# Absolute continuity of the Lebesgue integral

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?

• 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.
Commented Mar 24, 2022 at 4:35
• can you check if my prof works as well , if so upvote it plz. Commented Jun 30, 2022 at 3:15
• @D.R. ^^^^^^^^^^ Commented Jun 30, 2022 at 3:15

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

• 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
– user67133
Commented Oct 22, 2013 at 3:05
• 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
Commented Oct 22, 2013 at 3:13
• @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
Commented Oct 22, 2013 at 3:14
• 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!
– user67133
Commented Oct 22, 2013 at 3:23
• @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
– user67133
Commented Dec 13, 2014 at 14:09

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}

• Easiest and cleanest proof. Thanks! Commented Aug 24, 2017 at 11:33
• jdods: I put up another proof below. Would you care to verify it?
– Hans
Commented Oct 1, 2017 at 6:41
• 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
Commented Apr 28, 2021 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. Commented Apr 28, 2021 at 17:01
• Can you check if my proof is correct or not, if so vote it up plz! Commented Jun 30, 2022 at 3:14

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

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

• 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. Commented Jun 30, 2022 at 17:50
• @jdods BIG brains fart on my end, Edit made! Commented Jun 30, 2022 at 17:53
• True, I used a highfalutin theorem like MCT. @jdods Commented Jun 30, 2022 at 17:54
• 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. Commented Jun 30, 2022 at 18:03
• @jdods true, though it can be proven using Fatou's Lemma Commented Jun 30, 2022 at 19:51

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.

• 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. Commented Oct 2, 2017 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
Commented Oct 2, 2017 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. Commented Oct 3, 2017 at 3:20
• 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. Commented Oct 3, 2017 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
Commented Oct 3, 2017 at 23:38