Anonymous at Sat, 27 Jul 2024 23:02:06 UTC No. 16299617

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Anonymous at Sat, 27 Jul 2024 23:43:37 UTC No. 16299668

>>16299517
whats the definition of the integral of |f|

Anonymous at Sat, 27 Jul 2024 23:56:00 UTC No. 16299688

>>16299668
The riemann integral of |f| is the common limit of the upper and lower Riemann sums of |f|. The lebesgue integral of |f| would presumably be the infinite sum of the riemann Integrals of some sequence of functions g_k, the sum of which converges to |f| almost everywhere.

That's the thing though. I don't see how we know such functions exist, just because such functions (f_k) exist for f.

Anonymous at Sun, 28 Jul 2024 00:44:17 UTC No. 16299751

I ask this because a few pages in the book after defining the Lebesgue integral they say this, and I don't see how it follows.

Here I would've just argued that because the sum of the integrals of the absolute values of the sinx/x isn't convergent, then by definition the lebesgue integral doesn't exist. But then, what if another sequence of functions exists that satisfies the requirements?

Anonymous at Sun, 28 Jul 2024 08:06:50 UTC No. 16300069

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Anonymous at Sun, 28 Jul 2024 10:39:36 UTC No. 16300199

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Anonymous at Sun, 28 Jul 2024 13:06:11 UTC No. 16300320

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Anonymous at Sun, 28 Jul 2024 13:28:25 UTC No. 16300344

Any function [math]f[/math] can be split into the difference of the positive and negative part
[eqn]f(x) = f^+(x) - f^-(x) \\
f^+(x) = \max(f(x),0) \\
f^-(x) = \max(-f(x),0)[/eqn]
The positive and negative parts of a riemann integrable function are also riemann integrable.

If
[eqn]f = \sum_{k=0}^\infty f_k = \sum_{k=0}^\infty (f^+_k - f^-_k)[/eqn]
then
[eqn]|f| = \sum_{k=0}^\infty (f^+_k + f^-_k)[/eqn]

Also
[eqn] \sum_{k=0}^\infty \int_{\mathbb{R}^n}|f_k(x)| |d^n x| = \sum_{k=0}^\infty \int_{\mathbb{R}^n} (|f^+_k(x)| + |f^-_k(x)|) |d^n x|[/eqn]

Do you see it now?

Anonymous at Sun, 28 Jul 2024 14:59:30 UTC No. 16300439

Posting for more interest

Anonymous at Sun, 28 Jul 2024 15:52:57 UTC No. 16300479

>>16300344
Your if then statement doesn't follow surely? You've written that

[math]|f| = \sum_{k=0}^\infty (|f_k|)[/math]
Which is only true if the support of the f_k never overlap, or only overlap on a set of measure zero.

Anonymous at Sun, 28 Jul 2024 16:31:12 UTC No. 16300504

>>16300344
Maybe the authors simply forgot to define the lebesgue integral of |f| as [math]\sum_{k=0}^\infty \int_{\mathbb{R}^n}|f_k(x)| |d^n x|[/math]?

Besides that, how do we know that if one sequence of R-integrable functions h_k, which is equal to f almost everywhere, has a divergent series of integrals of |h_k|, then f is not L-integrable. What if another sequence exists that is also equal to f almost everywhere, but has a convergent series of integrals of absolute values?

Anonymous at Sun, 28 Jul 2024 19:28:16 UTC No. 16300744

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Anonymous at Sun, 28 Jul 2024 19:48:07 UTC No. 16300779

>>16299517
That has to be one of the most convoluded definitions of the Lebesque integral I have ever seen, just use Stein and Shakarchi (vol 3) instead if you're primarily interested in integration on Euclidean or sigma finite spaces and Cohn if you're interested in general Lebesque integrals.

Anonymous at Sun, 28 Jul 2024 19:54:03 UTC No. 16300791

>>16300779
Yes, but the rest of the book has been really good so far, and I think it is generally reputable. "Vector Calculus, Linear Algebra and Differential Forms - A Unified Approach, 5th edition" by Hubbard & Hubbard, by the way.

It seems like it should be a really trivial thing to prove that "f L-integrable implies |f| L-integrable" and yet it is proving so troublesome.

Anonymous at Sun, 28 Jul 2024 20:47:26 UTC No. 16300877

>>16300791
>Hubbard & Hubbard
You're not meant to learn about Lebesque integration from that kind of book...

