Omissions in Mathematics Education: Gauge Integration - Comments

[micromass]

micromass submitted a new PF Insights post

Omissions in Mathematics Education: Gauge Integration

Continue reading the Original PF Insights Post.

 

[fresh_42]

Enlightening! I always thought of Lebesgue as Riemann with running over the null-sets. This article cleared this wrong view.

 

[strangerep]

@micromass: Coincidentally, I recently read a bit about the "Henstock--Kurzweil" integral, which I understand is the same thing as the Gauge Integral? (Btw, is there any reason you didn't mention that name?)

Regarding the integral $$\int_0^\infty \frac{\sin(x)}{x}\, dx ~~,$$I've only ever seen that performed by contour integration (i.e., interpreted as a Cauchy PV integral). How is it done by Gauge Integration? (Maybe you could post an online link to the details?)

And speaking of contour integration, one thing I like about Cauchy PV integrals is their relationship to Lebesgue integrals. But I get the impression from your article that one cannot in general achieve such a close relationship with Gauge Integrals, since the domain is $\mathbb C$, not $\mathbb{R}$ ?

 

[micromass]

@micromass: Coincidentally, I recently read a bit about the "Henstock--Kurzweil" integral, which I understand is the same thing as the Gauge Integral? (Btw, is there any reason you didn't mention that name?)

Yes, it is the same thing. I'll add the various names to the article. Thanks

Regarding the integral $$\int_0^\infty \frac{\sin(x)}{x}\, dx ~~,$$I've only ever seen that performed by contour integration (i.e., interpreted as a Cauchy PV integral). How is it done by Gauge Integration? (Maybe you could post an online link to the details?)

One approach is to differentiate under the integral sign. Consider $F(t) = \int_0^{+\infty} e^{-tx}\frac{\sin(x)}{x}dx$. Then
F'(t) = -\int_0^{+\infty} e^{-tx} \sin(x)dx = -\frac{1}{1+t^2}
So $F(t) = \frac{\pi}{2} - \text{arctan}(t)$. Letting $t\rightarrow 0$ gives us the value of $\pi/2$.
Here you can find more details: http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf This document doesn't deal with the gauge integral, but everything in the document applies to that setting as well.

And speaking of contour integration, one thing I like about Cauchy PV integrals is their relationship to Lebesgue integrals. But I get the impression from your article that one cannot in general achieve such a close relationship with Gauge Integrals, since the domain is $\mathbb C$, not $\mathbb{R}$ ?

Right. It is possible though to define the gauge integral on $\mathbb{C}$, but this is technical. In any case, it's not something I've looked into.

 

[A. Neumaier]

The main reason why the standard courses about integration treat the Lebesgue integral rather than the Henstock integral is that the former has much stronger properties required in all applications to measure theory and functional analysis.

One cannot do most of modern mathematics without the Lebesgue integral, while one can do most of it without the Henstock integral. Being maximally general is simply something very different from being maximally useful.

 

[A. Neumaier]

to differentiate under the integral sign

This is a heuristic recipe that requires justification. Why is one allowed to do that in this particular case? Surely there are conditions on the integrand needed to make it work as it is known yhat the trick may fail to give the correct answer.

The justification must be based on proven properties of the integral. Are these properties satisfied for the Lebesgue integral? For the Henstock integral?

 

[strangerep]

One approach is to differentiate under the integral sign. [...]

Oh, ok. (I knew about DUI already, of course, but hadn't seen it applied to sin(x)/x.) In any case, I now realize I misinterpreted your article. I was replying to your statement:

Furthermore, integrals like $\int_0^\infty \sin(x)/x \; dx$ that can not be found by Lebesgue, can be found by the gauge integral and yield sensible answers

but forgot that you had said earlier that it's Riemann-integrable.

DUI tends to be used quite freely in theoretical physics, so it would be useful to know if there's any easy ways to tell where it doesn't work. Perhaps a subject for another insights article?

