A claim in measure theory which seems flawed to me

  • #1
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The claim states the following:
Let ##(X,\mathcal{A},\mu)## be a measurable space, ##E## is a measurable subset of ##X## and ##f## is a measurable bounded function which has a bounded support in ##E##.
Prove that: if ##f\ge 0## almost everywhere in ##E##, then for each measurable subset ##F\subset E##, we have:
$$0\le \int_F fd\mu \le \int_E fd\mu$$

Now, for the proof of the claim: they write in the book that ##f\ge 0## in ##E##, thus by the additive and non-negativity properties of the integral we get: $$\int_E f d\mu =\int_F fd\mu + \int_{E-F} f d\mu \ge \int_F f d\mu + 0 =\int_F fd\mu \ge 0$$

Now I believe in the rhs they used the "fact" that ##f\ge 0## also in ##F##, but the assumption in the claim is that ##f\ge 0## a.e in E, i.e for ##\exists F': F'\subset E## which has zero measure thus ##\int_F' f d\mu = 0##.

I believe this last statement is missing in the proof, i.e that also for zero measured subset of ##E## where ##f<0## the integral is zero.
Other than this little qualm everything seems right.

Am I right?
 

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  • #2
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You are technically correct. It is ok to skip some steps in a proof if the steps would be correct and not too hard to fill in. Whether a person is allowed to skip steps depends on how well he is trusted to thoroughly understand what is being left out. So a student should skip very few steps, but a professionally published math article might skip a lot. After a while, professional mathematicians do not want to see every simple detail of a proof.
 
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  • #3
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You are right. Another way to circumvent the problem is to replace ##f## by ##g:=f\chi_{E\cap \{f \geq 0\}}##. Since ##f## and ##g## agree almost surely on ##E##, their integrals coincide on all relevant subsets and thus you see that no generally is lost by assuming ##f\geq 0## everywhere, because this is true for ##g##.
 
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  • #4
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You are right. Another way to circumvent the problem is to replace ##f## by ##g:=f\chi_{E\cap \{f \geq 0\}}##. Since ##f## and ##g## agree almost surely on ##E##, their integrals coincide on all relevant subsets and thus you see that no generally is lost by assuming ##f\geq 0## everywhere, because this is true for ##g##.
The issue that you are pointing out is, IMO, more serious than just skipping steps. If they make a change from "##f \ge 0## a.e." to "##f \ge 0##" everywhere without mentioning it, that is careless and a bad habit.
 
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  • #5
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The issue that you are pointing out is, IMO, more serious than just skipping steps. If they make a change from "##f \ge 0## a.e." to "##f \ge 0##" everywhere without mentioning it, that is careless and a bad habit.
I agree. The proof should at least say: "WLOG, we may assume ##f\geq 0## everywhere on ##E##". Understanding why this can be done is then a good exercise for the reader.
 
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  • #6
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You are technically correct. It is ok to skip some steps in a proof if the steps would be correct and not too hard to fill in. Whether a person is allowed to skip steps depends on how well he is trusted to thoroughly understand what is being left out. So a student should skip very few steps, but a professionally published math article might skip a lot. After a while, professional mathematicians do not want to see every simple detail of a proof.
The problem with skipping some details that some might not see them as trivial (well all of maths is trivial when done correctly). Take for example the "infamous" attempt of proving the abc conjecture in the last decade or so.
https://www.quantamagazine.org/tita...h-over-epic-proof-of-abc-conjecture-20180920/
 
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  • #8
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If you say that all math is trivial when done correctly, you have not done enough math.
I've done enough to hold this view.
I don't say that it cannot take a lot of steps to prove something, but each step is trivial.
The complexity of maths is in the length of the arguments and the ideas on which step to use.
 
  • #9
martinbn
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Which book is this?
 
  • #10
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Which book is this?
It's a book in Hebrew of the Open University in Israel.
It was adapted from notes of Amnon Jakimovski.
 
  • #11
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I've done enough to hold this view.
I don't say that it cannot take a lot of steps to prove something, but each step is trivial.
The complexity of maths is in the length of the arguments and the ideas on which step to use.
Care to tell what math level you are at?
 
  • #12
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Care to tell what math level you are at?
I've taken quite a lot of graduate courses while attending for MSc studies in both pure and applied maths.
Never finished my thesis, found it hard to understand my adviser, thought of changing my adviser and to change my thesis topic but the school didn't approve.

Anyhow, I obviously have a Bsc in maths and physics.
Obviously mathematics is difficult since there are quite a lot of steps in the difficult proofs, I don't argue otherwise. But each step in the proof if it's correct eventually is indeed trivial if you have the adequate piece of knowledge to understand it.

I could argue that if a claim follows from the definitions and axioms then it does so trivially, if the proof of the claim is false then the error is either trivially spotted or difficult to spot it since there are so many steps in the proof.
 
  • #13
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Published math articles will skip long derivations if the details are routine and not of research interest. The page length is very important to keep down and the skipped details can be very long and tedious.
 
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  • #14
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Published math articles will skip long derivations if the details are routine and not of research interest. The page length is very important to keep down and the skipped details can be very long and tedious.
So perhaps Scholze and Styx just didn't understand the proof of Mochizuki even after he explained them their misunderstanding.
It's a possibility... :oldbiggrin:

Edit:
And Mochizuki could argue that to write all the details is tedious and it's not done in math research.
And his papers are already quite long.
 
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