Anonymous at Sun, 28 Jul 2024 20:56:15 UTC No. 16300893

>>16300877
I'm not specifically trying to learn lebesgue integration. I just wanted to lear vector calculus, linear algebra, and differential forms, but I don't want to just pass over what is an apparently trivial statement in the book and not understand it.

I'm not sure it will help, but these propositions immediately follow the example here >>16299751

Anonymous at Sun, 28 Jul 2024 22:44:27 UTC No. 16301032

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Anonymous at Mon, 29 Jul 2024 00:27:25 UTC No. 16301173

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Anonymous at Mon, 29 Jul 2024 08:39:12 UTC No. 16301542

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Anonymous at Mon, 29 Jul 2024 11:26:37 UTC No. 16301672

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Anonymous at Mon, 29 Jul 2024 12:41:34 UTC No. 16301737

>>16299517
Would this work:
Let [math]h_k = \left| \sum_{i=1}^k f_i \right| - \left| \sum_{i=1}^{k-1} f_i \right| [/math]

Anonymous at Mon, 29 Jul 2024 13:00:15 UTC No. 16301750

>>16301737
Because [math]h_k = \left| \sum_{i=0}^k f_i \right| - \left| \sum_{i=0}^{k-1} f_i \right| \leq \left| \sum_{i=0}^{k} f_i - \sum_{i=0}^{k-1} f_i \right| = \left| f_k \right| [/math]

We also have that [math]h_k[/math] are R-integrable because the sum of R-integrable functions are R-integrable and the absolute value of an R-integrable function is R-integrable.

Thus, [math]\sum_{k=0}^\infty \int_{\mathbb{R}^n}|h_k(\textbf{x})| |d^n \textbf{x}| \leq \sum_{k=0}^\infty \int_{\mathbb{R}^n}|f_k(\textbf{x})| |d^n \textbf{x}|[/math]

Furthermore, [math]\left| \sum_{k=0}^{N} f_k \right| = \sum_{k=0}^{N} h_k[/math] so [math]\left| \sum_{k=0}^{\infty} f_k \right| = \sum_{k=0}^{\infty} h_k[/math]

Thus [math]\int_{\mathbb{R}^n}|f(\textbf{x})| |d^n \textbf{x}| = \sum_{k=0}^\infty \int_{\mathbb{R}^n}h_k(\textbf{x}) |d^n \textbf{x}|[/math]

🗑️ Anonymous at Tue, 30 Jul 2024 03:41:18 UTC No. 16302801

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Anonymous at Wed, 31 Jul 2024 00:10:15 UTC No. 16303905

>>16300344
There are a few mistakes in here but yes this is the idea

Anonymous at Wed, 31 Jul 2024 09:12:16 UTC No. 16304431

Does anyone know how to prove that if we have one sequence [math]f_k[/math] that the sum of converges to [math]f[/math] almost everywhere, and [math]\sum_{k=0}^\infty \int_{\mathbb{R}^n}|f_k(\textbf{x})| |d^n \textbf{x}| = \infty[/math] then for any other sequence [math]g_k[/math] whose sum converges to [math]f[/math] almost everywhere must also have [math]\sum_{k=0}^\infty \int_{\mathbb{R}^n}|f_k(\textbf{x})| |d^n \textbf{x}| = \infty[/math], i.e. the series can't converge.

Anonymous at Wed, 31 Jul 2024 14:56:50 UTC No. 16304663

Bump >>16304431

Anonymous at Wed, 31 Jul 2024 16:04:55 UTC No. 16304717

>>16299517
What's this book?

Anonymous at Wed, 31 Jul 2024 16:23:21 UTC No. 16304732

>>16304717
"Vector Calculus, Linear Algebra, and Differential- A Unified Approach, 5th Edition" by John H. Hubbard and Barbara Burke Hubbard.

Anonymous at Wed, 31 Jul 2024 16:24:21 UTC No. 16304733

>>16304732
*And Differential Forms

Anonymous at Wed, 31 Jul 2024 20:58:10 UTC No. 16304984

Bump >>16304431

Anonymous at Thu, 1 Aug 2024 00:52:55 UTC No. 16305234

Bump >>16304431

Anonymous at Thu, 1 Aug 2024 09:16:00 UTC No. 16305652

Bump >>16304431