 

[pwsnafu]

This is a heuristic recipe that requires justification. Why is one allowed to do that in this particular case? Surely there are conditions on the integrand needed to make it work as it is known yhat the trick may fail to give the correct answer.

The justification must be based on proven properties of the integral. Are these properties satisfied for the Lebesgue integral? For the Henstock integral?

From
Necessary and Sufficient Conditions for Differentiating Under the Integral Sign by Erik Talvila

Theorem Let $f:[\alpha,\beta] \times [a,b] \to \mathbb{R}$. Suppose that $f(\cdot, y)$ is $ACG_*$ on $[\alpha, \beta]$ for almost all $y \in (a,b)$. Then $F:= \int_a^b f(\cdot,y) \, dy$ is $ACG_*$ on $[\alpha,\beta]$ and $F'(x) = \int_a^b f_1(x,y) \, dy$ for almost all $x \in (\alpha,\beta)$ if and only if
$\int_{x=s}^{t} \int_{y=a}^b f_1 (x,y) dy dx = \int_{y=a}^{b} \int_{x=s}^t f_1(x,y) dx dy$ for all $[s,t]\subset [\alpha,\beta]$.

Recall, a function is said to be absolutely continuous in the restricted sense on $E\subset [a,b]$ ($AC_*$) if for all $\epsilon > 0$ there exists $\delta > 0$ such that $\Sigma_{i=1}^{N} \sup_{x,y \in [x_i, y_i]} | F(x) - F(y)| < \epsilon$ for all finite sets of disjoint open intervals with endpoints in E and $\Sigma_{i=1}^N (y_i-x_i) < \delta$.
A function is said to be generalised absolutely continuous in the restricted sense on E ($ACG_*$) if it is continuous and E is a countable union of sets on which it is $AC_*$

 

[micromass]

The main reason why the standard courses about integration treat the Lebesgue integral rather than the Henstock integral is that the former has much stronger properties required in all applications to measure theory and functional analysis.

One cannot do most of modern mathematics without the Lebesgue integral, while one can do most of it without the Henstock integral. Being maximally general is simply something very different from being maximally useful.

Almost all of the theorems of Lebesgue integration also hold for the gauge integral. This includes stuff like dominated convergence and integration under the integral sign.

The justification must be based on proven properties of the integral. Are these properties satisfied for the Lebesgue integral? For the Henstock integral?

Yes, they are.

Oh, ok. (I knew about DUI already, of course, but hadn't seen it applied to sin(x)/x.) In any case, I now realize I misinterpreted your article. I was replying to your statement:
but forgot that you had said earlier that it's Riemann-integrable.

DUI tends to be used quite freely in theoretical physics, so it would be useful to know if there's any easy ways to tell where it doesn't work. Perhaps a subject for another insights article?

That's a good idea!

 

[micromass]

Here is the theorem of differentiation under the integral sign for the gauge integral:

Theorem: Let $f:[a,b]\times [c,d]\rightarow \mathbb{R}$ (where $b$ can be infinity) such that for each $t\in [c,d]$, the function $x\rightarrow f(x,t)$ is measurable on $[a,b]$.
Suppose that:
1) There exists $t\in [c,d]$ such $x\rightarrow f(x,t)$ is gauge integrable.
2) The partial derivative $\frac{\partial f}{\partial t}$ exists on $[a,b]\times [c,d]$.
3) There are gauge integrable functions $\alpha$ and $\omega$ such that
\alpha(x)\leq \frac{\partial f}{\partial t}(x,t)\leq \omega(x)
for all $x\in [c,d]$ and $t\in [a,b]$.
1) Then $x\rightarrow f(x,t)$ is gauge integrable for each $t\in [c,d]$.
2) The function $x\rightarrow \frac{\partial f}{\partial t}$ is gauge integrable for each $t\in [c,d]$
3) We have
\frac{d}{dt}\int_a^b f(x,t)dx = \int_a^b \frac{\partial f}{\partial t}dx

In particular, if $f(x,t) = e^{-tx}\frac{\sin(x)}{x}$, then this is clearly measurable in $x$ since it is continuous.
Setting $t=1$ gives us $e^{-x}\frac{\sin(x)}{x}$ which is gauge integrable by the following theorem.
Theorem: A measurable function $g$ is gauge integrable iff there are gauge integrable functions $g_1$, $g_2$ such that $g_1\leq g\leq g_2$.

