43

This is an exercise that I am having trouble with. Not for a grade just for practice. Its an obvious result intuitively but I am having trouble making a rigorous argument.

Assume $f$ is Lebesgue integrable on $E$. Prove that for all $\varepsilon>0$ there exists a $\delta>0$ such that if the Lebesgue measure of $A$ is less than $\delta$, the integral of $|f|$ over $A$ is less than $\varepsilon$. Here $A$ is a subset of $E$.

Anyone have any ideas?

Mike Pierce
  • 18,938
rmh52
  • 1,136
  • 1
    The answers given are already excellent, but I just wanted to provide the following "summary" or "intuitive slogan" for the proof: basically, the problem is easy for bounded functions (if bounded by $B$, then integral over set of measure $\delta$ is bounded by $\delta \times B$); and $L^1$ functions are "almost bounded", up to a tiny bit of extra $L^1$ mass. – D.R. Mar 24 '22 at 04:35
  • 1
    can you check if my prof works as well , if so upvote it plz. – homosapien Jun 30 '22 at 03:15
  • @D.R. ^^^^^^^^^^ – homosapien Jun 30 '22 at 03:15

4 Answers4

54

Note that, by the Lebesgue dominated convergence theorem, we have that $$\lim_{\lambda \to\infty}\int_{\{|f| > \lambda\}}|f|\ d\mu = 0.$$ This follows easily since $\chi_{\{|f| > \lambda\}}|f| \le |f| \in L^1$ and $\chi_{\{|f| > \lambda\}}|f| \to 0$ since $f$, being integrable, is finte almost everywhere.

Let $\epsilon > 0$, then there exists $\lambda > 0$ such that $$\int_{\{|f| > \lambda\}}|f|\ d\mu < \frac{\epsilon}{2}.$$

Choose $\delta \le \frac{\epsilon}{2\lambda}$ and take any measurable set $A$ such that $\mu(A) < \delta$. Then we have $$\int_A|f|\ d\mu = \int_{A \cap \{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}|f|\ d\mu \le$$ $$\le \int_{\{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}\lambda\ d\mu$$

note that this last inequality follows from the fact that $A \cap \{|f| > \lambda\} \subset \{|f| > \lambda\}$ and the fact that $|f| \le \lambda$ on $A \cap \{|f| \le \lambda\}$. Then we are done since $$\int_{\{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}\lambda\ d\mu \le \frac{\epsilon}{2} + \delta \lambda \le \epsilon.$$ This concludes the proof! :D

anonymous
  • 1,019
  • I see your point but actually, since you prove the monotone convergence theorem before linearity for the Lebesgue integral, I think that this method is sufficiently elementary.. it uses nothing that you didn't prove if you have already defined the $L^1$ space, right? :D –  Oct 22 '13 at 03:05
  • 3
    It looks like this solution is correct (+1), though I prefer a more elementary one. In particular, simply choosing a simple function $\phi$ such that $\int f - \int \phi < \epsilon$ (by definition of integration as supremum over all simple functions $\phi$ : $0 \leq \phi \leq f$) will give the result potentially even faster. Also, out of interest note that monotone convergence could've been used for the first statement. – snar Oct 22 '13 at 03:13
  • @user01... interesting! you are right, I forgot one uses monotone convergence to prove linearity. I don't see the method with simple functions using that though -- linearity for those follows directly from the definition of integration, which you presumably define before MCT, ya? – snar Oct 22 '13 at 03:14
  • yes, sure! my point is more that you don't even care about something like this statement if you haven't proved MTC yet, since it's (one of) the reason(s) you defined the Lebesgue integral. Anyway, I am sure that your proof works as well! –  Oct 22 '13 at 03:23
  • yeah, makes sense. cheers. – snar Oct 22 '13 at 04:02
  • @user011235813213455891 Hi I have a question, regarding the first step of the proof , how did you apply the dominated convergence theorem here? I mean as $f$ is integrable on the set $E$ we have that $f \mid_E < \infty$ a.s but then how does that imply that $\exists g$ s.th $f_n(x) \leq g(x) \forall x,n$? I understand all the other arguments besides why the limit goes inside the integral – user3503589 Dec 13 '14 at 13:01
  • 1
    @user3503589: your answer is right below the integral, "This follows easily since...". we are bounding $f\chi_{\lambda}$ with $f$ itself, which is an integrable function :) I hope it is clear now! :D –  Dec 13 '14 at 14:09
  • @user011235813213455891 That's exactly my point just because f is integrable does not imply its bounded, it just means that $f < \infty$ a.s . I asked a question a few weeks back related to this http://math.stackexchange.com/questions/1040191/a-basic-question-on-the-existence-of-expectation. – user3503589 Dec 13 '14 at 14:19
  • @user011235813213455891 My bad . It makes perfect sense. Thanks a ton for the clarification. I am grateful. It was a very stupid question as you had already explained it in the solution. I am sorry for the lame question – user3503589 Dec 13 '14 at 14:31
  • No worries, I am glad you got it! ;) –  Dec 13 '14 at 15:48
  • Isn't the very definition of Lebesgue integrable $f\geq0$ to be measurable and such that its decreasing rearrangement $t\mapsto |{f>t}|$ is (Riemann-Stieltjes) integrable? (See, e.g., Lieb and Loss (2001).) – Oskar Limka Feb 26 '16 at 22:15
  • So complete and undisputable! – Behnam Esmayli Jun 08 '16 at 19:50
  • How do you prove $\int_{{|f| > \lambda}}|f|\ d\mu < \frac{\epsilon}{2}$? – Ufos Apr 28 '21 at 16:03
  • can you check if my prof works as well , if so upvote it plz. – homosapien Jun 30 '22 at 03:14
35

