Questions regarding Gauge and Darboux Integrals?

Question

Can the Gauge Integral exist on a function defined on a countable set? What would it equal?

I was wondering because if we took the Darboux Integral of $f:C\cap[a,b]\to\mathbb{R}$ where $C$ is a countable dense set and $f$ is continuous in $C$, the integral is the same as the Darboux Integral of $f:[a,b]\to\mathbb{R}$.

The definition of all known integrals state $f$ has to be defined in $[a,b]$ ? Why so? What are they trying to avoid?

Darboux integral (if it exists) equals Lebesgue integral, assuming measure is usual measure on real line. In that case Lebesgue integral over a countable set is zero. — herb steinberg, Jul 26 '19 at 02:31
@herbsteinberg What if we took the Darboux sum of $x$ for $x\in\mathbb{Q}\cap[0,1]$ without the “$f:[a,b]\to\mathbb{R}$ criteria”? If we divided $[0,1]$ into 10 partitions what is the supremum of $[0,1/10]$? Isn’t it $1/10$. What is the Infimum? Isn’t it zero. If we apply this to the other $9$ partitions we get the same supremums and infimums as $g:[0,1]\to\mathbb{R}$ for $g(x)=x$. As the partitions get smaller we get the same results as the Darboux Integral of $g:[0,1]\to\mathbb{R}$....... — Arbuja, Jul 26 '19 at 15:07
@herbsteinberg ..........However the Lebesgue Integral and (presumably the Gauge Integral) say the integral of $s:[0,1]\cap\mathbb{Q}\to\mathbb{R}$ for $s(x)=x$ is zero. — Arbuja, Jul 26 '19 at 15:08
@herbsteinberg Check the comment where I talk about the Darboux sum. The Lebesgue Integral does not equal the limit of the Darboux sum and hence the Darboux Integral. If we ignore the criteria $f:[a,b]\to\mathbb{R}$ and replace it with $f:C\cap[a,b]\to\mathbb{R}$ where $C$ is a countable set dense in $[a,b]$, $f$ has the same Darboux sum limit, and integral, as $f:[a,b]\to\mathbb{R}$. — Arbuja, Jul 26 '19 at 17:32
@herbsteinberg In the case of $x$ for $x\in\mathbb{Q}\cap[0,1]$ the limit of the darboux sum gives $1/2$, while the Lebesgue integral gives zero.There is a conflict. Read all my comments to see why. — Arbuja, Jul 26 '19 at 20:20
I am unable to follow your details, but you seem to have a lower Darbous sum of $0$. What is $f$ outside of $C$? Are you integrating over $[a,b]$ or over $C\cap [a,b]$? If the latter, what is the measure you are using for the countable set? — herb steinberg, Jul 27 '19 at 00:04
@herbsteinberg I'm not using a measure, just darboux sum. For $f(x)=x$ and $f:\mathbb{Q}\cap[0,1]\to\mathbb{R}$, $\inf_{x\in[0,1/5]}x+\inf_{x\in[1/5,2/5]}x+\inf_{x\in[2/5,3/5]}x+\inf_{x\in[3/5,4/5]}x+\inf_{x\in[4/5,1]}x=0 \times 1/5+1/5 \times 1/5+2/5 \times 1/5+3/5 \times 1/5+4/5 \times 1/5=10/25=2/5$ and $\sup_{x\in[0,1/5]}x+\sup_{x\in[1/5,2/5]}x+\sup_{x\in[2/5,3/5]}x+\inf_{x\in[3/5,4/5]}x+\inf_{x\in[4/5,1]}x=1/5 \times 1/5+2/5 \times 1/5+3/5 \times 1/5+4/5 \times 1/5+1 \times 1/5=3/5$. Remember focus on defined points and discard undefined. Note that $\frac{2}{5}<\frac{1}{2}<\frac{3}{5}$. — Arbuja, Jul 27 '19 at 00:29
@herbsteinberg If you saw the calculations, the darboux sum of $f:\mathbb{Q}\cap[0,1]\to\mathbb{R}$ is the same as the sum of $f:[0,1]\to\mathbb{R}$. — Arbuja, Jul 27 '19 at 00:36
I got what you are doing. To say the least, what you are doing is unusual. I am puzzled as to why. As for your last question - I guess it has to with the fact that integral in its simplest form is used to get an area under a curve, so the domain is an interval. — herb steinberg, Jul 27 '19 at 02:04
@herbsteinberg I keep posting because I'm afraid I will never graduate from college. In case that happens, I need someone to look into this. I want to publish a paper in a magazine or undergraduate journal on these functions, their graph, derivatives, intuition behind the integrals, and various measures to apply. I have other posts hidden in this website. — Arbuja, Jul 27 '19 at 02:30
@herbsteinberg Sambo’s answer gives a good reason. Additivity of the domains would not work. However, the Integral gives an appropriate average of $f$. My argument is the additivity of the domains is only needed when $f[a,b]\to\mathbb{R}$; otherwise the relationship between the integral and the average is more important. — Arbuja, Jul 27 '19 at 17:10

