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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?

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  • $\begingroup$ 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. $\endgroup$ – herb steinberg Jul 26 at 2:31
  • $\begingroup$ @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 $g:[0,1]\to\mathbb{R}$....... $\endgroup$ – Arbuja Jul 26 at 15:07
  • $\begingroup$ @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. $\endgroup$ – Arbuja Jul 26 at 15:08
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    $\begingroup$ 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? $\endgroup$ – herb steinberg Jul 27 at 0:04
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    $\begingroup$ 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. $\endgroup$ – herb steinberg Jul 27 at 2:04
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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).

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  • $\begingroup$ 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. $\endgroup$ – Arbuja Jul 27 at 4:20
  • $\begingroup$ 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. $\endgroup$ – Arbuja Jul 27 at 4:53
  • $\begingroup$ @Arbuja Are you trying to integrate or take an average? These are not the same thing. $\endgroup$ – Sambo Jul 27 at 17:18
  • $\begingroup$ Shouldn’t the average use integration? It many not be the same thing but there is a relationship. $\endgroup$ – Arbuja Jul 27 at 17:23
  • $\begingroup$ 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. $\endgroup$ – Arbuja Jul 27 at 17:44

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