It is given that $f$ is integrable, so $$\int_E|f|d\mu=L<\infty.$$ Choose a simple function $0\leq g\leq|f|$ with $$g=\sum_{i=1}^Ng_i\mathbf{1}_{A_i}$$ such that $$\int_E(|f|-g)d\mu<\frac{\epsilon}{2}.$$

Let $G=\max (g_i)$ and choose $\delta<\frac{\epsilon}{2GN}$.

For any $A$ with $\mu(A)<\delta$ we have $$ \begin{aligned} \int_A|f|d\mu&=\int_A(|f|-g)d\mu+\int_Agd\mu\\ &\leq\int_E(|f|-g)d\mu+\int_Agd\mu\\ &<\frac{\epsilon}{2}+\sum_{i=1}^Ng_i\mu(A_i\cap A)\\ &\leq\frac{\epsilon}{2}+GN\mu(A)\\ &<\frac{\epsilon}{2}+GN\delta\\ &<\frac{\epsilon}{2}+GN\frac{\epsilon}{2GN}=\epsilon. \end{aligned} $$

jdods
  • 6,248
  • 1
    Easiest and cleanest proof. Thanks! – Dr Richard Clare Aug 24 '17 at 11:33
  • jdods: I put up another proof below. Would you care to verify it? – Hans Oct 01 '17 at 06:41
  • I understand that simple functions approximate measurable functions (that's how we get the lebesgue integral), does that mean there exists an N such that $\int_E(|f|-g)d\mu<\frac{\epsilon}{2}$, or is there more at play? – Ufos Apr 28 '21 at 16:07
  • @Ufos That $|f|$ is Lebesgue "integrable" requires its integral to be finite. When its integral is infinite, then this may not work. – jdods Apr 28 '21 at 17:01
  • Can you check if my proof is correct or not, if so vote it up plz! – homosapien Jun 30 '22 at 03:14
4

For an approach using MCT:

Theorem: If $f$ is Lebesgue integrable, then for any $\epsilon>0$, ther exists a $\delta>0$ such that if $m(E)<\delta$, then $\int_E \vert f \vert < \epsilon.$

Proof:

We can safely assume that $f \geq 0$. Then define

$$f_N(x):=f(x) \cdot \chi_{E_N}$$

where

$$E_N:=\{x: f(x) \leq N\}$$

Then the $f_N$ are measurable and furthermore one has for every $N \in \mathbb{N}$ that

$$f_N \leq f_{N+1}$$

Then by the Monotone Convergence Theorem one has for any given $\epsilon >0$ that there exists an $N \in \mathbb{N}$ such that

$$\int_{\mathbb{R}^d} f - f_N < \frac{\epsilon}{2} \space \space \space \space (*)$$