score 1 · Accepted Answer · answered Jul 27 '19 at 03:46

1

I will assume that in your context, $f$ is a function defined (and continuous, for simplicity) on all of $[a,b]$. If this is not a fair assumption let me know and I will edit (or delete) my answer.

Let $C$ be a countable dense subset of $\mathbb{R}$. As you noted, applying the definition of the Darboux integral (in particular with Darboux sums) to $f$ while only considering the domain $C \cap [a,b]$ yields the same result as the Darboux integral of $f$. For lack of better notation, I will denote the limiting Darboux sum of $f$ when restricted to $C \cap [a,b]$ by:

\begin{align*} \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} \end{align*}

So, what you have noticed is that:

\begin{align*} \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} = \int_{[a,b]} f \end{align*}

(I have not checked that this is always true, but I suspect that it is if $f$ is continuous on $[a,b]$.)

The reason, then, that we don't use this as a definition for the integral of $f$ over $C \cap [a,b]$ is that we lose a nice property of integration: additivity of the domain.

Let $g_1 : [a,b] \rightarrow \mathbb{R}$ and $g_2 : [c,d] \rightarrow \mathbb{R}$ be such that $[a,b] \cap [c,d] = \varnothing$. Define $g : [a,b] \cup [c,d] \rightarrow \mathbb{R}$ as follows: \begin{align*} g(x) = \left\{ \begin{matrix} g_1(x) & \text{if } x \in [a,b] \\ g_2(x) & \text{if } x \in [c,d] \end{matrix} \right. \end{align*} Then we have the following nice property:

\begin{align*} \int_{[a,b]\cup[c,d]} g = \int_{[a,b]} g_1 + \int_{[c,d]} g_2 = \int_{[a,b]} \left. g \right|_{[a,b]} + \int_{[c,d]} \left. g \right|_{[c,d]} \end{align*}

As it turns out, with your definition, this no longer holds. Consider the following example. Let $C_1 = \mathbb{Q}$, and let $C_2 = \mathbb{Q} + \sqrt{2} = \{ q + \sqrt{2} : q \in \mathbb{Q}\}$. Then both $C_1$ and $C_2$ are countable dense sets, and their intersection is $\varnothing$. Therefore their union, $C = C_1 \cup C_2$, is also countable and dense.

But then by your definition, we should have:

\begin{align*} \int_{C_1 \cap [a,b]} \left. f \right|_{C_1 \cap [a,b]} = \int_{C_2 \cap [a,b]} \left. f \right|_{C_2 \cap [a,b]} = \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} = \int_{[a,b]} f \end{align*}

And so, as long as $\int_{[a,b]} f \neq 0$, we end up with:

\begin{align*} \int_{C_1 \cap [a,b]} \left. f \right|_{C_1 \cap [a,b]} + \int_{C_2 \cap [a,b]} \left. f \right|_{C_2 \cap [a,b]} \neq \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} \end{align*}

Since additivity of the domain is one of the most important features of the integral, we choose not to use the definition you proposed. Instead, we agree that the integral over a countable dense set is either undefined (as per the Darboux integral) or zero (as per the Lebesgue integral).

