R}^{+}$). 
Now suppose that for each $j\in\mathbb{N}$, ${{(n_{k}^{j})}_{k}}$ is a strictly increasing subsequence of $\mathbb{N}$ such that for all natural numbers $i,j$ with $i\ne j,$ we have ${n_{k}^{i}}\ne{n_{m}^{j}},$ for all $k,m\in \mathbb{N},$
and that the series $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{{n_{k}^{j}}}}}}{{z}_{{n_{k}^{j}}}}$ is not absolutely convergent.
% and for different j the related subsequences are disjoint.
 Now for each $k\in\mathbb{N}$ put $B_k=(\frac{1}{2^k},\frac{1}{2^{k-1}}]$. Then for all $j\ne k$, $m(B_i\cap B_k)=0$,  and $m({\cup _k}{B_k})=1$, where $m$ denotes the Lebesgue measure. For each $j$ assume ${{f}_{j}}:=\sum\limits_{k=1}^{{\infty}}{{{\chi }_{{{B}_{n_{k}^{j}}}}}{{z}_{n_{k}^{j}}}}$, thus ${{f}_{j}}\in\mathcal{G}$, moreover by the disjointness of the supports, ${{\{{{f}_{j}}\}}_{j}}$ is linearly independent in $X$. (We do the same for $f:[0,1]\to [0,\infty)$ and $\mathcal{L}$).
% Note that by Theorem 1.8 without loose of generality we can assume that $f:[a,b]\to \mathbf{R}^{+}$.

%Note that for ${{\left\| f \right\|}_{x}}:[0,1]\to [0,\infty )$, we define ${{\left\| f \right\|}_{x}}(x):={{\left\| f(x) \right\|}_{x}}$.
\begin{theorem}
${{\overline{\Span({{\{{{f}_{j}}\}}_{j}})}}}\subset \mathcal{G}\cup \{0\}$. In particular, $\mathcal{G}$ is spaceable. 
\end{theorem}
 \begin{proof}
 Let ${{m}_{{}}},n\in \mathbb{N}$ such that $m<n$, and ${{a}_{1}},...,{{a}_{n}}\in \mathbb{R}$, so $\left\| \sum\limits_{j=1}^{m}{{{a}_{j}}{{f}_{j}}} \right\|_X\le \left\| \sum\limits_{j=1}^{n}{{{a}_{j}}{{f}_{j}}} \right\|_X,$ and hence ${{({{f}_{j}})}_{j}}$ is a basic sequence, and therefore a Schauder basis for ${{\overline{\Span({{\{{{f}_{j}}\}}_{j}})}}}$. Noticing that the ${{\left\{ {{B}_{n_{k}^{j}}} \right\}}_{k}}$ are pairwise disjoint, we get:
 \begin{align*}
 {{\left\| \sum\limits_{j=1}^{m}{{{a}_{j}}{{f}_{j}}} \right\|}_{X}}&=\sup \left\{{{\left\| \sum\limits_{j=1}^{m}{{{a}_{j}}{{f}_{j}}}(x) \right\|}_{X}}: x \in [0,1]\right\}\\
 &=
 \sup \left\{{{\left\| \sum\limits_{k=1}^{{\infty}}{\sum\limits_{j=1}^{m}{{{a}_{j}}{{z}_{n_{k}^{j}}}{{\chi }_{_{{{B}_{n_{k}^{j}}}}}}(x)}} \right\|}_{X}}: x \in [0,1]\right\}\\
 &\le  \sup \left\{{{\left\| \sum\limits_{k=1}^{{\infty}}{\sum\limits_{j=1}^{n}{{{a}_{j}}{{z}_{n_{k}^{j}}}{{\chi }_{_{{{B}_{n_{k}^{j}}}}}}(x)}} \right\|}_{X}}: x \in [0,1]\right\}\\
  &=\sup \left\{{{\left\| \sum\limits_{j=1}^{n}{{{a}_{j}}{{f}_{j}}}(x) \right\|}_{X}}:x\in [0,1]\right\}= {{\left\| \sum\limits_{j=1}^{n}{{{a}_{j}}{{f}_{j}}} \right\|}_{X}}.
