Insights Omissions in Mathematics Education: Gauge Integration - Comments

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1. Jul 22, 2016

micromass

Staff Emeritus
2. Jul 22, 2016

Staff: Mentor

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

3. Jul 22, 2016

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 $\mathbb{R}$ ?

4. Jul 23, 2016

micromass

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

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.

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.

5. Jul 23, 2016

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.

6. Jul 23, 2016

A. Neumaier

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?

7. Jul 23, 2016

strangerep

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?

8. Jul 23, 2016

pwsnafu

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 $AC_*$

9. Jul 23, 2016

micromass

Staff Emeritus
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.

Yes, they are.

That's a good idea!

10. Jul 23, 2016

micromass

Staff Emeritus
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 guage 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.

11. Jul 24, 2016

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 $f$?

12. Jul 24, 2016

micromass

Staff Emeritus
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.

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

13. Jul 24, 2016

wrobel

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

14. Jul 24, 2016

micromass

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

Last edited: Jul 24, 2016
15. Jul 24, 2016

A. Neumaier

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?

16. Jul 24, 2016

micromass

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

Yes.

17. Jul 24, 2016

A. Neumaier

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?

18. Jul 24, 2016

micromass

Staff Emeritus
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".

19. Jul 24, 2016

A. Neumaier

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

20. Jul 24, 2016

micromass

Staff Emeritus
No, every gauge integrable function is measurable.