We have $\frac{\partial f}{\partial t} = -e^{-tx}\sin(x)$ exists on $[0,+\infty]\times [\varepsilon,1]$ and is easily seen to be gauge integrable by using the above Theorem. So the theorem applies, at least for $t\in [\varepsilon, 1]$. So for those $t$, we can indeed et
\int_0^{+\infty} e^{-tx}\frac{\sin(x)}{x}dx = \frac{\pi}{2} - \text{arctan}(t)
Then we would need to switch limit and integral to conclude that $\int_0^{+\infty}\frac{\sin(x)}{x}dx = \frac{\pi}{2}$. This is provided in the document I linked.

 

[wrobel]

Does the gauge integral allow to define spaces analogous to $L^p$? If the gauge integral $\int|f|=0$ then what can we say about $f$?

 

[micromass]

Does the gauge integral allow to define spaces analogous to $L^p$?

The situation is the same as with convergent and absolutely convergent series. Absolutely convergent series allow us to define $\ell^p$. Convergent series is general do not. In the same way, the absolutely convergent integrals can be used to define $L^p$. Note that $|f|$ is absolutely convergent iff $f$ is Lebesgue integral.

If the gauge integral $\int|f|=0$ then what can we say about $f$?

We can say that $f=0$ a.e. just as with the Lebesgue integral.

 

[wrobel]

Thanks.
So if $|f|$ is gauge integrable then $f$ it is Lebesgue integrable?

 

[micromass]

Thanks.
So if $|f|$ is gauge integrable then $f$ it is Lebesgue integrable?

That is essentially correct. We just want $f$ to be measurable/gauge integrable to exlcude pathological cases having to do with nonmeasurable functions.

 

[A. Neumaier]

That is essentially correct. We just want $f$ to be measurable/gauge integrable to exclude pathological cases having to do with nonmeasurable functions.

What do you mean by ''essentially correct"? The question was a precise mathematical statement, so it is either known to be true or not known to be true. And is the inverse also true, is every Lebesgue integrable function also gauge integrable?

 

[micromass]

What do you mean by ''essentially correct"? The question was a precise mathematical statement, so it is either known to be true or not known to be true.

Read the rest of the post for the exact statement that is true. It also says that the statement as posted is false.

And is the inverse also true, is every Lebesgue integrable function also gauge integrable?

Yes.

 

[A. Neumaier]

Read the rest of the post for the exact statement that is true. It also says that the statement as posted is false.
Yes.

The two answers don't quite match. It seems to me that you wanted to say that ''$|f|$ is gauge integrable iff $f$ is measurable and Lebesgue integrable''. Is this the correct assertion?

 

[micromass]

The two answers don't quite match. It seems to me that you wanted to say that ''$|f|$ is gauge integrable iff $f$ is measurable and Lebesgue integrable''. Is this the correct assertion?

Lebesgue integrable implies measurable, so I don't know why you put in a condition "measurable and Lebesgue integrable".

So no, it is not the correct assertion. It is "$f$ is Lebesgue integrable iff $f$ is measurable and $|f|$ is gauge integrable".

 

[A. Neumaier]

Lebesgue integrable implies measurable, so I don't know why you put in a condition "measurable and Lebesgue integrable".

OK, thanks. Are there nonmeasurable functions $f$ for which both $f$ and $|f|$ are gauge integrable? (assuming ZFC)

 

[micromass]

OK, thanks. Are there nonmeasurable functions $f$ for which both $f$ and $|f|$ are gauge integrable? (assuming ZFC)

No, every gauge integrable function is measurable.

