In the present section we investigate the Δ-limit point and Δ-cluster point concepts for a function defined on a time scale \(\mathbb{T}\). The results of this section coincide with the statistical limit point and the statistical cluster point in the case \(\mathbb{T}=\mathbb{N}\). In other words, these new notions give us a progression of generalizations of results for statistical limit points and statistical cluster points, introduced by Fridy in [8].
Definition 2.1
(Δ-Limit point)
A real number L is called a Δ-limit point of a function \(f:\mathbb{T}\rightarrow\mathbb{R}\) if there exists a subset K of \(\mathbb{T}\) with a non-zero Δ-density or if it does not have a Δ-density such that \(f(t)\rightarrow L\) whenever \(t\rightarrow\infty\) in K.
Note that in Definition 2.1 the measurable set K may have a positive Δ-density or may not have even a Δ-density. For describing this situation we will use the Δ-non-thin subset notation. This notation can be considered as a modified Fridy non-thin term defined for subsequences. Detailed information as regards the classical thin or non-thin concepts can be found in [8]. We proceed with the next definition.
Definition 2.2
(Δ-Cluster point)
A real number L is called a Δ-cluster point of a measurable function \(f:\mathbb{T}\rightarrow\mathbb{R}\) if for all \(\varepsilon>0\) the set \(\{t\in\mathbb{T}:|f(t)-L|<\varepsilon \}\) is a Δ-non-thin set.
We denote the set of Δ-limit points and Δ-cluster points of f by \(\Lambda_{f}\) and \(\Gamma_{f}\), respectively.
Definition 2.3
(Δ-Boundedness)
A measurable function \(f:\mathbb{T}\rightarrow \mathbb{R}\) is called Δ-bounded if there exists a real number r such that \(\delta_{\Delta}(\{ t\in\mathbb{T}: \vert f(t)\vert\leq r\}) =1\).
Definition 2.4
(Δ-Monotone increasing)
A function \(f:\mathbb{T}\rightarrow \mathbb{R}\) is called Δ-monotone increasing if there exists a subset K of \(\mathbb{T}\) with \(\delta_{\Delta}(K)=1\) such that f is monotone on K. That is, for each pair \(t_{1},t_{2}\in K\), \(t_{1}< t_{2}\) implies \(f(t_{1})\leq f(t_{2})\).
Proposition 2.5
Let
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
be a measurable function, then
\(\Lambda_{f}\subset\Gamma_{f}\).
Proof
Let \(L\in\Lambda_{f}\). Then there exists a Δ-non-thin set \(K\subset\mathbb{T}\) such that
$$\mathop{\lim_{s\rightarrow\infty}}_{s\in K} f(s)=L $$
and
$$\limsup_{s\rightarrow\infty}\frac{\mu_{\Delta} ( K(s) ) }{\sigma(s)-a}=d>0. $$
Let an arbitrary \(\varepsilon>0\) be given. The set \(\{t\in K:|f(t)-L|\geq \varepsilon\}\) is measurable and bounded. From
$$\bigl\{ t\in\mathbb{T}: \bigl|f(t)-L \bigr|< \varepsilon \bigr\} \supset K- \bigl\{ t\in K: \bigl|f(t)-L \bigr|\geq \varepsilon \bigr\} , $$
we have
$$\frac{\mu_{\Delta} ( \{ t\in\mathbb{T}:|f(t)-L|< \varepsilon \} (s) ) }{\sigma(s)-a}\geq\frac{\mu_{\Delta} ( K(s) ) }{\sigma(s)-a}-\frac{O(1)}{\sigma(s)-a}. $$
Finally we have
$$\limsup_{s\rightarrow\infty}\frac{\mu_{\Delta} ( \{ t\in\mathbb{T}:|f(t)-L|< \varepsilon \} (s) ) }{\sigma (s)-a}\geq d>0, $$
which means the set \(\{ t\in\mathbb{T}:|f(t)-L|<\varepsilon \} \) is a Δ-non-thin set; \(L\in\Gamma_{f}\). □
We will proceed some special cases of the above concepts. We also should emphasize that the sets \(\Lambda_{f}\) and \(\Gamma_{f}\) are not equal in general. Details are in the following example.
Example
(i) Let the time scale be \(\mathbb{T}=\mathbb{N}\). This case is called the discrete case and it is easy to see that all definitions above coincide with the definition of a limit point and cluster point in the classical statistically convergence theory.
