\documentclass{article}1\title{The Impact of Functions on Disconnected Sets in Real Analysis}2\author{Bharath Krishnan}3\usepackage{amssymb}4\usepackage{amsmath}5\usepackage{ragged2e}6\newcommand{\dd}[1]{\mathrm{d}#1}78\begin{document}9\maketitle1011\subsection{Introduction}1213Mathematicians have analyzed specific cases of functions of disconnected functions (functions on disconnected sets) but seldom analyzed such functions in general. With careful analysis, one would find such functions can be continuous, differentiable and integrable. The concept is implied in definitions of analysis textbooks but there are no examples to reinforce these implications. Moreover, usually given examples of disconnected functions are "no-where continuous" Dirichlet function (the characteristic function of rationals), defined $1$ on rationals and $0$ on irrationals, and Thomae's function defined $1/q$ on $x=p/q$ ($\gcd(p,q)=1$) and $0$ on irrationals.1415Such functions can be useful in textbooks for students to thoroughly understand definitions of continuous functions, derivatives and integrals. Moreover, they clear misconceptions students have on these definitions. Such functions are also interesting for people who see functions as a formula that is ambiguous.1617Throughout this article, we will explain the rigorous definitions of limits, continuous functions and derivatives and see how these definitions apply to disconnected functions. Then we will observe the impact these issues have on concepts of differentiation and integration theorems.18\subsection{Using ${f(x)}^{g(x)}$}1920We could nicely constructed example to understand this general concept. Rather than using a piece-wise function, we will apply this concept to "naturally-made functions" such as ${f(x)}^{g(x)}$. The function would be useful in textbooks and is interesting to analyze in the real and complex plane. The complex function $z^z$ was already analyzed by Mark D. Meyerson in "The $X^X$ spindle". Instead the focus of this article will be strictly on ${f(x)}^{g(x)}$ in the real plane.2122Although analyzing ${f(x)}^{g(x)}$ on the cartesian plane is straight-forward, mathematicians are either unaware or dispute the notion that defined points exist at points outside of integers when $f(x)<0$. They would convert the following into a complex multi-valued function.2324\subsection{Domain of ${f(x)}^{g(x)}$}2526Generally when $a$ is negative, $a^b$ is thought to make sense only if $b$ is an integer. One reason $a^b$ is not continuous on the real plane in terms of the complex plane is that $\lim_{\epsilon \to 0}{(a+\epsilon i)}^b$, where $b \not \in \mathbb{Z}$, does not approach $a^{b}$. Also mathematicians find it crucial that $a^{b}=e^{b\log(a)}$, for all of $x$.2728One solution is to convert $f(x)^{g(x)}$ into a complex function and find the point where each branch of ${f(z)}^{g(z)}$ intersects the real plane. The function ${f(x)}^{g(x)}$, when $f(x)<0$, extends to the complex plane as $e^{g(x)(\text{log}|f(x)|+(2n+1)i\pi)}$, where $n\in \mathbb{N}$ represents all branches of the complex logarithm. The complex function is equivelent to $|f(x)|^{g(x)}(\cos((2n+1)\pi g(x))+i\sin((2n+1)\pi g(x)))$.2930In order to get the point where $|f(x)|^{g(x)}(\cos((2n+1)\pi g(x))+i\sin((2n+1)\pi g(x)))$ intersects the real plane plane for each $n\in\mathbb{N}$, $((2n+1)\pi g(x))$ must equal $k\pi$ for $k\in\mathbb{Z}$ so the imaginary component is zero.3132Say $g(x)\in{\mathbb{R} \backslash \mathbb{Q}}$, then we need $(2n+1)\pi(g(x))=k\pi$ but since $g(x)$ is irrational $(2n+1)\pi(g(x)){\neq}k\pi$. Hence the $x$ values that make the output of $g(x)$ irrational would make $f(x)^{g(x)}$ undefined in the real plane.3334Now we could focus on the values of $g(x)$ that gives rationals. Say we find that $g(x)=\left\{\left.\frac{2p+1}{2q}\right|p,q \in \mathbb{Z}\right\}$ or $x=\left\{\left.g^{-1}\left(\frac{2p+1}{2q}\right)\right|p,q \in \mathbb{Z}\right\}$. Then if $(2n+1)\pi\left(\frac{2p+1}{2q}\right)=k\pi$, $2n+1=m(2q)$ for any $m\in{\mathbb{N}}$ in order for $(2n+1)\pi\left(\frac{2p+1}{2q}\right)$ to cancel into $k\pi$. So if $m=\frac{2n+1}{2q}$, then $m\not\in{\mathbb{N}}$. However, this is a contradiction since $m$ must be an integer and hence ${f(z)}^{g(z)}$ has no branches that intersect the real plane $x=\left\{\left.g^{-1}\left(\frac{2p+1}{2q}\right)\right|p,q \in \mathbb{Z}\right\}$.3536If $g(x)=\left\{\left.\frac{2p+1}{2q+1}\right|p,q \in \mathbb{Z}\right\}$ where $p\text{,}q \in \mathbb{N}$ than $(2n+1)\pi\left(\frac{2p+1}{2q+1}\right)=k\pi$. In order for to happen $2n+1=m(2q+1)$ where $m\in \mathbb{N}$. Since $m=\frac{2n+1}{2q+1}$, $m$ must be an odd integer. Thus we can rewrite $\left\{\left. m \right|m\in{2\mathbb{N}+1} \right\}$ as $\left\{\left. {2m+1}\right|m \in \mathbb{N}\right\}$ and get $2n+1=(2m+1)(2q+1)$ which by substitution gives $(2m+1)(2q+1)\pi\left(\frac{2p+1}{2q+1}\right)$. As a result we get $\left|f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)\right|^{\frac{2p+1}{2q+1}}\left(\cos((2m+1)(2p+1)\pi)+\right.$ $\left.i\sin((2m+1)(2p+1)\pi)\right)$. Since $(2m+1)(2p+1)$ is always odd we end up with ${f(x)}^{g(x)}=-{|f(x)|}^{g(x)}$ for $x=\left\{\left.g^{-1}\left(-\frac{2p+1}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ when $f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)<0$.3738However, if $g(x)$ is equal to $\frac{2p}{2q+1}$ where $p,q \in \mathbb{N}$ than $(2n+1)\pi\left(\frac{2p}{2q+1}\right)=k\pi$. In order for this to happen $2n+1=m(2q+1)$. Since $m=\frac{2n+1}{2q+1}$, $m$ must be odd. Hence we replace odd $\left\{\left.m\right| m\in 2\mathbb{N}+1\right\}$ with $\left\{\left.2m+1\right|m\in\mathbb{N}\right\}$ to get $2n+1=(2m+1)(2q+1)$. Using substitution, one gets $(2m+1)(2q+1)\pi(\frac{2p}{2q+1})$. Thus we end up with $|f(x)|^{g(x)}\left(\cos((2m+1)(2p)\pi)+i\sin((2m+1)(2p)\pi)\right)$. Since $(2m+1)(2p)$ is always even, we end up with $|f(x)|^{g(x)}$ for $x=\left\{\left.g^{-1}\left(\frac{2p}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ when $f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)<0$.3940Hence for $S(x)={f(x)}^{g(x)}$ we must find values where $g(x)=\left\{\left.