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Hello all mathematicians!! Again, I am struggling with solving the exercises in Lebesgue Integral for preparing the quiz. At this moment, I and my friend are handling this problem, but both of us agreed this problem is a bit tricky. The problem is following.

Let ($X,\mathcal{A},\mu$) be a measure space and suppose $\mu$ is $\sigma$-finite. Suppose $f$ is integrable. Prove that given $\epsilon$ there exist $\delta$ such that

$$ \int_A |f(x)|\mu(dx) \;\;<\;\;\epsilon $$

whenever $\mu(A) < \delta$.

Could anybody give some good idea for us? Think you very much for your suggestion in advance.

Did
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Block Jeong
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4 Answers4

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If $f$ is integrable there exists a simple function $0 \le \phi \le f$ with the property that $$\int f - \phi \, d\mu < \frac{\epsilon}{2}.$$ Write $$\phi = \sum_{k=1}^n c_k \chi_{C_k}.$$ Since $$ \int_A \phi \, d\mu = \sum_k c_k \mu(A \cap C_k) \le \mu(A) \sum_{k=1}^n c_k$$ you can take $$\delta = \frac{\epsilon}{2 \sum _{k=1}^n c_k}$$ to conclude $$\mu(A) < \delta \implies \int_A f \, d\mu \le \int (f-\phi) \, d\mu + \int_A \phi \, d\mu< \epsilon.$$

Umberto P.
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Hint: $$ \int_A|f|\leqslant x\mu(A)+\int_{\{|f|\gt x\}}|f|\qquad\&\qquad\lim_{x\to\infty}\int_{\{|f|\gt x\}}|f|=0$$

Did
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  • I understand the first part. However, what if x on the first part goes to $\infty$? And for the second part, I don't get the exact meaning of ${|f|>x }$. Could you expalain more in detail? Thank you for your answer! – Block Jeong Oct 03 '13 at 13:32
  • Ah! I am correcting the question for the second part. I want to ask how the second part does work. Thanks for the idea. – Block Jeong Oct 03 '13 at 14:02
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    @BlockJeong: $x$ does not have to go to $\infty$. You just have to take it fixed, but big enough, so that $\int_{{|f|>x}} |f|$ is small (i.e.: $<\varepsilon/2$). Then, take $\delta$ such that $x \delta < \varepsilon/2$. – André Caldas Oct 04 '13 at 17:42
  • +1. This is a very elegant solution but I have a question. How do you justify in a simple manner that $$\int_{|f|>x} |f| \to 0?$$ The proof that I know uses Chebyshev's inequality to establish $\mu({|f| > x}) \to 0$, and the absolute continuity of the integral--the very property the OP is trying to prove--to establish $\int_{|f| > x} |f| \to 0$. – Umberto P. Aug 20 '15 at 17:52
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    @UmbertoP. Lebesgue dominated convergence theorem. – Did Aug 20 '15 at 19:59
  • Of course. Thanks @Did – Umberto P. Aug 20 '15 at 20:17
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Define the sequence $A_n=\{|f|\geq n\}$ (the set of all the points in which the value of $|f|$ is greater or equal than $n$). Since $f$ is integrable, clearly $|f|$ is measurable and therefore $A_n$ is measurable for any $n$. $A_n$ is defined a.e. (as $f$ is); should it sound odd to you, since everything holds up to a.e. equivalence, you could replace $f$ with any representative of its equivalence class.

Obviously $A_n$ is decreasing ($A_n\supseteq A_{n+1}$ for any $n$), therefore there exists the limit $$ A_\infty\dot=\lim_{n\to\infty}A_n=\bigcap_{n=1}^\infty A_n $$ and $A_\infty$ is measurable. From the integrability of $f$ it follows that $\mu(A_\infty)=0$ (if $\mu(A_\infty)>0$ then $\int_X |f|{\rm d}\mu \geq \int_{A_\infty}|f|{\rm d}\mu = \infty \cdot \mu(A_\infty) = \infty$, so $f$ would not be integrable).

Therefore the sequence $$ f_n ~\dot=~ f\chi_{A_n} $$ converges to $0$ a.e. ($\chi_{A_n}$ is the indicator function of the set $A_n$). It is not difficult to argue that $f_n$ is measurable for any $n$.

Now apply Lebesgue's dominated convergence theorem to $f_n$ which is pointwise less or equal than $|f|$, which is integrable.

Therefore $$ \int_{A_n}f_n{\rm d}\mu ~\stackrel{n\to\infty}{\longrightarrow} 0 $$ It should be easy from here... :)

AndreasT
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  • Thanks for the answer! Does it hold $|f(x)| < \infty$ if $f$ is integrable? No possibility that $|f| = infty$ even though $f$ is integrable? – Block Jeong Oct 03 '13 at 14:25
  • I edited the answer. Remember that we are in terms of "almost everywhere"... in this sense, for instance, the function $\infty\cdot\chi_{\mathbb Q}$ is integrable wrt Lebesgue measure in $\mathbb R$, and $A_\infty=\mathbb Q$ which is non-empty but has null measure. – AndreasT Oct 03 '13 at 14:57
  • Your $A_n$ is not arbitrary. – user398843 Oct 14 '19 at 23:38
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This is my full solution based on the hint from the top answer:

Let $f$ be integrable and let $\epsilon > 0$ be arbitrary. We know that $$\lim_{\alpha\to\infty}\int_{\{|f|\geq \alpha\}}|f| d\mu=0$$ Thus, we can take $\alpha$ big enough so that $$\int_{\{|f|\geq \alpha\}}|f| d\mu < \frac{\epsilon}{2}$$ Fix $\delta = \dfrac{\epsilon}{2\alpha}$ and let $A$ be such that $\mu(A) < \delta$. We have that: \begin{align*} \int_{A}|f| d\mu &= \int_{A}|f| \cdot I_{\{|f|< \alpha\}} d\mu + \int_{A}|f| \cdot I_{\{|f|\geq \alpha\}} d\mu \\ &\leq \alpha \cdot \mu(A) + \int_{A \cap \{|f|\geq \alpha\}}|f| d\mu \\ &\leq \alpha \cdot \mu(A) + \int_{\{|f|\geq \alpha\}}|f| d\mu \\ &< \alpha \cdot \frac{\epsilon}{2\alpha} + \frac{\epsilon}{2} \\ &=\epsilon \end{align*}

marshvee
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