Is it true almost all Lebesgue measurable functions are nonintegrable?
I'm not sure how to state this rigorously but I have an intuition "almost all Lebesgue measurable functions are nonintegrable". Edit: According to answer in (this link) and (this paper) "almost all" can be defined without a measure on the set of measurable functions.
Is the intutuition in my title correct? (Sorry for the edits)
Last Edit: (In case I dispute the answer)
The person who answered my post assumes neither the set of nonintegrable and integrable functions have the "same size" but I do not quite agree. I may be incorrect that "almost all" Lebesguemeasurable functions are nonintegrable but I do believe nonintegrable functions have a "greater size" than integrable functions (see this response).
Note that almost all functions can be desribed as a set of pseudorandom points that are nonuniformly distributed in a subspace of $\mathbb{R}^2$.
Now assume we have that same function but it's defined on a lebesgue measurable set (e.g. defined on $[0,1]$). If we partition the functions? domain, note the subset of points in that function might have the largest preimage in a partition that is nonLebesgue measurable which makes the function nonintegrable. The chance that a random set is Lebesgue measurable is extremely small (see this this link).
Therefore, using the previous paragraph "almost all" functions or "most" functions are nonintegrable.
This may imply but not directly prove that I'm correct.

Hello can you be more specific ? Thank you

Like your word "almost" is not right here... Make your statement firm and then think you want to prove it or not ? If you say almost all that means you already know there are some functions which integrAble!

@Aman R I’m not sure how to. This is the best I can do?

I was wondering if someone can make my ideas more firm. I have this “concept” in my head but I don’t know how to state it clearly. I still haven’t completed Intro to real analysis but I’m too impatient to wait until I learned more.

@Aman R There are some functions that are Lebesgue integrable!

Almost all generally means all except a countable set. The set of integrable functions is not countable, so the statement "all Lebesgue measurable functions are nonintegrable" is indeed false. What you probably mean is that the set of nonintegrable functions is much larger than the set of integrable function. This could be rephrased as wether the cardinality of measurable functions is strictly larger than cardinality of the set of integrable functions.

NB: I don't agree, "almost all" rather / also(?) means : all but a set of measure zero. A set of measure 0 does not need to be countable.

@ M F M: You are right. But since Bharathk98 does not introduce a measure on the set of functions, a set of functions with measure zero does not make much sense unless we clearly define a measure.

@Niremberg The answer here https://mathoverflow.net/a/28114/87856 states that one does need to define a measure of set functions to define “almost all”.

Very interesting. Thank you for the reference :)

I have an issue with the way you (seemingly) use " almost all" and "nowhere dense" almost interchangeably or as equivalent to each other (conceptually). The rationals have measure zero on the real line but they are dense. It is very hard to tackle this question because it seems too open ended/ open to interpretation.

@Mathe You are right. I got rid of the dense/nowhere dense. Is the question more understandable. How much more should I pay?

@Bharathk98 It is a very interesting question but I would have to do research on the extensions to 'Lebesgue almost every’ to even attempt solving it. I can't guarantee I can come up with an answer, but since reading a paper (at least one) seems to be a prerequisite, I would increase the bounty to $120 at least. Even then I can't guarantee to solve it since I haven't even read the paper yet.

@Mathe What do you think the answer to this question is? (Right now someonelse answered this but I disputed their answer.). They stated the “sizd” of integrable and nonintegrable functions are “large”. Is this true? See my last edit in my post.


@Bharathk98 I can also help you with you, but it will take some back and forth to discuss details and exchange ideas. I also can't guarantee to come up with a firm answer, but I will try my best to spend enough time and back and forth and make you satisfied. Let me know.

@Kav10 Any help is appreciated.

Alright, I'll reserve the question. Please remember I cannot guarantee I can come up with the exact answer you are looking for. Sounds good?

Sure. If you give the answer I’m looking for, I will accept. Otherwise, I will wait for an answer from @Mathe.

I don't think it will work that way. If I reserve the question, Mathe won't be able to answer it. Not sure what you have in mind. Please explain a bit more.

@Kav10 Is it possible for others to check the answer. I don’t think I myself will be able to know if it’s correct or not.

Well, in any case I'd say the question is now veryclearly posed, namely, whether the ("natural") measure of the subspace of (Lebesgue) integrable functions within the space of Lebesgue measurable functions is zero. Highly nontrivial at first sight (but I'm really not an expert on that).

I don't think they will be able to see the answer. They can see it after you accept it. And with me, most likely, it will be a back and forth discussion, until we get to something that you are happy with (hopefully).

@Kav10 Ok, I’ll let you answer.

@Kav10 Sorry if I sounded rude. I have a habit of doing that.

Oh, no you are totally fine. I didn't feel any rudeness.

So, just to confirm. I will try to answer your question, but cannot guarantee you'll be happy with it. But, I will definitely spend time to discuss this, back and forth.

@Kav10 Sure. We can do that.

OK, I will provide an answer by tomorrow and if you are not happy with it, we can start discussing it in the comments below (not public).

Sure, we can discuss this privately, but given the fact that I haven’t completed Intro to Real Analysis I wouldn’t know if it’s correct, but I will accept your answer for your effort.

Sounds good. Don't worry, I will try to provide a simple and reasonable answer as much as possible.

Not sure, where the comments sections under my answer went, I cannot access that. Here is Dave's response: Keep in mind that cardinality is a VERY rough measure of size  in xyz 3space, lines, and planes, and all of space, and small intervals on the xaxis, and small Cantor sets on the xaxis, and many other seemingly differentsized sets all have the same cardinality. It's like measuring animals by how many legs they have, whereas the things above would be weight, volume, height, etc. (cont

My guess is that for most such smallness/largeness notions, nonintegrable functions are large. The mathoverflow question/answers Situations where “naturally occurring” mathematical objects behave very differently from “typical” ones may be of interest to you. https://mathoverflow.net/questions/398387/situationswherenaturallyoccurringmathematicalobjectsbehaveverydifferen

@Kav10 I’m a little confused. He says, “The set of nonintegrable functions are large but he never states how large the set of integrable functions is?” Is it possible to ask him. Perhaps we ask him which one is larger.

The answer is neither one is larger. That confirms my response and explanations.


I don’t know if you saw my comments under the response, here. It’s gone and I cannot see those (not sure why!). But please go ahead and read all the explanations, links, and details I provided and make a decision. Like you said, I am expecting you to accept but you can do whatever you want.

@Apolgies Kav10 but I need to know from another source. I was hoping to at least hear from DaveL. but if you aren’t willing to contact him I can understand.

I contacted him multiple times as you requested. Added your questions multiple times as you requested. But, like I said before, you did what I expected/felt you’ll do :)

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