\frac{p}{2q+1}\right|p,q\in\mathbb{Z}\right\}$ where $p$ is an odd or even integer and $q$ is an odd integer. Depending on if $p$ is odd or even it may end up in the rule $|f(x)|^{g(x)}$ or $-|f(x)|^{g(x)}$.4142In piece-wise notation, we can express ${f(x)}^{g(x)}$ as the following.4344\begin{equation}45{f(x)}^{g(x)}=\begin{cases} {|f(x)|}^{g(x)} & \left\{x= g^{-1}\left(\frac{2p}{2q+1} \right) |\ p, q \in \mathbb{Z}\right\}\land \left\{f(x)<0\right\}\\ -|f(x)|^{g(x)} & \left\{x=g^{-1} \left( \frac{2p+1}{2q+1} \right)| p, q \in \mathbb{Z}\right\}\land \left\{f(x)<0\right\}\\ {f(x)}^{g(x)} & f(x)>0\\ \end{cases}46\end{equation}4748Here we find when $f(x)>0$, the function is defined on connected sets but when $f(x)<0$, the function is defined on disconnected sets that forms a totally disconnected space.4950This can also be generalized to all disconnected functions.5152\begin{equation}53S(x)=\begin{cases} I_1(x)& x\in A_1\\ I_2(x) & x\in A_2\\ I_3(x)& x\in A_3\\ {}\quad \text{.} & \text{}\quad{.}\\{}\quad \text{.}& {}\quad\text{.}\\I_n(x) & x\in A_n \end{cases}54\end{equation}5556Where $I_1(x),I_2(x),..,I_n(x)$ are "rules" which are continuous almost everywhere and $A_1,A_2,A_3,..,A_n$ ($B_n=\bigcup_{i=1}^{n}A_i$ are disjointed. (Note that two or more rules may be equal to one another)5758If we were to set $S(x)={f(x)}^{g(x)}$ then $I_1(x)={|f(x)|}^{g(x)}$, $I_2(x)={|f(x)|}^{g(x)}$, $A_1=\left\{\left.g^{-1} \left( \frac{2p+1}{2q+1}\right| p,q \in \mathbb{Z} \right)\right\}$ and $A_2=\left\{\left. g^{-1} \left( \frac{2p+1}{2q+1} \right)\right| p,q \in \mathbb{Z}\right\}$596061\subsection{Limit Points and Continuity}62In standard Calculus, the definition of a limit is $\lim_{x\to a}f(x)=L$, which means that for every $\epsilon>0$ there is corresponding $\delta>0$ where $|x-a|<\delta$ such that $|f(x)-L|<\epsilon$.6364This definition is true for a function defined on the interval. However, totally disconnected functions require specific sets of real numbers to approximate $a$. Hence a more precise definition is required.6566"Bartles Elements of Real Analysis" states the following definition.6768\subsubsection{Definition}69Let $B_n\subseteq\mathbb{R}$ ($B_n$) and let $a$ be a cluster point of B. For a function $f:B_n\to \mathbb{R}$ a real number $L$ is said to be the limit of $f$ at $a$ if, given any $\epsilon>0$ there exists $\delta>0$ such that if $x\in{B_n}$ and $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.7071According to definition, we can choose any $\epsilon$ and $\delta$ as long as $B$ has a cluster point at $a$. This means $B_n$ can closely approximate $a$. (Note that $a$ may not belong to $B_n$).7273Inorder to determined the point where the limit exists we must find the values that are cluster points of $A_1$ and $A_2$.7475We will begin by assuming that any $a\in\mathbb{R}$ in ${f(x)}^{g(x)}$ is a cluster point of $A_1$ and $A_2$. In order to prove this assumption, $A_1$ and $A_2$ must be dense in $\mathbb{R}$. If $g$ is continuous, analytic and bounded then if $\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ and $\left\{\left.\frac{2p}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ are dense everywhere then so would $A_1=\left\{\left.g^{-1}\left(\frac{2p}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ and $A_2=\left\{\left.g^{-1}\left(\frac{2p+1}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$.7677There is a rigorous way of proving that $\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ and $\left\{\left.\frac{2p}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ is dense everywhere; but I will give an informal proof.787980For example to prove $\left\{\left.\frac{2p+1}{2q+1}\right|p,q \in \mathbb{Z}\right\}$ is dense. Show the set can approximate $a$ from the values lesser than or greater than $a$.8182We could then solve for the integer $p$ to get8384\begin{equation}85p=\frac{1}{2}\left\lfloor{a(2q+1)}\right\rfloor-\frac{1}{2}86\end{equation}8788The floor function is used since $p$ must be an integer.8990Then substitute $p$ into (3) to get which leads to an inequality9192\begin{equation}93\frac{2\left\lfloor\frac{1}{2}{a(2q+1)}-\frac{1}{2}\right\rfloor+1}{2p+1}<a<\frac{2\left\lceil\frac{1}{2}{a(2q+1)}-\frac{1}{2}\right\rceil+1}{2p+1}94\end{equation}9596Note that $\left\lfloor{a}\right\rfloor=a \pm \epsilon$, where $\epsilon$ is the error, and $\left\lceil{a}\right\rceil=\left\lfloor{a}\right\rfloor+1$9798\begin{equation}99\frac{2\left(\frac{1}{2}{a(2q+1)}-\frac{1}{2}\pm\epsilon\right)+1}{2q+1}<a<\frac{2\left(\frac{1}{2}{a(2q+1)}-\frac{1}{2}\pm\epsilon\right)+2}{2q+1}100\end{equation}101102\begin{equation}103a\pm\frac{\epsilon}{2q+1}<a<a\pm\frac{\epsilon+1}{2q+1}104\end{equation}105106As $q$ becomes larger, both sides of the inequality approach $a$.107108We can apply a similar proof to show $\left\{\left.\frac{2p}{2q+1}\right|p,q \in \mathbb{Z}\right\}$ is dense everywhere. Since $\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ and $\left\{\left.