Lebesgue integral question: integrable on [0,1] with lambda in (0,1)

[Zermelo]

The task is:

Let f be an Lebesgue integrable function on [0, 1], and let [imath]\lambda \in (0,1)[/imath]. If [imath]\int_Efd\mu = 0[/imath] for every set [imath]E \subset [0, 1][/imath] such that [imath]\mu E = \lambda[/imath], prove that [imath]f = 0[/imath] almost everywhere.

This question would be relatively easy to prove if f is non-negative. But in the general case, I can't figure out the proof.

I think it would be best to re-use some results, like

Or


The problem here is that every statement [imath]\int_Efd\mu = 0[/imath] doesn't have to mean that f = 0, instead it can mean that [imath]\int_{E_+}f^+d\mu = \int_{E_-}f^-d\mu,\ E_+ U E_- = E[/imath]. The fact that this holds for every E with measure lambda is quite powerful, and my idea is to construct one such set E1 that contains E+, which would then mean that [imath]\int_{E_+}f^+d\mu = \int_{E_1 \setminus E_+}f^-d\mu[/imath]. I'm hoping that there's a smart way of constructing many of those sets E1, E2, ... (even maybe a countable infinity of them), each containing E+, and each with measure lambda. Then, that would mean that [imath]\int_{E_+}f^+d\mu = \int_{E_n \setminus E_+}f^-d\mu[/imath], for all n in N. This is quite a strong statement, and if I construct these sets in a proper way, I might be able to prove that this can't hold.

Does anyone have a better idea, or maybe can help with finishing up my idea?

 

[blamocur]

One can probably prove that for arbitrary [imath]x[/imath] and large enough [imath]n[/imath] [imath]\int_x^{x+\frac{\lambda}{n}} fd\mu = 0[/imath]
Then one can represent [imath][x,y][/imath] as a countable union of chunks of size [imath]\frac{\lambda}{n}[/imath] and thus prove that [imath]\int_x^y fd\mu = 0[/imath] for arbtrary [imath]x,y[/imath].
Applying Lemma 2 would complete the proof.
Am I making sense?

 

[Zermelo]

One can probably prove that for arbitrary [imath]x[/imath] and large enough [imath]n[/imath] [imath]\int_x^{x+\frac{\lambda}{n}} fd\mu = 0[/imath]
Then one can represent [imath][x,y][/imath] as a countable union of chunks of size [imath]\frac{\lambda}{n}[/imath] and thus prove that [imath]\int_x^y fd\mu = 0[/imath] for arbtrary [imath]x,y[/imath].
Applying Lemma 2 would complete the proof.
Am I making sense?

That's exactly the idea that I have, but the first part is the biggest problem

I know that on [x, x + lambda], the integral has to be 0. But I don't know that for [x, x + lam / 2]. I know that it's zero for [x, x + lam / 2] U E, where E has the measure of lam / 2. This would still mean that f can behave in a very wonky way on [x, x + lam / 2] and on E, but somehow the integrals on these 2 sets cancel each other out and give 0.

I am a bit skeptical about Lema 2, because it uses segments [a, b], and here it states that the integral is 0 over any set with measure lambda. This is a much more powerful statement than any segment with measure lambda...

An interesting idea would be "sliding" the interval [x, x + lambda], with x going from 0 to 1 - lambda. This moving interval would cover the whole [0, 1] segment, for different xs. The hope would be fixing the first interval [0, lambda], and concluding that f has to be either 0 or non 0 there. If it's 0, the proof is over, if it's different than 0, sliding the interval to the right, [x, x + lambda], would maybe disprove that claim.

Also, there are 2 equivalents to the statement f = 0 almost everywhere.
It's that the integral over any set E in [0, 1] is 0, or that the integral of |f| on [0, 1] is zero.

I have that the integral is 0 for every set of measure lambda, not for every set. Could this imply that it works for every set E in [0, 1], not just ones with measure lambda?

 

[blamocur]

While writing down proofs of my statements I think I came up with stronger ones -- let me know what you think.

Lemma A: For any [imath]d[/imath] such that [imath]0 < d < \min\left(\lambda,1-\lambda \right)[/imath] and all subsets [imath]S=[a,b][/imath] such that [imath]\mu(S) = b-a = d[/imath] the integral below does not depend on [imath]S[/imath]:
[math]\int_S fd\mu = C_d[/math]Lemma B: [imath]C_d = 0[/imath].