 \end{align*}
 It follows by Lemma 2.6 that, for each $g\in{{\overline{\Span({{\{{{f}_{j}}\}}_{j}})}}}$ with $g\ne 0$, there is a nonzero sequence ${{({{\alpha }_{j}})}_{j}}$ of real numbers satisfying $g=\sum\limits_{j=1}^{{\infty}}{{{\alpha }_{j}}{{f}_{j}}}$. The proof is now complete. 
 \end{proof}
 Note that by the definition of lineablity and spaceability, we can conclude the lineability of $\mathcal{G}$.  
%\begin{theorem}[{\cite[Theorem 3.41]{R}}]
%Given two sequences ${{\{a_n}\}}$, ${{\{b_n}\}}$, put ${{A}_{n}}=\sum\limits_{k=0}^{n}{{{a}_{k}}}.$ If $n\ge 0$; put $A_{-1}=0$. Then, if $0\le p\le q$, we have
%\begin{align}
%\sum\limits_{n=p}^{q}{{{a}_{n}}{{b}_{n}}=\sum\limits_{n=p}^{q-1}{{{A}_{n}}({b_n}-{b_{n+1}})+}}{{A_q}{b_q}-{{{A}_{p-1}}{{b}_{p}},}}
%\end{align}
%and the last expression on the right is clearly equal to the right side of (5).
%\end{theorem}
%Note that by this theorem we have the following:
%\begin{align}
%\sum\limits_{n=p}^{q}{{{a}_{n}}{{b}_{n}}}=\sum\limits_{n=p}^{q}{({{A}_{n}}-{{A}_{n-1}}){{b}_{n}}}=\sum\limits_{n=p}^{q}{{{A}_{n}}{{b}_{n}}}-\sum\limits_{n=p-1}^{q-1}{{{A}_{n}}{{b}_{n+1}}}.
%\end{align}
%Now if for each $n\in \mathbb{N},$ $a_n\ge 0$ and $b_n\ge 0$, by (5) we have
%\begin{align}
%\sum\limits_{n=p}^{q}{{{a}_{n}}{{b}_{n}}\ge \sum\limits_{n=p}^{q}{{{A}_{n}}{{b}_{n}}=\sum\limits_{n=p}^{q}{({{a}_{1}}+...+{{a}_{n}}){{b}_{n}}\ge \sum\limits_{n=p}^{q}{{{a}_{0}}{{b}_{n}}.}}}}
%\end{align}


 Now we are ready to show the \textit{strong algebrability} of $\mathcal{L}$.  Let $f$ and $\{{{z}_{k}}\}_{k\in \mathbb{N}}^{{}}$ be as above for $[0,\infty)$ in place of $X$. Due to Theorems 2.1, 2.2, 2.3 and Proposition 2.4 the function $f$ is non strongly McShane (product) integrable. 
 %and hense by Theorem 2.3 without loss of generality we can assume that $f:[0,1]\to [0,\infty)$.
Now for each $\alpha <\mathfrak{c}$
we choose a subsequense ${{\left( z_{\alpha,k } \right)}_{k\in \mathbb{N}}}$ of ${{\left( z_{k} \right)}_{k\in \mathbb{N}}}$ such that $\left\{ \ln ({{ z_{\alpha,k } })}:\alpha <\mathfrak{c} \right\}$ is linearly independent over the rational numbers and for each $k\in \mathbb{N},$ $( {{z}_{\alpha ,k}} )\ge 1$. For each $\alpha<\mathfrak{c}$ put ${{l}_{\alpha }}:=\sum\limits_{k=1}^{\infty }{{{z }_{\alpha,k }}}{{\chi }_{{{B}_{k}}}}$.  Note that for each $\alpha<\mathfrak{c}$, $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z_{\alpha,k }}}$ is not absolutely convergent.