\frac{2p}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ is dense in $\mathbb{R}$ it follows that $A_1$ and $A_2$ are dense.109110Since both sets are dense, for a limit to exist at any $a\in\mathbb{R}$, as the inputs of each set approach $a$, the outputs of each set must approach the same value. Hence the following is required.111112\begin{equation}113\underset{x\in A_1}{\underset{x\to a}\lim}{f(x)^{g(x)}}=\underset{x\in A_2}{\underset{x\to a}\lim}{f(x)^{g(x)}}114\end{equation}115116We know from (1) the equality in (9) can never be true unless the limit is at a point where ${|f(x)|^{g(x)}}$ and ${-|f(x)|}^{g(x)}$ intersect. Both rules can never intersect when $f(x)<0$ since $|f(x)|>0$ making ${|f(x)|}^{g(x)}>0$ and $-{|f(x)|}^{g(x)}<0$. However ${f(x)}^{g(x)}$, ${|f(x)|}^{g(x)}$ and ${-|f(x)|^{g(x)}}$ can intersect when $f(x)=0$.117118On the other hand, if we substitute ${f(x)}^{g(x)}$ into $P(x)$ such that $P\left(|f(x)|^{g(x)}\right)=P\left(-|f(x)|^{g(x)}\right)$ then the $\underset{x\to a}{\lim}{f(x)}^{g(x)}$ can exist for any $a\in\mathbb{R}$. The limit would be in the form of $P\left(|{f(a)|}^{g(a)}\right)$.119120Mathematicians might argue one could extend $P(f(x)^{g(x)})$ into $P(|f(x)|^{g(x)})$. However, for the purpose of analysis we will see if continuity, differentiability or integrability can exist for this disconnected function.121122Using the analysis definition of a limit we could apply a similar principal for continuity of ${f(x)}^{g(x)}$.123124\subsubsection{Definition}125Let $B_n\subseteq\mathbb{R}$, let $f:B_n \to \mathbb{R}$ and let $a \in {B}$. We say that $f$ is continuous at $a$ if, given any number $\epsilon>0$ such that $x$ is any point of $B$ satisfying $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.126127This rigorous definition states that as the $\epsilon$ band narrows and approaches $L$ the $\delta$ band should narrows and approaches $a$, $L$ should equal $f(a)$.128129Informally, the limit at $a$ which is a cluster point of $A_1$ and $A_2$ must equal ${f(a)}^{g(a)}$. Hence the following is required.130131\begin{equation}132\underset{x\in A_1}{\underset{x\to a}\lim}{f(x)^{g(x)}}=\underset{x\in A_2}{\underset{x\to a}\lim}{f(x)^{g(x)}}={f(a)}^{g(a)}133\end{equation}134135We already know from (8), the limit can exist only at $f(x)=0$. If ${0}^{g(f(0))}$ is defined, then the point is continuous.136137For $P({f(x)}^{g(x)})$ the limit can exist anywhere but continuous points exist where $P\left({f(a)}^{g(a)}\right)$ is defined. The continuous points exist in the domain of $P({f(x)}^{g(x)})$ and hence the function is continuous.138139Some may have misconceptions that continuous function should not have discontinuous, undefined "breaks". However, these "breaks" are actually removable singularities: points where the limit is finite and undefined. Generally, continuities and discontinuities are determined in the domain making removable singularities neither continuities nor discontinuities. The only exceptions are at vertical asymptotical discontinuities, which splits a function into several parts.140141Mathematician tend to replace removable singularities with continuous points as singularities are viewed as unnecessary; however, mathematical textbooks state functions need only be continuous in the domain. Here is an analogy, picture a steel wire made up of atoms which are made of particles that are made of even smaller particles and so on. Now matter how small the particles are there is always space between each one, making the steel wire disconnected. However, we see the steel wire as connected since we only see the "defined particles" instead of the "undefined space". Replacing the undefined space with a defined space is merely an illusion.142143Hence $P(f(x)^{g(x)})$ is continuous if there are no discontinuities at vertical asymptotes.144145Generalizing the following case to $S(x)$, if $S(x)$ has sets defined by more than one rule, then $\underset{\left\{x\in{B_n}\right\}\to a} \lim S(x)=\bigcap_{i=1}^{n}I_{i}(a)$ and is continuous at $x=a_1$ where $\left\{a_1\right146\}$ is the set of values of $a$ where $\bigcup_{i=1}^{n}I_n(a)$.147148If $S(x)$ is defined by the same rule, then $\underset{{\left\{x\in A_1\right\}}\to a} \lim S(x)={I_1}(a)=I_2(a)=...=I_n(a)$ and $S(x)$ is continuous at $x=B_n$.149150\subsection{The Existence of a Derivative}151In this section we will show that the derivative of a disconnected function can exist.152153If ones takes the newton-quotient of the derivative.154155\begin{equation}156\underset{x\in B_n}{\underset{h\to0}{\lim}}\frac{S(x+h)-S(x)}{h}157\end{equation}158159Since $S(x)$ is disconnected, we must restrict $h$ to values $f(x+h)$ is defined. Finding such an $h$ would be tedious.160161In most real analysis textbooks, the derivative is not shown as a newton quotient. Instead they use the following notation.162163\begin{equation}164\lim_{x\to a}\frac{S(x)-S(a)}{x-a}165\end{equation}166167This is preferred since we can focus properties of $f(x)$ as opposed to $f(x+h)$.168169In "Elementary Real Analysis", the author states a similar but rigorous definintion. He says, "Let $B_n\subset \mathbb{R}$ be an interval, let $S:B_n\to \mathbb{R}$, and let $a \in B_n$. We say that the real number $L$ is the derivative of $f$ at $a$ if any given $\epsilon>0$ there exists $\delta(\epsilon)>0$ such that if $x\in B_n$ satisfies $0<|x-a|<\delta(\epsilon)$.170171\begin{equation}172\left|\frac{S(x)-S(a)}{x-a}-L\right|<\epsilon173\end{equation}174175The definition states as the epsilon band $L-\epsilon<\frac{S(x)-S(a)}{x-a}<L+\epsilon$ narrows to $L$, the delta band $\delta-a<x<\delta(\epsilon)+a$ must narrow to $a$ in a poportional amount to epsilon band. If all of $x\in B_n$ are included in epsilon band than the derivative at $a$ can exist.176177The definition implies a function that is defined by two or more distinct rules cannot have a derivative and that function defined by one rule can have a derivative.178179However, this is simpler to understand using the informal definition. From the notation in (11) we have180181\begin{equation}182S'(a)=\underset{x\in B_n}{\underset{x\to a}\lim}\frac{S(x)-S(a)}{x-a}183\end{equation}184185where $S(x)={f(x)}^{g(x)}$. In order for a derivative to exist the following must be equal.186187\begin{equation}188\underset{x\in A_1}{\underset{x\to a}\lim} \frac{{f(x)}^{g(x)}-{f(a)}^{g(a)}}{x-a}=\underset{x\in