answered Jul 27 '19 at 03:46

Sambo

6,243

Good point, but what is the average? If we took the average in terms of the Lebesgue Integral, it is zero. But intuitively the domain could be greater. – Arbuja Jul 27 '19 at 04:20
It seems the average in terms of the integral wouldn’t work in this case. We could just extend the countable set to all real numbers. But then again what is the average if $f(x)$ in $[0,1]$ is $2$ at $x\in\mathbb{Q}\setminus\mathbb{Q}^2$ and $1$ at $x\in\mathbb{Q}^2$? We can’t use the Lebesgue Integral, intuitively if the average exists, it has to be greater than zero. The Darboux Integral says it does not exist but does not give an average for the Derichlet function. We need something different. – Arbuja Jul 27 '19 at 04:53
@Arbuja Are you trying to integrate or take an average? These are not the same thing. – Sambo Jul 27 '19 at 17:18
Shouldn’t the average use integration? It many not be the same thing but there is a relationship. – Arbuja Jul 27 '19 at 17:23
Ok...they are not the same thing. But wouldn’t it be easier to compute the average if we used an integral? The average of $f$ is the same as $F:[a,b]\to\mathbb{R}$ where $f=F$ at $x=C$. In my opinion, integration with additivity works if the measure of the domain is $1$ but not otherwise. Otherwise, it is best to use an Integral that gives an appropriate average. – Arbuja Jul 27 '19 at 17:44
What is your goal here? To define the average of a function defined on a countable dense set? – Sambo Jul 27 '19 at 17:56
The Integral is the “area under the curve”. The countable set has points so infinitesimally close to each other we can ignore the undefind singularities. So taking the integral of $f$ is like taking the integral on the extension of its domain to all real numbers. – Arbuja Jul 27 '19 at 18:00
If you have $f$ defined only on $C \cap [a,b]$, how do you extend it to be defined on $[a,b]$? – Sambo Jul 27 '19 at 18:05
I’m not sure. However, if you take the Darboux sum of $f$ and replace $f:[a,b]\to\mathbb{R}$ criteria with $f:C\cap[a,b]\to\mathbb{R}$ it’s the same as the sum of $F:[a,b]\to\mathbb{R}$. – Arbuja Jul 27 '19 at 18:40
Where $f=F$ for $x\in C$ – Arbuja Jul 27 '19 at 19:10
@Arbuja This is not true for any extension $F$ you choose. For example, suppose I have $f : \mathbb{Q} \cap [0,1] \rightarrow \mathbb{R}$ defined as $f(x) = x$, and I set $F:[0,1] \rightarrow \mathbb{R}$ to be $F(x) = f(x)$ when $x \in \mathbb{Q}$ and $F(x) = 0$ otherwise. Then the Darboux sum of $F$ on $[0,1]$ is certainly not the same as the Darboux sum of $f$ on $\mathbb{Q} \cap [0,1]$. – Sambo Jul 27 '19 at 19:13
Perhaps you believe that if $f$ is continuous on $C \cap [a,b]$, there is a unique Darboux integrable function $F : [a,b] \rightarrow \mathbb{R}$ such that $F(x) = f(x)$ on $C \cap [a,b]$. I don't know whether this is true. – Sambo Jul 27 '19 at 19:15
I confused you. $F=f$ such that $\lim\limits_{x\to a} f(x) =F(a)$ for $a\in\mathbb{R}$. If not than the Darboux sum of $f$ does not equal $F$. – Arbuja Jul 27 '19 at 19:20
You seem to be assuming that such a function $F$ will always exist if $f$ is continuous on a countable dense set (e.g., $\mathbb{Q}$). This is not the case. – Sambo Jul 27 '19 at 19:40
Right. $F=f$ such that $\lim_{x\to a}f(x)=f(a)$ for $a\in T$ and the Lebesgue measure of $\mathbb{R} \setminus T$ is zero. Sorry for the trouble. – Arbuja Jul 27 '19 at 19:58
So we are no longer assuming that the dense set is countable? I'm quite confused. – Sambo Jul 27 '19 at 20:01
When $\mu$ is the Lebesgue measure, $f:C\cap[a,b]\to\mathbb{R}$, $\mu(C)=0$, $C$ is dense in $\mathbb{R}$, $F:[a,b]\to\mathbb{R}$, $\lim\limits_{x\to a}f(x)=F(a)$, $a\in A$ and $\mu(\mathbb{R}\setminus A)=0$, we get $\int F=\int f$ – Arbuja Jul 27 '19 at 20:20