To prove Lemma A: consider two sets [imath]S_1[/imath] and [imath]S_2[/imath] satisfying the conditions in Lemma A. Then [imath]\mu(S_1 \cup S_2) \leq \mu(S_1) + \mu(S_2) = 2d < 1[/imath].
From the definition of [imath]d[/imath] we can show that [imath]1-2d > \lambda-d[/imath].
This means that [imath]\mu([0,1]-S_1-S_2) > \lambda-d[/imath], i.e, we can find a subset [imath]G[/imath] such that [imath]\mu(G) = \lambda-d[/imath], and [imath]G[/imath] does not intersect with either [imath]S_1[/imath] or [imath]S_2[/imath]. But [imath]\mu(S_1\cup G) = \mu(S_2\cup G) = \lambda[/imath] and thus (because [imath]G[/imath] does not intersect either [imath]S[/imath]):
[math]\int_{S_1} fd\mu + \int_G fd\mu = \int_{S_2} fd\mu + \int_G fd\mu = 0[/math]Which means that
[math]\int_{S_1} fd\mu = \int_{S_2} fd\mu[/math]To prove Lemma B we find large enough [imath]n[/imath] such that [imath]d=\frac{\lambda}{n}[/imath] satisfies lemma A, which proves that for such [imath]d[/imath]'s we have [imath]C_d = 0[/imath]. To prove this for arbitrary [imath]d[/imath] we notice that any interval can be represented by a countable union of non-intersecting small intervals of lengths [imath]\frac{\lambda}{n}[/imath].

Note: we can probably prove Lemma A for arbitrary measurable [imath]S[/imath], not just for intervals, but proving the existence of [imath]G[/imath] would be trickier.

 

[blamocur]

It just crossed my mind that we can use arbitrary measurable [imath]S[/imath], not just intervals, because I believe I can prove
Lemma C: for any measurable [imath]H\in [0,1][/imath] and any [imath]b \leq \mu(H)[/imath] there exists a measurable [imath]G\subset H[/imath] such that [imath]\mu(G) = b[/imath].

 

[Zermelo]

While writing down proofs of my statements I think I came up with stronger ones -- let me know what you think.

Lemma A: For any [imath]d[/imath] such that [imath]0 < d < \min\left(\lambda,1-\lambda \right)[/imath] and all subsets [imath]S=[a,b][/imath] such that [imath]\mu(S) = b-a = d[/imath] the integral below does not depend on [imath]S[/imath]:
[math]\int_S fd\mu = C_d[/math]Lemma B: [imath]C_d = 0[/imath].

To prove Lemma A: consider two sets [imath]S_1[/imath] and [imath]S_2[/imath] satisfying the conditions in Lemma A. Then [imath]\mu(S_1 \cup S_2) \leq \mu(S_1) + \mu(S_2) = 2d < 1[/imath].
From the definition of [imath]d[/imath] we can show that [imath]1-2d > \lambda-d[/imath].
This means that [imath]\mu([0,1]-S_1-S_2) > \lambda-d[/imath], i.e, we can find a subset [imath]G[/imath] such that [imath]\mu(G) = \lambda-d[/imath], and [imath]G[/imath] does not intersect with either [imath]S_1[/imath] or [imath]S_2[/imath]. But [imath]\mu(S_1\cup G) = \mu(S_2\cup G) = \lambda[/imath] and thus (because [imath]G[/imath] does not intersect either [imath]S[/imath]):
[math]\int_{S_1} fd\mu + \int_G fd\mu = \int_{S_2} fd\mu + \int_G fd\mu = 0[/math]Which means that
[math]\int_{S_1} fd\mu = \int_{S_2} fd\mu[/math]To prove Lemma B we find large enough [imath]n[/imath] such that [imath]d=\frac{\lambda}{n}[/imath] satisfies lemma A, which proves that for such [imath]d[/imath]'s we have [imath]C_d = 0[/imath]. To prove this for arbitrary [imath]d[/imath] we notice that any interval can be represented by a countable union of non-intersecting small intervals of lengths [imath]\frac{\lambda}{n}[/imath].

Note: we can probably prove Lemma A for arbitrary measurable [imath]S[/imath], not just for intervals, but proving the existence of [imath]G[/imath] would be trickier.

Thank you very much blamocur!
I just have one question: why is Cd = 0?

The way I see it, we see that[imath]\int_S fd\mu = - \int_G fd\mu[/imath], where G has measure [imath]\lambda - \frac{\lambda}{n}.[/imath] If we let n tend to infinity, we see that this integral is indeed 0, but then d = 0, aka S is a 0 length interval. Not sure how I would construct [0, x] from these

Now another theorem comes to mind, which states that the Lebesgue integral is absolutely continuous, for every eps > 0 I can find a delta > 0 such that the integral of any function over a set of measure delta, is equal to epsilon. That means that I could have a d that isn't zero, with an integral that is epsilon, but still not 0...
Just trying to figure out how to make it go to 0

 

[blamocur]

I just have one question: why is Cd = 0?