 

[A. Neumaier]

No, every gauge integrable function is measurable.

Then the optimal statement should be ''$|f|$ is gauge integrable iff $f$ is Lebesgue integrable'', shouldn't it?

 

[micromass]

Then the optimal statement should be ''$|f|$ is gauge integrable iff $f$ is Lebesgue integrable'', shouldn't it?

No, that would be incorrect. Can you show why you think your statement is true?

 

[A. Neumaier]

No, that would be incorrect. Can you show why you think your statement is true?

Well if (i) Lebesgue integrable implies measurable and (ii) every gauge integrable function is measurable and (iii) $f$ is Lebesgue integrable iff $f$ is measurable and $|f|$ is gauge integrable" (three assertions taken from your posts) then my statement follows.

 

[micromass]

How would "$|f|$ is gauge integrable" imply "$f$ is measurable"? From my posts follows only that $|f|$ is measurable.

 

[A. Neumaier]

How would "$|f|$ is gauge integrable" imply "$f$ is measurable"? From my posts follows only that $|f|$ is measurable.

Ok; I see the subtle difference. Thus ''$f$ is Lebesgue integrable iff $f$ and $|f|$ are gauge integrable".

 

[Stephen Tashi]

The Insight article has nice pictures that give intuition about the Riemann integral and the Legesgue integral. Is there a useful picture that explains the gauge integral ?

The arguments given in favor of the gauge integral focus on the nice implications it provides - if f is gauge integrable then ... To use such implications in a specific setting, one would need to establish the "f is gauge integrable" clause. I don't get any intuitive understanding of how to do that, except in a trivial case where we can define \delta(x) to be a constant function, reverting it to the ordinary \delta.

For example, I gather that \int_0 ^ {\infty} \frac{\sin{x}}{x} can be defined as an "extended" gauge integral in the usual way, by taking the limit of a gauge integrals over a finite intervals. So, technically, we need the existence of a different function \delta(x) for each of the finite intervals.

Does the "gauge" in gauge integral has something to do with the concept of "gauge" in physics?

(Unfortunately for me, the concept of a "gauge" in physics isn't very intuitive. There are explanations such as
https://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/ , but I need someone to explain the explanation!)

 

[jedishrfu]

minor typo in the first paragraph look for integratal

 

[jedishrfu]

Liked your article too.

 

[disregardthat]

Interesting. So do you as for improper Riemann integrals define for example the gauge integral $\int^{\infty}_a f(x) dx$ as the limit of the gauge integrals $\lim_{t \to \infty} \int^t_a f(x) dx$ (whenever the limit exists or is $\pm \infty$) where $f : [a, \infty) \to \mathbb{R}$ is gauge integrable on $[a,t]$ for all $t \geq a$? As far as I can see, you only gave a definition of a gauge integrable function on a closed interval.

Assuming the above, what's stopping us from simply defining $\int^{\infty}_a f(x) dx$ as the limit of the Lebesgue integrals $\lim_{t \to \infty} \int^t_a f(x) dx$ (whenever the limit exists or is $\pm \infty$) for functions $f : [a,\infty) \to \mathbb{R}$ which are Lebesgue integrable on $[a,t]$ for all $t \geq a$? Using the same idea for all the different improper integrals would give us a notion of improper Lebesgue integrals, and improperly Lebesgue integrable functions.

It seems to me that perhaps the theorem you gave about the equivalence between Lebesgue integrable functions and gauge integrable functions such that $\int |f| < \infty$ implies that the definition above of a Lebesgue integrable function (proper and improper) is equivalent with the definition of a gauge integrable function.

Maybe I'm wrong and missing something here. What exactly can gauge integration do which Lebesgue integration and improper Lebesgue integration like I defined above cannot?

Also, what do you do about gauge integrals $\int_A f(x) dx$ for arbitrary sets $A$? For which sets and functions are such integrals defined?