Again, you will need to show that such an $A$ and $F$ exist. – Sambo Jul 27 '19 at 20:33
Above, you said "taking the integral of $f$ is like taking the integral on the extension of its domain to all real numbers". You amended this to say it is like taking the integral of $F$ over $A$, where $F$ has the properties you described. But you have not proven that such an F always exists. – Sambo Jul 27 '19 at 22:11
Take the Darboux sum of $f(x)=x$ for $x\in\mathbb{Q}\cap[0,1]$. Set $\mathbb{Q}$ is countable and dense in $\mathbb{R}$ meaning it can approximate arbitrarily close to any real number, so $\lim\limits_{x\to p}f(x)=f(p)$ where $p\in[a,b]$. Variable $p$ can be seen as an input and $f(p)$ is the same as $f(x)=x$ for $x\in[a,b]$. Hence if we partition $[a,b]$ into $\cup_{i=1}^{n}[x_i,x_{i+1}]$, $\sum\limits_{i=1}^{n}\lim_{x\to p}\inf\limits_{x\in\mathbb{Q}\cap[x_i,x_{i+1}]}x(x_{i+1}-x_{i})=\sum\limits\inf\limits_{p\in[x_i,x_{i+1}]}p(x_{i+1}-x_{i})<\int f(p)<$ – Arbuja Jul 27 '19 at 23:21
...........$\sum\limits_{i=1}^{n}\lim_{x\to p}\inf\limits_{x\in\mathbb{Q}\cap[x_i,x_{i+1}]}x(x_{i+1}-x_{i})=\sum\limits\inf\limits_{p\in[x_i,x_{i+1}]}p(x_{i+1}-x_{i})$ Hence $\int f(x)=\int f(p)$ – Arbuja Jul 27 '19 at 23:23
Actually $p\notin[a,b]$. $p\in A\cap[a,b]$. As we showed earlier the Darbox sum of $f(x)$ is the Darboux sum of $f(p)$. According to the Lebesgue criteria of Riemann integration, if the discontinuities of $f(p)$, outside the domain or at $\mathbb{R}\setminus p$, has a Lebesgue measure of zero, an integral is defined. – Arbuja Jul 27 '19 at 23:54
I understand that you're showing me how the example $f(x)=x$ works out, but this does not mean it holds for general $x$. – Sambo Jul 28 '19 at 00:04
See the whole proof. Replace $f(x)=x$ with a general $f(x)$ and $f(p)=p$ with a general $f(p)$. The proof should work. As long as $\mathbb{R}\setminus p$ has a Lebesgue measure of zero, the area of $f(p)$ exists and is the area of $f(x)$. – Arbuja Jul 28 '19 at 00:11
What does $\mathbb{R} \backslash p$ mean? You seem to be using $p$ for the variable of your function. – Sambo Jul 28 '19 at 00:15
$p\in A\cap[a,b]$ such that $\lim\limits_{x\to p}f(x)=f(p)$. If the set of discontinuities of $f(p)$ or $[a,b]\setminus A$ has a Lebesgue measure of zero, the Darboux integral of $f(p)$ exists and is the same as the Darboux integral of $f(x)$. (Remember $x\in{C}$ and $C$ is dense in $\mathbb{R}$ with a Lebesgue measure of $0$ and check the proof above) – Arbuja Jul 28 '19 at 01:17
That doesn't answer my question – Sambo Jul 28 '19 at 02:23
Correction, $\mathbb{R}\setminus{p}$ should have been $[a,b]\setminus A$ where $p\in A\cap[a,b]$. This is the set of discontinuities of $f(p)$ – Arbuja Jul 28 '19 at 15:48
Your proof is not clear about distinguishing between your functions being restricted to different domains. From what I understand, you write "$f(x)$" to denote the original function $f : C \cap [a,b] \to \mathbb{R}$, and you write "$f(p)$" to denote a new function $F : [a,b] \backslash A \to \mathbb{R}$ such that $F(x) = f(x)$ for every $x \in C \cap [a,b]$. Is this correct? – Sambo Jul 28 '19 at 15:56
$f(x)$ is $f:C\cap[a,b]\to\mathbb{R}$ but $f(p)$ is $F:A\cap[a,b]\to\mathbb{R}$. I made this correction in the third comment of my proof. By the Lebesgue criteria for Riemann Integration if $[a,b]\setminus A$ (which is the set of discontinuities of $F$ ) has a Lebesgue measure of zero, a Reimman Integral exists. As the proof shows, if $F$ has a Riemann Integral so does $f(x)$. – Arbuja Jul 28 '19 at 16:21
And what if $[a,b] \backslash A$ doesn't have a lebesgue measure of zero? – Sambo Jul 28 '19 at 16:24
Then the Riemann integral of $F$, and hence $f(x)$, doesn’t exist. – Arbuja Jul 28 '19 at 16:30

Questions regarding Gauge and Darboux Integrals?

1 Answers1