Then choose your $\delta>0$ such that $\delta<\frac{\epsilon}{2N}$. Then if $m(E)< \delta$ one can compute

\begin{equation} \begin{split} \int_E f &= \int_E f- f_N + \int_E f_N && \text{adding 1 trick}\\ &\leq \int_{\mathbb{R}^d} f - f_N + \int_E f_N && \text{monotonicity of Leb integral}\\ &\leq \int_{\mathbb{R}^d}f-f_N+Nm(E) && \text{by definition}\\ &< \frac{\epsilon}{2} + N\frac{\epsilon}{2N} && \text{since $m(E) < \delta$ and by $(*)$}\\ &= \epsilon. \end{split} \end{equation} And since $\epsilon$ was arbitrary, the proof is complete. $\blacksquare$

homosapien
  • 4,157
  • I believe you want $\delta<\frac{\epsilon}{2N}$, and your last inequality could be strict also. As far as I can tell, your argument is fine. I think mine became popular since it is very basic in substance and notation and doesn't use anything other than the definitions involved, e.g. no advanced results like monotone convergence. But I think monotone convergence is fine to use as well. Your proof is essentially identical to mine but using a not-necessarily-simple (but specified) approximation $f_N$ instead of my (unspecified) simple $g$, but one could argue your approach is a little cleaner. – jdods Jun 30 '22 at 17:50
  • @jdods BIG brains fart on my end, Edit made! – homosapien Jun 30 '22 at 17:53
  • True, I used a highfalutin theorem like MCT. @jdods – homosapien Jun 30 '22 at 17:54
  • 1
    Nothing wrong with that. Like I said, it has the nice feature that it doesn't "choose" an arbitrary simple function and instead specifies $f_N$. Though the proof of MCT I found does indeed choose an arbitrary sequence of functions. So this feature of "arbitrary choice" is simply hidden by using MCT. I think your variation is a good addition here though. – jdods Jun 30 '22 at 18:03
  • 1
    @jdods true, though it can be proven using Fatou's Lemma – homosapien Jun 30 '22 at 19:51
2

Since $f$ is Lebesgue integrable, without loss of generality, we assume $f\ge 0$. Assume the contrary, $\big(\exists \epsilon_0>0$ and a sequence of measurable subsets $(A_n)_{n=1}^\infty\big) \ni \big(\mu(A_n)\le\frac1{2^n}\ {\large\land}\ \int_{A_n}f>\epsilon_0,\forall n\in\mathbf N\big)$. $\Big(\int_{\cup_{n=m}^\infty A_n}f\ge\int_{A_k}f>\epsilon_0, \forall \big(k\ge m \land (k,m\in\mathbf N)\big)\Big){\large\land}\big(\int_{\cup_{n=m}^\infty A_n}f \text{ decreases with }m\big)\implies \int_{\cap_{m=1}^\infty\cup_{n=m}^\infty A_n}f=\lim\limits_{m\to\infty}\int_{\cup_{n=m}^\infty A_n}f\ge\epsilon_0.$ However, by Borel Cantelli lemma $\sum_{n=1}^\infty \mu(A_n)\le1<\infty \implies\mu(\cap_{m=1}^\infty\cup_{n=m}^\infty A_n)=0\implies \int_{\cap_{m=1}^\infty\cup_{n=m}^\infty A_n}f=0<\epsilon_0.$ A contradiction.

