Is integration theory supposed to be this hard?

[lugita15]

Fredrik, there's another approach to integration that may interest you, the one that is developed in Wheeden and Zygmund. In this approach you essentially get the definition of the integral for "free", and a lot of the theorems that you have to struggle with in other approaches also come for "free", but then you have to prove some theorems that are pretty trivial in the other approaches, like the dominated convergence theorem and the result that the integral is equal to the limit of the integrals of simple functions. It goes like this: for any nonnegative function f:ℝ^{n}\rightarrowℝ we define the integral of f over a region E\subsetℝ^{n} to be the Lebesgue measure of the region under the graph of the function, considered as a subset of ℝ^{n+1}. That's it! (And of course you have to do the usual definition of the integral for arbitrary functions using positive and negative parts, but that's trivial.) Altogether, I like this geometric approach better, and of course you ultimately get the same theorems and properties of the integral as from the step-function approach.

 

[Fredrik]

Thanks for the tip. That sounds like a good approach if we're only interested in integrals on $\mathbb R^n$. If we want to work with other measures than the Lebesgue measure, or do integrals on something else, like topological groups, then I think we need one of the approaches discussed above.

 

[lugita15]

Thanks for the tip. That sounds like a good approach if we're only interested in integrals on $\mathbb R^n$. If we want to work with other measures than the Lebesgue measure, or do integrals on something else, like topological groups, then I think we need one of the approaches discussed above.

Actually, I think this approach can be easily generalized to abstract integration: if (X,\mu) is a measure space, just use the product measure on the space X\timesℝ. Now as for other things you might want to do, like integrals with respect to operator-valued measures, that might be a bit more iffy because I don't know whether you can meaningfully form a product measure out of an operator-valued measure.

 

[Fredrik]

Anyone see how to prove that the limit definition "implies" the supremum definition?

Limit definition: An a.e. real-valued measurable function f is said to be integrable if there's a sequence $\langle f_n\rangle$ of integrable simple functions that's Cauchy in the mean and such that $f_n\to f$ a.e. The integral of an integrable function is defined by
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
\int f\dmu=\lim_n\int f_n\dmu,$$ where the $f_n$ are the terms of any sequence of integrable simple functions that's Cauchy in the mean and such that $f_n\to f$ a.e.

Supremum definition: An a.e. real-valued measurable function f is said to be integrable if the sets
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
I_f^\pm=\bigg\{\int\phi\dmu\bigg|\phi\text{ is simple, }0\leq\phi\leq f^\pm\bigg\}$$ are bounded from above. The integral of an integrable function is defined by
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
\int f\dmu=\sup I_f^+-\sup I_f^-.$$
I was able to prove that if f is integrable according to the supremum definition, then it's integrable according to the limit definition, and the two definitions agree about the value of the integral. But I don't see how to prove that if f is integrable according to the limit definition, then it's integrable according to the supremum definition. I have proved that f is integrable according to the limit definition if and only if f⁺ and f^- are integrable according to the limit definition. I'm thinking that I should try to derive a contradiction from the assumption that one of $I_f^+$ and $I_f^-$ is not bounded from above. If $I_f^+$ isn't bounded from above, there's a sequence $\langle\phi_n\rangle$ of integrable simple functions such that for all n, $0\leq\phi_n\leq f^+$ and $\newcommand{\dmu}{\ \mathrm{d}\mu}\int\phi_n\dmu>\lim_n\int f^+_n\dmu+1$, making it impossible for the sequence $\newcommand{\dmu}{\ \mathrm{d}\mu}\langle\int\phi_n\dmu\rangle$ to have the right limit, but unless I can find such a seqeuence that's Cauchy in the mean, this doesn't prove anything.

Reminder: A sequence $\langle f_n\rangle$ of a.e. real-valued measurable functions is said to be Cauchy in the mean, or to be a Cauchy sequence in the mean, if for all $\varepsilon>0$ there's an $N\in\mathbb Z^+$ such that for all $n,m\in\mathbb Z^+$,
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
n,m\geq N\ \Rightarrow\ \int|f_n-f_m|\dmu<\varepsilon.$$

 

[lugita15]

Reminder: A sequence $\langle f_n\rangle$ of a.e. real-valued measurable functions is said to be Cauchy in the mean, or to be a Cauchy sequence in the mean, if for all $\varepsilon>0$ there's an $N\in\mathbb Z^+$ such that for all $n,m\in\mathbb Z^+$,
$$\newcommand{\dmu}{\ \mathrm{d}\mu}
n,m\geq N\ \Rightarrow\ \int|f_n-f_m|\dmu<\varepsilon.$$

Is that the same as Cauchy in the L1 norm?