In the proof of the following theorem we have inspired by the proof of Theorem 2.5 \cite{GPKP2}.
\begin{theorem}
$\{{l_\alpha:\alpha<\mathfrak{c}\}}$ is a set of free generators, and the algebra generated by this set is contained in $\mathcal{L}\cup\{0\}$. In particular, $\mathcal{L}$ is strongly $\mathfrak{c}$-algabrable.
\end{theorem}
%Note that by a definition of lineablity and spaceability, we can conclude the lineability of $\mathcal{G}$.  
\begin{proof}
Let $(r_{ij}:i=1,..m,j=1,..n)$ be an arbitrary matrix with non-negative entries and non-zero and distinct rows. It suffices to show that, 
for all $\alpha_1,...,\alpha_n<\mathfrak{c}$ and for all $\beta_1,...,\beta_m\in\mathbb{R}$ which do not vanish simultaneously, the function 
\begin{align*}
l:={{\beta }_{1}}l_{{{\alpha }_{1}}}^{{{r}_{11}}}...l_{{{\alpha }_{n}}}^{{{r}_{1n}}}+\cdots+{{\beta }_{m}}l_{{{\alpha }_{1}}}^{{{r}_{m1}}}...l_{{{\alpha }_{n}}}^{{{r}_{mn}}},
\end{align*}
is in $\mathcal{L}$. First we note that the numbers 
$\ln{{( z_{{\alpha _1},k}^{r_{i1}}...{ z_{{\alpha_n},k}^{r_{in}}} )}}$ are dinstinct for $i=1,...,m$. Indeed
\begin{align*}
\ln{{( z_{{\alpha _1},k}^{r_{i1}}...{ z_{{\alpha_n},k}^{r_{in}}} )}}={{r}_{i1}}\ln {{( z_{{\alpha _1},k}^{r_{i1}}{{}} )}}+\cdots+{{r}_{in}}\ln {{( z_{{\alpha_n},k} )}},
\end{align*} 
and $\ln {{(z_{{{\alpha }_{1,j}}}^{{}} )}},...,\ln {{( z_{{{\alpha }_{n,j}}}^{{}} )}}$ are $\mathbb{Q}$-linearly independent.
 On the other hand the ${B_k}$ are pairwise disjoint. So for each $i=1,...,m$ and each $j=1,...,n$, we have 
\begin{align*}
{{\left( \sum\limits_{k=1}^{\infty }{{{z}_{{{\alpha }_{i,k}}}}{{\chi }_{{{B}_{k}}}}} \right)}^{{{r}_{ij}}}}=\sum\limits_{k=1}^{\infty }{z_{{{\alpha }_{i,k}}}^{{{r}_{ij}}}}{{\chi }_{B_k}}\,
\end{align*} 
 and hence,
\begin{align*}
l_{{{\alpha }_{1}}}^{{{r}_{i1}}}...l_{{{\alpha }_{n}}}^{{{r}_{in}}}={{\left( \sum\limits_{k=1}^{\infty }{{{z}_{{{\alpha }_{1,k}}}}{{\chi }_{{{B}_{k}}}}} \right)}^{{{r}_{i1}}}}...{{\left( \sum\limits_{k=1}^{\infty }{{{z}_{{{\alpha }_{n,k}}}}{{\chi }_{{{B}_{k}}}}} \right)}^{{{r}_{in}}}}
=\sum\limits_{k=1}^{\infty }{\overset{{{r}_{i1}}}{\mathop{{{z}_{{{\alpha }_{1}},k}}...}}\,}\overset{{{r}_{in}}}{\mathop{{{z}_{{{\alpha }_{n}},k}}}}{{\chi }_{B_k}}\,.
\end{align*}
To simplify the notation put $w_{i,k}:=z _{{{{\alpha }_{1}},k}}^{{{r}_{i1}}}...z _{{{\alpha }_{n}},k}^{{{r}_{in}}}$, so
\begin{align*}
l:=\sum\limits_{k=1}^{\infty }({{\beta }_{1}}w_{1,k}+\cdots+{{\beta }_{m}}w_{{m,k}}){{\chi }_{{B}_{k}}}.