(ii) Let \(q>1\) be a fixed integer and \(\mathbb{T}=\{q^{m}:m\in\mathbb {N}\}\). Consider the sequence of natural numbers \((k_{n})\) such that \(k_{n+1}-k_{n}>1\). If we take the subset \(K=\{q^{k_{n}}:\in\mathbb{N}\}\) then we can easily show that K is a Δ-non-thin set. Let \(k_{n}\leq k< k_{n+1}\) and \(t=q^{k}\), then we have
$$\frac{\mu_{\Delta}(K(t))}{\sigma(t)-q}=\frac{q-1}{q(q^{k}-1)}\sum_{i=1}^{n}q^{k_{i}}. $$
If one can take the limit \(t\rightarrow\infty\) in the above equality we can conclude that the limit does not exist. That means K can be considered as a Δ-non-thin set.
(iii) Let us consider the continuous case \(\mathbb{T}=[0,\infty)\) and the function \(f:\mathbb{T}=[0,\infty)\rightarrow\mathbb{R}\) defined by
$$ f(t)=\left \{ \begin{array}{@{}l@{\quad}l} 0, & t=0, \\ \frac{1}{n+1}, & t\in\bigcup_{k=0}^{\infty} ( (2k+1)2^{n}-1,(2k+1)2^{n} ], \end{array} \right . $$
for \(n\in\mathbb{N}\). Since for each \(n=0,1,2,\ldots \) , we have
$$ \delta_{\Delta} \biggl( \biggl\{ t\in\mathbb{T}:f(t)=\frac{1}{n+1} \biggr\} \biggr)=\frac {1}{2^{n+1}}>0, $$
and we have \(1/(n+1)\in\Lambda_{f}\). Moreover,
$$ \delta_{\Delta} \biggl( \biggl\{ t\in\mathbb{T}:f(t)\geq \frac{1}{n+1} \biggr\} \biggr) =\frac{1}{2}+ \frac{1}{2^{2}}+ \cdots+\frac {1}{2^{n+1}}=1-\frac{1}{2^{n+1}} $$
implies that
$$ \delta_{\Delta} \biggl( \biggl\{ t\in\mathbb{T}:f(t)< \frac {1}{n+1} \biggr\} \biggr) =1- \biggl( 1-\frac{1}{2^{n+1}} \biggr) =\frac{1}{2^{n+1}} . $$
(2.1)
This means that \(0\in\Gamma_{f}\). The set of Δ-cluster points is exactly the set \(\{1,1/2,1/3,\ldots\}\cup\{0\}\). Now we shall show that \(0\notin\Lambda_{f}\). For our purpose, we can consider a measurable subset \(A\subset\mathbb{T}\) with
$$ \mathop{\lim_{t\rightarrow\infty}}_{t\in A} f(t)=0. $$
We claim that \(\delta_{\Delta}(A)=0\). Let \(\varepsilon>0\) be given. There exists a natural number m such that \(2^{-m-1}<\varepsilon/3\). Then there exists \(s_{1}\in\mathbb{T}\) such that
$$ \frac{\mu_{\Delta} ( \{ t\in A:f(t)\geq\frac{1}{m+1} \} (s) ) }{s}< \frac{\varepsilon}{3} $$
(2.2)
holds for all \(s>s_{1}\). On the other hand, from (2.1), \(s_{2}\in \mathbb{T}\) can be chosen such that
$$\begin{aligned} \frac{\mu_{\Delta} ( \{ t\in A:f(t)< \frac{1}{m+1} \} (s) ) }{s} \leq&\frac{\mu_{\Delta} ( \{ t\in\mathbb {T}:f(t)<\frac{1}{m+1} \} (s) ) }{s} \\ <&\frac{\varepsilon}{3}+\frac{1}{2^{m+1}} \\ <&\frac{2\varepsilon}{3} \end{aligned}$$
(2.3)
holds for all \(s>s_{2}\). Let us define \(s_{0}=\max\{s_{1},s_{2}\}\). From (2.2) and (2.3) we have
$$\begin{aligned} \frac{\mu_{\Delta} ( A(s) ) }{s} =&\frac{\mu_{\Delta} ( \{ t\in A:f(t)\geq\frac{1}{m+1} \} (s) ) }{s}+\frac{\mu _{\Delta} ( \{ t\in A:f(t)< \frac{1}{m+1} \} (s) ) }{s} \\ <&\frac{\varepsilon}{3}+\frac{2\varepsilon}{3}=\varepsilon \end{aligned}$$
for all \(s>s_{0}\).
Proposition 2.6
Let
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
be a measurable function with
\(\Delta\mbox{-}\!\lim_{t\rightarrow\infty}f(t)=L\), then
\(\Lambda _{f}=\Gamma_{f}= \{ L \} \).