Hans
  • 9,804
  • There's a bit of abusing indices that needs to be cleaned up, seems like $n$ is simultaneously fixed and runs from $m$ to infinity in the initial construction. Seems odd to construct $A_n$ with self reference to integral over itself. Maybe a mistake or I am missing something. You should check to make sure no mistake. – jdods Oct 02 '17 at 12:45
  • @jdods: Thank you. I cleaned up the fixed indices. Regarding the self-reference issue, are you talking about the first statement following "Assume the contrary"? If so, it is only a statement to the contrary of the conclusion. I do not see any problem. Would you mind to review it again? – Hans Oct 02 '17 at 17:00
  • Honestly, it's a bit over my head! I'd have to work really hard to understand it. I don't really understand all the wedges. – jdods Oct 03 '17 at 03:20
  • Oh I think I see, the first wedge is the logical operation, I thought it was for minimum. That's why I thought A_n construction was self referential. My mistake. – jdods Oct 03 '17 at 23:07
  • @jdods: Oh, hahaha, yes. The first time I saw that appearing in probability book, I thought it was logical. It confused the hell out of me. – Hans Oct 03 '17 at 23:38
  • @jdods: So do you agree with my proof? – Hans Oct 15 '17 at 08:25
  • It seems correct, although it's not clear how it proves the statement in the OP. Where is your $\delta$ or $\int |f|$. Your argument appears related but will need some explanation on the connection. I find the "thick symbolic notation" a bit hard to read. Don't get me wrong, I definitely can appreciate the aesthetic of that, but rarely if ever encounter that style, so I could be misinterpreting something. A few questions: are we assured the existence of such $A_n$ with precise measure $1/2^n$? Maybe need to specify that $f>0$? – jdods Oct 16 '17 at 16:41
  • @jdods: With "thick symbolic notation" I just want it to be more compact. Apologize for making the reading more difficult. I will use it more sparingly next time. As for the logic of the proof, I am using a contrapositive argument. So there must exist an $\epsilon_0$ such that there is a sequence of offending sequence of measurable set $(A_k)$ such that $\mu(A_k)\to 0^+$. I choose a subsequence and denote it by $(A_n)$ such that $\mu(A_n)\le \frac1{2^n}$ --- you are right I should not have used $=$ but $\le$ here and also I assume $f\ge0$. I have fixed my answer. Please review. Thank you. – Hans Oct 17 '17 at 05:45
  • It's not perfectly clear to me what the negation of your assumption is. Maybe (1) $\forall \epsilon >0$ and for any sequence of sets ${A_n}{n=1}^\infty$ there is at least one $n$ such that either or both $\int{A_n} f \leq \epsilon$ or $\mu(A_n)>1/2^n$, or is it (2) $\forall \epsilon >0$ and for any $n$ there is an $A_n$ such that either or both $\int_{A_n} f \leq \epsilon$ or $\mu(A_n)>1/2^n$? The latter seems to imply the result, the former probably does too, but again it's just not obviously clear to me. It may be to others. – jdods Oct 18 '17 at 00:46
  • Or maybe the correct negation is: $\forall \epsilon >0$ there is a sequence ${A_n}$ such that $\mu(A_n)>1/2^n$ and/or $\int_{A_n} f \leq \epsilon$ for at least one $n$. – jdods Oct 18 '17 at 00:50
  • @jdods: No, that is wrong. $\neg\big(\forall\epsilon>0, \exists\delta>0 \ni p\big) \Longleftrightarrow \big(\exists\epsilon>0\ni(\forall\delta>0\implies\neg p)\big)$. – Hans Oct 18 '17 at 06:47
  • Sure, I agree with that, and it's clear to you how that translates to your proof (and probably many others), but it isn't so clear to me. Again though, maybe just personal preference. I can definitely understand the idea behind your argument though. Your proof would at least benefit from a final concluding statement: "therefore for any delta, there is an $A_n$..." – jdods Oct 18 '17 at 12:49
  • In plain language: $``$Assume the contrary, $\exists \epsilon>0$ such that $\forall n\in\mathbb{N}$ we can find an $A_n$ with $\mu(A_n)<\frac{1}{2^n}$ but $\int_{A_n} f\geq \epsilon."$ So we have a sequence ${A_n}$... leads to your contradiction. The conclusion is that for each $\epsilon>0$, there is at least one $n\in\mathbb{N}$ such that we are unable to find an $A_n$ with $\mu(A_n)<\frac{1}{2^n}$ that also satisfies $\int_{A_n} f\geq\epsilon$. So just pick any $\delta\leq \frac{1}{2^n}$. Thus any set $A$ with measure less than $\delta$ must satisfy $\int_A f <\epsilon$. – jdods Oct 18 '17 at 14:48
  • @jdods: You got it! Mind to upvote my answer? ;-) – Hans Oct 19 '17 at 17:36