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 \(\Delta\mbox{-}\!\lim_{t\rightarrow\infty}f(t)=\beta\). □