A_2}{\underset{x\to a}\lim} \frac{{f(x)}^{g(x)}-{f(a)}^{g(a)}}{x-a}189\end{equation}190191From (14), the derivative cannot exist unless ${f(a)}^{g(a)}$ defined. This makes sense in real analysis since the derivative can be defined in the functions' domain if there are no discontinuities or sharp corners.192193When ${f(a)}^{g(a)}$ is defined we have to find the rule where ${f(x)}^{g(x)}$ is defined. From (1), $A_1$ is defined by the rule $|f(x)|^{g(x)}$ and $A_2$ is defined by the rule $-|f(x)|^{g(x)}$. Now we could change (14) into the following.194195\begin{equation}196\underset{x\in A_1}{\underset{x\to a}\lim}\frac{{f(x)}^{g(x)}-{f(a)}^{g(a)}}{x-a}= \underset{x\to a}\lim\frac{{|f(x)|}^{g(x)}-{f(a)}^{g(a)}}{x-a}197\end{equation}198199\begin{equation}200\underset{x\in A_2}{\underset{x\to a}\lim}\frac{{f(x)}^{g(x)}-f(a)^{g(a)}}{x-a}=\lim_{x\to a} \frac{-{|f(x)|}^{g(x)}-{f(a)}^{g(a)}}{x-a}201\end{equation}202203We do not have to worry about the values of ${f(x)}^{g(x)}$ that is undefined as this is also determined by ${f(a)}^{g(a)}$204205\begin{equation}206\underset{x\to a}\lim\frac{{|f(x)|}^{g(x)}-{f(a)}^{g(a)}}{x-a}=\underset{x\to a}\lim\frac{{-|f(x)|}^{g(x)}-f(a)^{g(a)}}{x-a}207\end{equation}208209From (12), the equality exist unless $|f(x)|^{g(x)}=-|f(x)|^{g(x)}$ intersect at x-values in the domain of ${f(x)}^{g(x)}$.210211On the other hand, if we take the derivative of $P\left({f(x)}^{g(x)}\right)$ we have212213\begin{equation}214S'(a)=\underset{x\in B_2}{\underset{x\to a}\lim}\frac{P\left(f(x)^{g(x)}\right)-P\left(f(a)^{g(a)}\right)}{x-a}215\end{equation}216217To find the derivative of that function we must find the derivative of the rule and restrict the that domain of that derivative to totally disconnected space as the orginal function.218219\begin{equation}220S'(a)=\frac{d}{da}P^{'}\left(|f(a)|^{g(a)}\right)=P^{'}\left(|f(a)|^{g(a)}\right)\left(|f(a)|^{g(a)}\right)\left(\ln|f(a)|g'(a)+\frac{\text{sgn}(f(a))g(a)f'(a)}{|f(a)|}\right)221\end{equation}222223\begin{equation}224S'(x)=P^{'}\left({\left|f(x)\right|}^{g(x)}\right)\left(|f(x)|^{g(x)}\right)\left(225\ln|f(x)|g'(x)+\frac{g(x)f'(x)}{f(x)}\right)226\end{equation}227228Finally we can remove the absolute value so the function is defined on a disconnected set.229230\begin{equation}231P^{'}\left({f(x)}^{g(x)}\right)\left({f(x)}^{g(x)}\right)\left(\ln|f(x)|g'(x)+\frac{g(x)f'(x)}{f(x)}\right)232\end{equation}233234The idea of a disconnected derivative was mentioned by Bartle in "Introduction to Real Analysis" where he states, "It is possible for the definition of a domain more general than an interval (since point $c$ need only be an element of the domain and also a cluster point)". However, he divert attention from this concept stating, "The significance of this function is more naturally apparent for functions defined on intervals. Consequentally we will limit our attention to such functions".235236But there are problems with having a derivative defined on dense sets and undefined on an uncountable number of points. The main issue is that defined points of the derivative can approximate infinitesimally close to undefined holes. This raises the questions whether a derivative should be defined strictly on its domain. The second problem is that this kind of derivative could bring issues to the fundamental theorem of Calculus which will explain later.237238In general if $S(x)$ coincides with two or more functions than the derivative is the following.239240\begin{equation}241S'(x)=\begin{cases} \left\{\bigcap_{i=1}^{n}I_{n}^{'}(x)\right\}& x=\left\{\text{Values of}\ a \ \text{where}\ \bigcap_{i=1}^{n}I_{n}^{'}(a)\right\}\\242\end{cases}243\end{equation}244245In $S(x)$, if all the dense sets of $B_n$ are defined by the same rule $(I_1(x)=I_2(x)=...=I_n(x))$ than the derivative of $S(x)$ is the following.246247\begin{equation}248S'(x)=\begin{cases} I_1^{'}(x)& x\in B_n\\249\end{cases}250\end{equation}251252\begin{equation}253\forall x\in B_n,S'(x)=I_1'(x)\ni{S}^{'}:{B_n}\to\mathbb{R}\land{I_1}^{'}:E\to\mathbb{R}254\end{equation}255256The set $E$ is defined almost everywhere. (Points that are not continuous have measure of zero.)257258\subsection{Maxima and Minima}259260261According to Fermat's Theorem262263\subsubsection{Definition}264If $F:(a,b)\to\mathbb{R}$ has a maximum of $(x_0,F(x_0))$, where $x_0{\in}(a,b)$, then $F'(x_0)=0$.265266The definition applies to functions defined on the interval. However, a similar definition can be applied to $S(x)$ if $x_0$ belongs in $B_n$ such that $S'(x_0)=0$. We can extend the previous definition.267268\subsubsection{Definition}269If $S:(a,b)\to{B_n}$, where $B_n\subset{\mathbb{R}}$, has a maximum of $(x_0,S(x_0))$ with $x_0{\in}(a,b)$, then $S'(x_0)=0$.270271However in most cases, $x_0$ may not be in $B_n$ and $S'(x_0)$ is undefined. But we can closely approximate $x_0$ using $B_n$.272273We know a set dense in all real numbers can approximate arbitrarily close to any $x\in\mathbb{R}$ including $x_0$ such that as $\left\{x\in B_n\right\}\to{x_0}$, $S(x)\to0$. The same can be applied to $P\left(|f(x)|^{g(x)}\right)$274275If we substitute $x\to{x_0}$ into $P\left(|f(x)|^{g(x)}\right)$ then we get $P\left({f(x)}^{g(x)}\right)\to P\left({|f(x_0)|}^{g(x_0)}\right)$. Note that we are not stating thos limit is the local maximum of ${f(x)}^{g(x)}$ but we can approximate infinitesimally close to the local maximum or minimum of $P\left(|f(x)|^{g(x)}\right)$. We will call these elements the approaching maximal ($\text{apprmax}$) or approaching minimal elements ($\text{apprmin}$).276277Here is the precise definition.278279\subsubsection{Definition}280If $f:(a,b)\to{B_n}$, where $B_n\subset{\mathbb{R}}$ and $S'(x_0)$ is undefined then the maxima or minima is approaching maximum and approaching minimum element such that $\lim_{x\to{x_0}}S'(x)=0$.281282Note