Proof
Assume that \(\Delta\mbox{-}\!\lim_{t\rightarrow\infty}f(t)=L\). From Theorem 1.1 there exists a measurable function \(g:\mathbb{T}\rightarrow \mathbb{R}\) such that \(\lim_{t\rightarrow\infty}g(t)=L\), and \(f(t)=g(t)\) holds for \(\Delta\mbox{-}a.a.~t.\) Let us define \(K= \{ t\in\mathbb {T}:f(t)=g(t) \} \). From the definition of \(\Delta\mbox{-}a.a.~t.\), \(\delta_{\Delta}(K)=1\) and
$$ \mathop{\lim_{t\rightarrow\infty}}_{t\in K} f(t)=\mathop{\lim _{t\rightarrow\infty}}_{t\in K}g(t)=L. $$
That is, \(L\in\Lambda_{f}\). By Proposition 2.5, \(L\in\Gamma_{f}\).
Now we will show that L is a unique element of \(\Gamma_{f}\). Let \(\varepsilon >0\) be given. Since \(\Delta\mbox{-}\!\lim_{t\rightarrow\infty}f(t)=L\), we have \(K_{1}=\{t\in\mathbb{T}:|f(t)-L|<\varepsilon/2\}\) with \(\delta_{\Delta }(K_{1})=1\). If \(L^{\prime}\) is another element of \(\Gamma_{f}\) then we can define \(K_{2}=\{t\in \mathbb{T}:|f(t)-L^{\prime}|<\varepsilon/2\}\); then
$$ \limsup_{s\rightarrow\infty}\frac{\mu_{\Delta} ( K_{2}(s) ) }{\sigma(s)-a}=d>0. $$
(2.4)
We claim that \(K_{1} \cap K_{2}\neq\emptyset\). Assume that \(K_{1}\cap K_{2}= \emptyset\) then \(K_{1}\subset ( K_{2} ) ^{c}\) and from \(\delta_{\Delta}(K_{1})=1\) we have \(\delta _{\Delta}( ( K_{2} ) ^{c})=1\) and so \(\delta_{\Delta }(K_{2})=0\). But this contradicts (2.4). For each \(t_{0}\in K_{1}\cap K_{2}\) we have
$$ \bigl\vert L-L^{\prime} \bigr\vert \leq \bigl\vert L-f(t_{0}) \bigr\vert + \bigl\vert f(t_{0})-L^{\prime} \bigr\vert < \varepsilon. $$
Since \(\varepsilon>0\) is arbitrary, we have \(L=L^{{\prime}}\). □
Proposition 2.7
Let
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
be a measurable function; then the set
\(\Gamma_{f}\)
is closed.
Proof
Let \((L_{n})\) be a sequence in \(\Gamma_{f}\) such that \(L_{n}\rightarrow L\) whenever \(n\rightarrow\infty\). For a given \(\varepsilon>0\), choose \(n_{0}\) large enough to make \(L-\varepsilon< L_{n_{0}}<L+\varepsilon\) and choose \(\varepsilon^{\prime}>0\) such that \((L_{n_{0}}-\varepsilon ^{\prime }, L_{n_{0}}+\varepsilon^{\prime})\subset(L-\varepsilon ,L+\varepsilon)\). In this case, from
$$ \bigl\{ t\in\mathbb{T}: \bigl\vert f(t)-L_{n_{0}} \bigr\vert < \varepsilon ^{\prime} \bigr\} \subset \bigl\{ t\in\mathbb{T}: \bigl\vert f(t)-L \bigr\vert <\varepsilon \bigr\} $$
and
$$ \limsup_{s\rightarrow\infty}\frac{\mu_{\Delta} ( \{ t\in\mathbb{T}:\vert f(t)-L_{n_{0}}\vert < \varepsilon ^{\prime} \} (s) ) }{\sigma(s)-a}=d>0 $$
we have
$$ \limsup_{s\rightarrow\infty}\frac{\mu_{\Delta} ( \{ t\in\mathbb{T}:\vert f(t)-L\vert < \varepsilon \} (s) ) }{\sigma(s)-a}=d>0. $$
That means the set \(\{ t\in\mathbb{T}:\vert f(t)-L\vert <\varepsilon \} \) is not a set that has zero Δ-density; \(L\in\Gamma_{f}\). □
Theorem 2.8
Let
\(f,g:\mathbb{T}\rightarrow\mathbb{R}\)
be measurable functions. If
\(f(t)=g(t)\)
for
\(\Delta\mbox{-}a.a.~t.\)
then
\(\Lambda_{f}=\Lambda_{g}\)
and
\(\Gamma _{f}=\Gamma_{g}\).
Proof
Let \(L\in\Lambda_{f}\). Then there exists a Δ-non-thin set K such that
$$\mathop{\lim_{t\rightarrow\infty}}_{t\in K}f(t)=L $$
and
$$\limsup_{t\rightarrow\infty}\frac{\mu_{\Delta} ( K(t) ) }{\sigma(t)-a}>0. $$
Since \(\delta_{\Delta}( \{ t\in\mathbb{T}:f(t)\neq g(t) \} )=0\) we have \(\delta_{\Delta}( \{ t\in K:f(t)\neq g(t) \} )=0\). This means that the set \(\{ t\in K:f(t)=g(t) \} \) is not a set that has zero-Δ-density and so \(L\in\Lambda_{g}\). Then we have \(\Lambda _{f}\subset\Lambda_{g}\). It is easy to see that \(\Lambda_{g}\subset\Lambda_{f}\) from symmetry. Finally we have \(\Lambda_{f}=\Lambda_{g}\). The equality \(\Gamma_{f}=\Gamma_{g}\) can be shown in a similar way. □
Proposition 2.9
Let
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
be a measurable function. If
f
is bounded on a Δ-non-thin set then
\(\Gamma_{f}\neq\emptyset\).