 

[Fredrik]

Is that the same as Cauchy in the L1 norm?

Yes, I think Friedman's avoiding that term because this is earlier in the book than the definition of the L^p spaces.

Edit: I thought about it some more, and I have to change my answer to "no". These concepts are almost the same, but the L¹ norm only applies to bounded real- or complex-valued functions, while these are extended real-valued functions (that are almost everywhere real-valued).

 

[Fredrik]

Lang's Analysis II (maybe now Real analysis), has a good strong statement of Fubini. (And the functions in Lang have values in any Banach space.)

Thanks for this tip. I just checked it out. It looks really good. The actual title is "Real and functional analysis". Lang is using the same definition as Friedman, but starts with complex-valued functions right away (and doesn't use any properties of ℂ other than the ones shared by all Banach algebras) This is how he explains his choice to use the limit definition in the introduction to the chapter:

A posteriori, one notices that the monotone convergence theorem and the "Fautou lemma" of other treatments become immediate corollaries of the basic approximation lemmas derived from Lemma 3.1. Thus it turns out that it is easier to work immediately with complex valued functions than to go through the sequence of many other treatments, via positive functions, real functions, and only then complex functions decomposed into real and imaginary parts. The proofs become shorter, more direct, and to me much more natural. One also observes that with this approach nothing but linearity and completeness in the space of values is used. Thus one obtains at once integration with Banach valued functions.​

I'm going to have to read more of it. It looks like a version of what Friedman did, that's just better organized and with proofs that are easier to follow.

 

[micromass]

Thanks for this tip. I just checked it out. It looks really good. The actual title is "Real and functional analysis". Lang is using the same definition as Friedman, but starts with complex-valued functions right away (and doesn't use any properties of ℂ that other than the ones shared by all Banach algebras) This is how he explains his choice to use the limit definition in the introduction to the chapter:

A posteriori, one notices that the monotone convergence theorem and the "Fautou lemma" of other treatments become immediate corollaries of the basic approximation lemmas derived from Lemma 3.1. Thus it turns out that it is easier to work immediately with complex valued functions than to go through the sequence of many other treatments, via positive functions, real functions, and only then complex functions decomposed into real and imaginary parts. The proofs become shorter, more direct, and to me much more natural. One also observes that with this approach nothing but linearity and completeness in the space of values is used. Thus one obtains at once integration with Banach valued functions.​

I'm going to have to read more of it. It looks like a version of what Friedman did, that's just better organized and with proofs that are easier to follow.

Aah, that is very interesting. This also shows that Friedman's treatment of the integral is superior than my idea of "take the supremum". Indeed, Friedman's definition can be generalized to Banach spaces, while my definition cannot. Indeed, there is no notion of supremum in a general Banach space. (you'll need a Banach lattice for that).

 

[pwsnafu]

Thanks for this tip. I just checked it out. It looks really good. The actual title is "Real and functional analysis". Lang is using the same definition as Friedman, but starts with complex-valued functions right away (and doesn't use any properties of ℂ other than the ones shared by all Banach algebras)

This is how I was taught.

 

[kai_sikorski]

Haha, I'm visiting MBI for the semester. Just realized he's in the office across the hall from me.

If the curiosity is so high, you can even try to ask him :)

http://www.math.osu.edu/~friedman.158/

Email: afriedman@mbi.osu.edu (shown in public)

Varför inte?

Note: he has so incredible CV with plenty publications!

 

[Fredrik]

It's amazing that he got his Ph.D. 56 years ago and is still active. No need to ask him any questions on my behalf though. I got most of it figured out by now. The book by Lang explains some of the things I was wondering about, and helped me figure out a few more. I will however retract my comment that Lang's presentation is better. Some things are clearer in Lang. Some things are clearer in Friedman. I'm glad I have access to both.