that the approaching maximal element of continuous $S:(a,b)\to{B_n}$ approaches the local maximum of the rule but is always less than that local maximum. And the approaching minimal element of the function approaches the local minimum of the rule but is always greater than the local minimum.283284\begin{equation}285\text{apprmax} P\left({f(x)}^{g(x)}\right)<\text{max} P\left({f(x)}^{g(x)}\right)286\end{equation}287288\begin{equation}289\text{min} P\left({f(x)}^{g(x)}\right)<\text{apprmin} P\left({f(x)}^{g(x)}\right)290\end{equation}291292Moreover, the local maximum of the rule is the limit of the maximal element of the disconnected function. Similarly, if the minimal element of the local minimum both exist at $x=x_0$, then the local minimum is the rule of the minimal element of the function.293294\begin{equation}295\lim \text{apprmax} P\left({f(x)}^{g(x)}\right)=\sup P\left({f(x)}^{g(x)}\right)=\max P\left({|f(x)|}^{g(x)}\right)296\end{equation}297298\begin{equation}299\lim \text{apprmin} P\left({f(x)}^{g(x)}\right)=\inf P\left({f(x)}^{g(x)}\right)=\min P\left({|f(x)|}^{g(x)}\right)300\end{equation}301302\subsubsection{Maximum and Minima when ${f(x)}^{g(x)}$ is defined by two distinct rules}303304If ${f(x)}^{g(x)}$ is defined by more than one rule, the function can still have a local maximum and minimum even if the derivative cannot exist. Using intuition, the process simple to figure out.305306We could find the critical points of each rule and substitute the critical points into the function to see if they are defined. If so, the output of the function at the critical points should be compared to the entire function as the interval around each critical point narrows. If the output of the function at the critical point is greater than all other defined points as the interval around the critical point narrow than this is the local maximum. If the output of the function at the critical point is less than all other defined points as the interval narrow than this is the local maximum.307308\subsubsection{Definition}309If $S(x)$ is defined by the rules $\bigcup_{i=1}^{n}I_i(x)$ and $\bigcup_{i=1}^{n}I_i:(a,b)\to{B_n}$ with $\bigcup_{i=1}^{n}{I_i}^{'}(x_0)=0$ and $x_0\in(a,b)$; then a maximum exist if $S(x_0)>S(x)$ and a minimum exist if $S(x_0)<S(x)$ for $x\in(a,b)$310311The approaching maximum and minimum also exist assuming the conditions in the previous definition stay the same except for $S'(x_0)$ being defined.312313Putting all this together, we can apply concepts similar maximal and minimal elements to the mean value theorem and Rolle's theorem for continuous disconnected functions. However, we may have to make adjustment to the intervals as well.314315\subsubsection{Modified Rolle's Theorem}316317If a real function $S$ is defined on $B_n$ dense in $\mathbb{R}$ in the closed interval $[a,b]$ and $\underset{x\in B_n}{\underset{x\to a^{+}}\lim}f(x)=\underset{x\in B_n}{\underset{x\to b^{-}}\lim}f(x)$, with $f$ differentable on all of $B_n$ on open interval $(a,b)$, there exists a $c$ such that318319\begin{equation}320\underset{x\in B_n}{\underset{x\to c}\lim}S'(x)=0321\end{equation}322323324\subsubsection{Modified Mean Value Theorem}325If $S$ is continuous on $B_n$, which is dense in $\mathbb{R}$, in $[a,b]$ with $a<b$ and $S$ is differentable in $B_n$ on the open interval $(a,b)$ then there exist an $F'(c)$ such that326327\begin{equation}328\underset{x\in B_n}{\underset{x\to c}\lim}f(x)=\frac{\underset{x\in B_n}{\underset{x\to b^{+}}\lim} F(x)-{ \underset{x\in B_n}{\underset{x\to a^{-}}\lim} F(x)}}{\underset{x\in B_n}{\underset{x\to b^{+}}\lim} x-{ \underset{x\in B_n}{\underset{x\to a^{-}}\lim} x}}329\end{equation}330331\subsection{Convexity and Inflection Points}332The convexity and concavity of disconnected functions depends on convexity and concavity of each rule. Using intuition, one can figure out why this is the case.333334If we take the derivative of each rule we can determined335336What is intersecting is this kind if function is defined by more than one rule, the function, all together, can be both convex and concave in the same interval. However, stating function is convex and concave does not help create the graph. Hence we should state the "the rule is convex or concave".337338339340341342Hence in ${f(x)}^{g(x)}$, if ${|f(x)|}^{g(x)}$ i343344\subsection{Tangent Lines}345If a derivative exists in its domain then so does the tangent line. A secant line can run from one defined point to another as long as ${f(x)}^{g(x)}$ has dense subsets defined by the same rule.346347As the secant line narrows takes points in $A_n$ that approach $x=a$, the secant line approaches the tangent line. Hence the tangent line could exist if the following is equal.348349\begin{equation}350\left(\underset{x\in A_1}{\underset{x\to a}\lim} \frac{S(x)-S(a)}{x-a}\right)(x-a)+S(a)=\left(\underset{x\in A_2}{\underset{x\to a}\lim}\frac{S(x)-S(a)}{x-a}\right)(x-a)+S(a)351\end{equation}352353Thus a tangent line can exists on all the defined points of ${f(x)}^{g(x)}$. However having a tangent line defined on specific points is not such a fruitful notion. Most textbook may extend the disconnected function into all of $\mathbb{R}$ before taking the tangent.354355If the tangent line is not strictly determined by domain of the derivative we could give a looser definition. This because in real life we will a totally disconnected space dense in $\mathbb{R}$ as a connected curve and it will seem as though the tangent line exists anywhere.356357\begin{equation}358\left(\underset{x\in B_n}{\underset{x\to a}\lim}\left(\underset{t\in B_n}{\underset{t\to a}\lim}\frac{S(x)-S(t)}{x-t}\right)\right)(x-t)+\underset{t\in B_n}{\underset{t\to a}\lim}S(t)359\end{equation}360361Setting $S(x)=P\left(\right)$ we get the following.362363\begin{equation}364\underset{x\in B_2}{\underset{x\to