Proof
Assume that f is bounded on K, which is a Δ-non-thin subset of \(\mathbb{T} \) and \(\Gamma_{f}=\emptyset\). Then for each \(t\in K\) there exists a neighborhood of \(f(t)\)
$$ \mathcal{N} \bigl( f(t) \bigr) = \bigl\{ y\in\mathbb{R}: \bigl\vert y-f(t) \bigr\vert < \varepsilon(t) \bigr\} $$
such that \(\delta_{\Delta} ( f^{-1} ( \mathcal{N} ( f(t) ) ) ) =0\). Moreover, it is clear that
$$ f(K)\subset\bigcup_{t\in K}\mathcal{N} \bigl( f(t) \bigr). $$
If \(f(K)\) is not closed then we can consider \(\overline{f(K)}\) by adding limit points \(\beta_{i}\) (\(i\in I\)) of \(f(K)\) and we have
$$ \overline{f(K)}\subset \biggl( \bigcup_{t\in K} \mathcal{N} \bigl( f(t) \bigr) \biggr) \cup \biggl( \bigcup _{i\in I}\mathcal{N} ( \beta_{i} ) \biggr). $$
We have
$$ \mathcal{N} ( \beta_{i} ) = \bigl\{ y\in\mathbb{R}:\vert y- \beta_{i}\vert < \varepsilon(i) \bigr\} $$
and note that positive real numbers \(\varepsilon(i)\) can be chosen such that \(\delta_{\Delta} ( f^{-1} ( \mathcal{N} ( \beta_{i} ) ) ) =0\). Since \(\overline{f(K)}\) is compact, we can find \(\mathcal{B}_{1},\mathcal {B}_{2},\ldots,\mathcal{B}_{n}\) which is a finite subcover of \(\{ \mathcal{N} ( f(t) ) \} _{t\in K}\cup \{ \mathcal{N} ( \beta_{i} ) \} _{i\in I}\) such that
$$ \overline{f(K)}\subset\bigcup_{k=1}^{n} \mathcal{B}_{k}. $$
Then we have
$$ K\subset f^{-1} \bigl(\overline{f(K)} \bigr)\subset\bigcup _{k=1}^{n}f^{-1}(\mathcal{B}_{k}). $$
Since \(\delta_{\Delta} ( \bigcup_{k=1}^{n}f^{-1}(\mathcal {B}_{k}) ) =0\) we can conclude that \(\delta_{\Delta}(K)=0\). But this conclusion contradicts the Δ-non-thin property of K. One can prove everything in a similar way if \(f(K)\) is closed. □
The following corollary can be obtained immediately from Proposition 2.9.
Corollary 2.10
If a measurable function
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
is Δ-bounded, then
\(\Gamma_{f}\neq\emptyset\).
Proposition 2.11
Let
\(f:\mathbb{T}\rightarrow\mathbb{R}\)
be a measurable function. If
f
is Δ-monotone increasing and Δ-bounded, then it is Δ-convergent.
Proof
Since f is Δ-monotone increasing, there exists a subset \(K_{1}\subset\mathbb{T}\) with Δ-density equal to one, and for every \(t_{1},t_{2}\in K_{1}\) and \(t_{1}< t_{2}\) implies \(f(t_{1})\leq f(t_{2})\). Since f is Δ-bounded there exists a real number A and \(K_{2}= \{ t\in\mathbb{T}:\vert f(t)\vert \leq A \} \) with \(\delta_{\Delta} ( K_{2} ) =1\). Setting \(K=K_{1}\cap K_{2}\), one has \(\delta_{\Delta} ( K ) =1\). We now consider the following set:
$$ f(K)= \bigl\{ f(t):t\in K \bigr\} $$
and let β be the supremum of \(f(K)\) over K. For an arbitrary \(\varepsilon>0\) there is a point \(t_{0}\in K\) that satisfies \(\beta-\varepsilon< f(t_{0})\leq\beta\). From the monotonicity we can write \(\beta-\varepsilon< f(t)<\beta+\varepsilon\) for every \(t\in K \) satisfying \(t\geq t_{0}\). The last statement implies that \(\Delta\mbox{-}\!\lim_{t\rightarrow\infty}f(t)=\beta\). □