# 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?

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 Oct 22 '13 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 Oct 22 '13 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 Oct 22 '13 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 Oct 22 '13 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 Dec 13 '14 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! – Richard Clare Aug 24 '17 at 11:33
• jdods: I put up another proof below. Would you care to verify it? – Hans Oct 1 '17 at 6:41

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. – jdods Oct 2 '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 2 '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 3 '17 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. – jdods Oct 3 '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 3 '17 at 23:38