a}\lim}\frac{P\left(f(x)^{g(x)}\right)-P\left(|f(a)|^{g(a)}\right)}{x-a}(x-a)+{P\left(|f(a)|^{g(a)}\right)}365\end{equation}366367In (27) we used a tangent line of defined points that approach arbitrarily close any point. If this is the case, the tangent line of $P\left(f(x)^{g(x)}\right)$ is always the tangent line of the rule which can be $ f(x)^{g(x)}$.368369If ${f(x)}^{g(x)}$ is defined by two distinct rules, then the equality cannot exist unless both function intersect at a continuous point. This has little use in mathematics since we are taking tangent lines of isolated points.370371On the other hand we could split ${f(x)}^{g(x)}$ into $|f(x)|^{g(x)}$ and $-|f(x)|^{g(x)}$ and take tangent lines defined on the subsets of each rule.372373\begin{equation}374\underset{x\in A_1}{\underset{x\to a}\lim}\frac{f(x)^{g(x)}-{f(a)}^{g(a)}}{x-a}(x-a)+{{f(a)}^{g(a)}}375\end{equation}376377\begin{equation}378\underset{x\in A_2}{\underset{x\to a}\lim}\frac{f(x)^{g(x)}-{f(a)}^{g(a)}}{x-a}(x-a)+{{f(a)}^{g(a)}}379\end{equation}380381In (28) and (29) each tangent line is defined on points in their corresponding subset. If an extension is required than use the looser definition on (26) for (28) and (29).382383\subsection{Diambiguity of the Anti-Derivative and Indefinite Integral}384The definition of anti-derivative, indefinite integral and definite integrals are different depending on the definitions in textbooks. This can be misleading to students and should be clarified.385386In most calculus textbooks, the anti-derivative of $f(x)$ is $F(x)$ where $F'(x)=f(x)$ and the indefinite integral is the shorthand notation for $F(x)=\int{f(x)}dx+C$. These textbooks say the constant $C$ was added to indefinite integrals in order to represent all solution to the anti-derivatives.387388In some Calculus textbooks, they mention the solution of $F'(x)=f(x)$ as either the indefinite integral or the anti-derivative.389390In most calculus textbooks, the definite integral is addressed as the Riemman integral. However in most analysis textbooks, the definite integral could be both the riemman or lebesque integral.391392For $\int_{a}^{x}f(t)dt$, some textbooks call this the "area function" while other textbooks call this the indefinite integral. In Bartle's elements of real analysis, he states the indefinite integral is in the form $\int_{a}^{x} f(t)dt=F(x)-F(a)$ where $F(a)$ is the ball point of indefinite integral function which conflict with textbooks stating $\int f(t)dt=F(t)+C$ where $C$ is any constant number.393394To avoid confusion, the anti-derivative should be viewed as the direct inverse of the derivative and indefinite integral should be viewed as the "indefinite version" of the definite integral with $x$ added to the integrals' upper limit. Hence in Calculus we should avoid using the constant $C$ when expressing an indefinite integral but it should be used when solving the anti-derivative.395396We should also address the difference between riemman integral and other kinds of integrals. For example, we should adress the Riemmen indefinite and definite integral as seperate from the Lebesgue indefinite and definite integral397398\subsection{Integration of ${f(x)}^{g(x)}$}399In order for an integral of any function to exist in an interval the function has to be defined (everywhere) in the interval. In rigorous terms if $S(x)={f(x)}^{g(x)}$ then $S:[a,b]\cap{\mathbb{R}}\to\mathbb{R}$ for an integral to exist; however, $S:[a,b]\cap{{B_2}}\to{\mathbb{R}}$.400401The following definition of Riemman integrability is given in "Elements of Real Analysis".402403Mathematicians require that function be defined everywhere so that404405As mentioned earlier, the function ${f(x)}^{g(x)}$ is not continuous nor differentable when $f(x)<0$. The function does not meet any condition for a Riemman integral to exist. Moreover, the function is defined on countable sets. There is currently no measure than compares countable set to one another which can be used in an integral.406407However, $P\left({f(x)}^{g(x)}\right)$ is continuous and differentable in its domain. There is possibility a modified integral can exist. For the time being, we cannot say such a function is integratable but we could ask, "Can an area under $P\left({f(x)}^{g(x)}\right)$ exist?".408409410\subsubsection{Darboux Sums}411412We will be use the Darboux sum because since they are easier to use the left or right Riemman sums and can be applied to functions with any type of discontinuity.413414For the function $F$ where $F:[a,b]\to\mathbb{R}$, the upper sum and lower sum is determined by taking the upper bound (suprenum) and lower bound (infinum) of each sub-interval $x\in[x_i,x_{i+1}]$ ($1\le i\le n-1$), in the partition of the interval. The partition of the interval is a finite sequence between $[a,b]$ which can be shown as the following.415416\begin{equation}417a=x_0<x_1<x_2<x_3<x_4<...<x_n=b418\end{equation}419420If we multiply the suprenum and infinum of each sub-interval with its length, we get the area of that sub-interval.421422\begin{equation}423\left(\inf_{[x_i,x_{i+1}]}S\right){(x_{i+1}-x_{i})}424\end{equation}425426\begin{equation}427\left(\sup_{[x_i,x_{i+1}]}S\right){(x_{i+1}-x_{i})}428\end{equation}429430Now we will sum the areas of each sub-interval. As the length of the sub-intervals decrease the upper and lower sum should converges to the definite integral. This can be described in the following notation.431432\begin{equation}433L(P,S)=\sum_{i=1}^{n-1}\left(\inf_{[x_j,x_{j+1}]}S\right){(x_{j+1}-x_{j})}434\end{equation}435436\begin{equation}437U(P,S)=\sum_{i=1}^{n-1}\left(\sup_{[x_j,x_{j+1}]}S\right){(x_{j+1}-x_{j})}438\end{equation}439440\begin{equation}441L(P_n,S)<\int_{a}^{b}f(x)dx<U(P_n,S)442\end{equation}443444445\subsection{Supremum and Infimum vs Minima and Maxima}446Supremum or Infimum may have