\end{align*}
Since the logarithmic function is strictly monotone, we may assume that ${( w_{1,k})}>...>{( w_{m,k})}$, and also it is possible to assume that ${{\beta }_{1}}\ne 0$. Hence we can find ${k_0}\in \mathbb{N}$, such that 
\begin{align*}
\left| {{\beta }_{2}} \right|{{w }_{2,k}}+\cdots+\left| {{\beta }_{m}} \right|{{w }_{m,k}}<\frac{1}{2}\left| {{\beta }_{1}} \right|{{w }_{1,k}},
\end{align*}
for each $k\geq {k_0}$. Then for those $k$ we get:
\begin{align}
\left| {{\beta }_{1}}{{w }_{1,k}}+\cdots+{{\beta }_{m}}{{w }_{m,k}} \right|&\ge \left| {{\beta }_{1}} \right|{{w }_{1,k}}-\left| {{\beta }_{2}}{{w \notag }_{2,k}}+\cdots+{{\beta }_{m}}{{w }_{m,k}} \right|\\
&\ge \left| {{\beta }_{1}} \right|{{w }_{1,k}}-(\left| {{\beta }_{2}} \right|{{w}_{2,k}}+\cdots+\left| {{\beta }_{m}} \right|{{w }_{m,k}})>\frac{1}{2}\left| {{\beta }_{1}} \right|{{w }_{1,k}}.
\end{align}
On the other hand we assume that ${{z}_{{{\alpha }_{1}},k}}>1$, so $\overset{{{r}_{11}}}{\mathop{{{z}_{{{\alpha }_{1}},k}}}}\,>{{z}_{{{\alpha }_{1}},k}}$ and hence
\begin{align*}
\frac{{{w}_{1,k}}}{{{2}^{k}}}=\frac{z_{{{\alpha }_{1}},k}^{{{r}_{11}}}...z_{{{\alpha }_{n}},k}^{{{r}_{1n}}}}{{{2}^{k}}}\ge \frac{z_{{{\alpha }_{1}},k}^{{}}}{{{2}^{k}}}.
\end{align*}
But $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z}_{{{\alpha }_{1}},k}}$ is assumed not to be absolutely convergent (more precisely, it is divergent). Thus by using the comparison test and (5) %But we assumed that for each $k\in \mathbb{N},$ $\left\| {{z}_{\alpha ,k}} \right\|\ge 1$, and for each $\alpha<\mathfrak{c}$, 
 %put ${{g}_{\alpha }}=\sum\limits_{k=1}^{\infty }{{{z }_{\alpha,k }}}{{\chi }_{{{B}_{k}}}}$.  Note that for each $\alpha<\mathfrak{c}$,
%the series $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z_{\alpha,k }}}$ is assumed not to be absolutely convergent, so the proof of claim is complete. By (10)
 we conclude that the series $\sum\limits_{k=1}^{\infty }\frac{{{\beta }_{1}}{{w }_{1,k}}+\cdots+{{\beta }_{m}}{{w }_{m,k}}}{2^k}$ is not an absolutely convergent, and hence by Proposition 2.4, $l\in \mathcal{L}\cup \{0\}.$
%But for each $\alpha<\mathfrak{c}$, series $\sum\limits_{k=1}^{\infty }{\frac{1}{{{2}^{k}}}}{{z_{\alpha,k }}}$ is assumed to not absolutely convergent, so $\sum\limits_{k=1}^{\infty }\frac{{{\beta }_{1}}{{w }_{1}}+...+{{\beta }_{m}}{{w }_{m}}}{2^k}$ is not absolutely convergent series, and by Proposition 2.4, $g\in \mathcal{G}\cup \{0\}.$
\end{proof}
%\begin{remark}:  One can continue this trend of reserch by proving the $2^{\mathfrak{c}}$-dense algebrableity of $\mathcal{L}$ or $\mathcal{G}$ and the other question that can be asked here is that is the set of Kurzweil exponential product integrable functions algebrable? (Note that the concept of Kurzweil exponential product integrable functions was introduced in {\cite{S3} }). 
%\end{remark}
% --------------------------------------------------------------- 

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\end{document}