Good question. The proof I had in mind initially turned out to be flawed. But: we can prove that [imath]C_{\frac{d}{2}} = \frac{1}{2} C_d[/imath] and thus [imath]C_{\frac{d}{2^k}} = \frac{1}{2^k} C_d[/imath] for any [imath]k[/imath]. I claim that we can represent [imath]\lambda[/imath] as a countable sum of intervals of lengths [imath]\frac{d}{2^k}[/imath]. Indeed, we can use the binary representation of [imath]\frac{\lambda}{d}[/imath]:
[math]\frac{\lambda}{d} = \sum_{j=0}^\infty a_j 2^{-j}[/math]where [imath]a_0[/imath] is a non-negative integer and [imath]a_j \in \{0,1\}[/imath] for [imath]j>0[/imath]. This means that an interval [imath]T[/imath] of length [imath]\lambda[/imath] can be represented as a countable union of non-intersecting interval of lengths [imath]d a_j 2^{-j}[/imath]. But for the [imath]j[/imath]-th interval [imath]T_j[/imath] we have
[math]\int_{T_j} fd\mu = a_j 2^{-j} C_d[/math]and for the whole interval [imath]T[/imath] of length [imath]\lambda[/imath] we get
[math]C_d \sum_j a_j 2^{-j} = \int_T fd\mu = 0[/math] which means that [imath]C_d = 0[/imath].

Hope I haven't screwed up this one

 

[Zermelo]

Good question. The proof I had in mind initially turned out to be flawed. But: we can prove that [imath]C_{\frac{d}{2}} = \frac{1}{2} C_d[/imath] and thus [imath]C_{\frac{d}{2^k}} = \frac{1}{2^k} C_d[/imath] for any [imath]k[/imath]. I claim that we can represent [imath]\lambda[/imath] as a countable sum of intervals of lengths [imath]\frac{d}{2^k}[/imath]. Indeed, we can use the binary representation of [imath]\frac{\lambda}{d}[/imath]:
[math]\frac{\lambda}{d} = \sum_{j=0}^\infty a_j 2^{-j}[/math]where [imath]a_0[/imath] is a non-negative integer and [imath]a_j \in \{0,1\}[/imath] for [imath]j>0[/imath]. This means that an interval [imath]T[/imath] of length [imath]\lambda[/imath] can be represented as a countable union of non-intersecting interval of lengths [imath]d a_j 2^{-j}[/imath]. But for the [imath]j[/imath]-th interval [imath]T_j[/imath] we have
[math]\int_{T_j} fd\mu = a_j 2^{-j} C_d[/math]and for the whole interval [imath]T[/imath] of length [imath]\lambda[/imath] we get
[math]C_d \sum_j a_j 2^{-j} = \int_T fd\mu = 0[/math] which means that [imath]C_d = 0[/imath].

Hope I haven't screwed up this one

What about this:

We pick large enough n, so d = lamda/n satisfies Lema A. Then, every integral on [0, lambda/n], [lambda/n, 2 lambda/n] … [(n-1)lambda/n, lambda] have the same value, let’s call it Cd. So, nCd = integral over [0, lambda] = 0 => Cd=0?

 

[blamocur]

What about this:

We pick large enough n, so d = lamda/n satisfies Lema A. Then, every integral on [0, lambda/n], [lambda/n, 2 lambda/n] … [(n-1)lambda/n, lambda] have the same value, let’s call it Cd. So, nCd = integral over [0, lambda] = 0 => Cd=0?

This does not prove it for an arbitrary [imath]d[/imath], but only for [imath]d=\frac{\lambda}{n}[/imath].

 

[Zermelo]

This does not prove it for an arbitrary [imath]d[/imath], but only for [imath]d=\frac{\lambda}{n}[/imath].

Thank you so much. I used your ideas, managed to prove them rigorously, and finally use Lemma 2.

But I have to ask, are you like a professor of measure theory? ?
Because your ideas are so wild, I would have never come up with them… how do you do it?

 

[blamocur]

But I have to ask, are you like a professor of measure theory? ?

I am a retired software engineer. But I do often amuse other math majors by still remembering many things from my college days.
As for the solutions, they did not come easy. But the problems looked interesting, then nostalgia for my college years kicked in