been used instead of Maxima or Minima is because the suprenum and infimum of sub-intervals with jump or removable discontinuites are always defined. As the sub-intervals $[x_i,x_{i+1}]$ become smaller, the total area of partitions at discontinuities should have a smaller impact on the convergence of the upper and lower sums. This gave rise to the lebesque criteria of riemman integrability.447448\subsubsection{Definition}449For a function $S:[a,b]\to\mathbb{R}$, where $B$ is dense in $\mathbb{R}$, if $S$ is bounded and the lebesque measure of discontinuities is zero than the the function is Riemman integratable.450451The lebesque measure of countable sets is zero. If the function is discontinuous on a countable set than the function is Riemman integrable.452453The Suprenum and Infimum was created to include discontinuties and extend the type of functions that are Riemmen integrable. We could take this the next step further and include functions on a disconnected domain.454455\subsection{Darboux Sum without restrictions}456Further arguments can be made that the Darboux sum should not have the restriction $F:[a,b]\to\mathbb{R}$. This prevents mathematicians with experimenting with all kinds of functions to see if the upper and lower sum exists.457458Instead we should let any function be used for Darboux sums. If supremum and infimum of all the sub-intervals are defined no matter how small the length of the sub-interval; if the upper and lower sums converge, the function should have an integral.459460Using this approach we will take the Darboux sum of $S(x)=P\left({f(x)}^{g(x)}\right)$ which only requires that $S(x)$ in $[a,b]$ is bounded.461462$$L(P,S)=\sum_{i=1}^{n-1}\left(\underset{[x_j,x_{j+1}]}{\text{Inf}}S\right)(x_{j+1}-x_j)$$463464$$U(P,S)=\sum_{i=1}^{n-1}\left(\underset{[x_j,x_{j+1}]}{\text{Sup}}S\right)(x_{j+1}-x_j)$$465466However, we will not state the definite integral of $S(x)$. Instead we will state, "Area of $S(x)$".467468\begin{equation}469L(P,S)<\text{Area of} S(x)<U(P,S)470\end{equation}471472The disconnected function is defined on sets dense in $\mathbb{R}$. Hence the sets can approximate any real number. As the partitions $[x_{j},x_{j+1}]$ becomes smaller there is always an infinite number of defined points. If a maxima or minima does not exist for any of the sub-intervals there will always be an upper and lower bound.473474Moreover, using (21) and (22) the suprenum and infimum each sub-interval of $P\left({f(x)}^{g(x)}\right)$ is the same as the maximum and minimum of $P\left(\left|f(x)\right|^{g(x)}\right)$. If a function has a maximum or minimum it must have a suprenum or infimum. Hence we could if set $S_1=P\left({f(x)}^{g(x)}\right)$ we have the following inequality.475476\begin{equation}477L(P,S_1)<\int_{a}^{b} S_1(x)<U(P,S_1)478\end{equation}479480Since $L(P,S)=L(P,S_1)$ and $U(P,S)=U(P,S_1)$ we can change (36) into481482\begin{equation}483L(P,S_1)<\text{Area of}S(x)<U(P,S_1)484\end{equation}485486Hence using the "unrestricted" Darboux sums an area under $P\left({f(x)}^{g(x)}\right)$ would be $\int_{a}^{b}P\left({\left|f(x)\right|}^{g(x)}\right)$.487488Another interesting analogy is finding the area under small peice of wire, the wire is composed of atoms which are made of sub-particles, sub-particles which are made of quarks and so on. It seems no matter how small the smallest particles are is a space between each one, making the the wire almost totally disconnected.489490If we look under the area under the wire we act as though the wire is a connected space even through is actually disconnected. This because we focus on the defined points and ignore the undefined points. Hence I argue if the domain of defined points can approximate any value in $\mathbb{R}$.491492\subsection{Defining the Area under $S(x)$}493If an extended integral is to exist we must redefine the area under $S(x)$. Points of a dense set can approximate close to any $x$-value, meaning we can ignore the undefined points. Instead of stating the area must exist under a curve defined everywhere, we must state an area exist if the curve is a curve: a string of infinite points that no matter far we zoom in, infinite defined points will exist.494495The definition of the curve is the following.496497\subsubsection{Definition}498If $S:[a,b]\cap{A_n}\to\mathbb{R}$, such that $S$ is continuous on $[a,b]$ and $A_n$ is a dense subset of $\mathbb{R}$, then $S$ is a curve.499500By this we can say area exist under a function, if the function is a curve.501502503I believe we must redefined a curve which is equivelant to a "rule".504505\subsection{The effect on an "Extended Riemman Integral"}506If we allow Riemman integral to be defined on a disconnected sets dense in $\mathbb{R}$ we have to observe the effects of such functions have on various theorem Riemman integrability and Fundamental Theorem of Calculus507508\subsubsection{Riemman Integrability}509The previous factors of Riemman integrability require only minor adjustments to be suitable for disconnected function such as $P\left({f(x)}^{g(x)}\right)$510511Two important factors for Riemman integrability is continuity and differentability. We found $P\left({f(x)}^{g(x)}\right)$ is continuous and differentable in its domain. As argued previously, the properties a function should be determined in its domain. A function that is differentable should be a differentable function and function that is continuous in its domain should be a continuous function.512513The third important criteria is the Lebesque criterion for Riemman integrability needs no adjustments. We know previously that $P\left({f(x)}^{g(x)}\right)$ has no discontinuities but removable singularities and points with removable singularities have two properties5145151. The limit at the points exist5165172. The points are undefined518519One reason the points are called removable singularities is if you replace them with defined points that are continuous the discontinuities stay the same.520521We can summarize the properties for the "Extended Riemman Integratable".522523If $S$ is bounded, $S:[a,b]\cap{B_n}\to\mathbb{R}$ where $B_n\subseteq{R}$, and the Lebesque measure of discontinuities is zero than the function is Riemman integratable.524525526\subsubsection{Riemman Indefinite Integral and Anti-derivative}527528With an extended Riemman integral, the anti-derivative and Riemman Indefinite integral of continuous functions are now slightly different. While the anti-derivative is the exact opposite of differentiation and is restricted by the domain, the Riemman indefinite integral can exist at any any point.529530Since the anti-derivative and indefinite integral are different both cannot have the same notation. The anti-derivative of $f$ should be shown as $\frac{d^{-1}}{d^{-1}x}f+C$ while the indefinite integral of $f$ should be shown as $\int f$ (without the constant $C$).531532We can make two statements about the relationship of the anti-derivative to the extended riemman indefinite integral.5335341. The anti-derivative is restricted by discontinuities and points outside the domain, while the indefinite integral exists anywhere.5355362. The anti-derivative is the subset of the indefinite integral. $\frac{d^{-1}}{d^{-1}x}f\subseteq \int f$537538The problem is that the Riemman indefinite integral is not always the direct inverse of differentiation which weakens the fundamental theorem of Calculus.539540The fundamental theorem of Calculus is supposed to show the connection between the derivative, anti-derivative, indefinite and definite integral. Fortunately we do not need any adjustment. As long as $S:[a,b]\to\mathbb{R}$ and continuous, the indefinite integral is the same as the anti-derivative and the indefinite integral is the opposite of differentiation.541542We could make an extended version of the Second Fundamental Theorem of Calculus.543544\subsection{Modified Second Fundamental Theorem Calculus}545546Let $s$ and $S$ be a real-valued function defined on $x\in{B_n}$, where $B_n\subseteq{R}$, in the closed interval $[a,b]$, such that $S$ is continuous in the domain and $S'(x)=s(x)$.547548If $s$ is riemmen integratable on $[a,b]$ then549550$$\int_{a}^{b}s(x)=S_1(b)-S_1(a)$$551552Where $S_1(x)$ extends the domain of $S(x)$ to the entire interval.553554This establishes the Riemman integral as the superset of the anti-derivative.555556557\subsection{An example}558If we have the function of $|(-2)^{x}|$, an example of ${f(x)}^{g(x)}$. We find that when $x=\frac{p}{2q+1}|m,n\in{\mathbb{N}}$. Since $|(-2)^{x}|$ coincides with $2^x$ and the domain is dense, continuity can exist. Hence a derivative and anti-derivative can exist. The derivative is $\ln(2)|(-2)^x|$ and an anti-derivative is $\frac{1}{\ln(2)}|(-2)^{x}|+C$. If we remove the holes in $\left\{\left.\frac{2p+1}{2q}\right|m,n\in \mathbb{Z}\right\}$ then the indefinite integral is $\frac{1}{\ln(2)}2^x-\frac{1}{\ln(2)}{2^a}$. As a result the definite integral $\int_{a}^{b}(-2)^{x}=\big|_{a}^{b}\frac{1}{\ln(2)}(2^x)$ which could be checked by using the modified Riemman sum $\sum_{i=1}^{n-1}\underset{[x_j,x_{j+1}]}\inf|(-2)^{x}|\left(x_{j+1}-x_j\right)<\int_{a}^{b}|(-2)^x|<\sum_{i=1}^{n-1}\underset{[x_j,x_{j+1}]}\sup|(-2)^{x}|\left(x_{j+1}-x_j\right)$.559560\subsection{Conclusion}561In conclusion, we can make the following summary of ${f(x)}^{g(x)}$. Although it is not that unique compared other elementary function it is suprising.5625631. (Proposition) The function has a domain which includes negative non-integers.5645652. An elementary function that is defined on a connected set when $f(x)>0$ and a disconnected set when $f(x)<0$.5665673. A group of elementary functions that is not Riemman integrable or Lebesque integrable.568569The summary of ${P\left({f(x)}^{g(x)}\right)}$5705711. Has a limit that can exist everywhere.5725732. Is "naturally" continuous and differentable on a domain that is a disconnected space(Otherwise discontinuous at vertical asymptotes)5745753. Has no Riemman or Lebesque integral. But can have an "Extended Riemman integral".576577578\subsection{Propositions}5795801. For the function to be a continuous, differentable or integratable function; the function must be continuous, differentable or integratable in the domain.5815822. The Darboux sum should not have the requirement $f:[a,b]\to\mathbb{R}$5835843. The Riemman integral should be extended to include functions that are not defined everywhere.5855864. Integrals should exist for functions not defined everywhere as the defined points belong to set dense in $\mathbb{R}$.587588\subsubsection{Conclusion on Properties of Disconnected functions}5891. A function can be continuous and differentiable but not defined almost everywhere.5905912. The anti-derivative and indefinite integral are not the same. The "usual" Riemmen indefinite integral is different from the anti-derivative at removable or jump discontinuities. The extended Riemmen integral may differ at removable singularities.5925933. If the sets $B_n$ are not defined everywhere then $S(x)$ is an example of a functions that is not Lebesque nor Riemman integrtable. A new or extended kind of an integral could be formed which compares defined points of one set to defined points of another set.594595\subsection{Allowing an extended Riemman integral}596597The extended Riemman integral is different from Indefinite integral at points that are not continuous (removable singularities and discontinuities)5985993. (Proposition) The indefinite integral should not have the $C$ sign.600601602603\end{document}604605606607608609610611612613%sagemathcloud={"latex_command":"pdflatex -synctex=1 -interact=nonstopmode 'DissconnectedSet.tex'"}614