# Questions of an exercise in Lebesgue integral

I'm doing exercises related to Lebesgue integral and get stuck by two of them. I can't figure out what do some steps in solutions mean.

Some definitions probably will be used:

1. Definition of measurable set: A set $E$ measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.

2. Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.

3. Definition of almost everywhere convergence: Suppose $f(x), f_1(x), f_2(x), \dots, f_k(x), \dots$ are extended real functions defined on a set $E \subset \mathbb R^n$ that is each $f_i: E \to [-\infty, +\infty]$. If $\exists Z \subset E$ such that $m(Z)=0$ and $$\lim_{k \to +\infty}f_k(x) = f(x), x\in(E-Z),$$ then $\{f_k(x)\}$ converges almost everywhere to $f(x)$ on $E$, namely $$f_k(x) \to f(x),\ \text{a.e.}\ x ∈ E.$$

4. Egorov Lemma: Suppose $f(x), f_1(x), f_2(x), \dots, f_k(x), \dots$ are measurable functions and finite almost everywhere($m(\{x \in E: |f(x)| = +\infty \}) = 0$) and $m(E)<\infty$. Suppose $f_k(x) \to f(x) \ a.e. x \in E$, then for $\forall \epsilon > 0$, let $E_k(\epsilon) = \{x \in E: |f_k(x) - f(x)| \ge \epsilon \}$, we have $$\lim_{j \to \infty}m(\bigcup_{k=j}^{\infty} E_k(\epsilon)) = 0.$$

5. Definition of Lebesgue integral of simple function: We say that a simple function $\psi$ is Lebesgue integrable if the set $\{\psi \ne 0\}$ has finite measure. In this case, we may write the standard representation for $\psi$ as $\psi = \sum_{i=0}^n a_i \chi_{A_i}$, where $a_0 = 0, a_1, \ldots , a_n$ are distinct real numbers, where $A_0 = \{\psi = 0\}, A_1, \ldots , A_n$ are pairwise disjoint and measurable, and where only $A_0$ has infinite measure, Once $\psi$ is so written, there is an obvious definition for $\int \psi$, namely, $$\int \psi = \int_{\mathbb R} \psi = \int_{-\infty}^{+\infty} \psi(x) \ \mathsf dx = \sum_{i=1}^n a_i m(A_i).$$ In other words, by adopting the convention that $0 \cdot \infty = 0$, we define the Lebesgue integral of $\psi$ by $$\int \sum_{i=0}^n a_i \chi_{A_i} = \sum_{i=0}^n a_i m(A_i).$$ Please note that $a_im(A_i)$ is a product of real numbers for $i \ne 0$, and it is $0 \cdot \infty = 0$ for $i = 0$; that is, $\int \psi$ is a finite real number.

6. Definition of Lebesgue integrable of non-negative function: If $f: \mathbb R \to [0, +\infty]$ is measurable, we define the Lebesgue integral of $f$ over $\mathbb R$ by $$\int f = \sup \left\{\int \psi: 0 \le \psi \le f, \psi\text{ simple function and integrable }\right\}.$$

Exercise1:

Suppose $\{f_k(x)\}$ is a sequence of Lebesgue measurable functions defined on $E$ and $$\lim_{k \to \infty} f_k(x) = f(x), a.e. x \in E.$$ If $\exists$ non-negative measurable and Lebesgue integrable function $g(x)$ on $E$ such that $|f_k(x)| \le g(x) (k=1,2,3,\dot)$, then show that $$\forall \epsilon > 0, \lim_{j \to \infty} m(\bigcup_{k \ge j}^{\infty} \{x \in E: |f_k(x) - f(x)| > \epsilon \}) = 0.$$

My trial was using Egorov's lemma. However, its solution hit me with carefully looking at the condition that there doesn't say $m(E) < +\infty$. Yes, that's my fault. I have no doubt about it.

Solution: Since we have $|f_k(x)| \le g(x)(k \in \mathbb N)$ and its Lebesgue integrable property, we will get for $\forall k \in \mathbb N$, $$E_k(\epsilon) = \{ x \in E: |f_k(x) - f(x)| > \epsilon \} \subset \{x \in E: g(x) > \frac{\epsilon}{2}\},$$ and $$\int_{\{x \in E: g(x) > \frac{\epsilon}{2}\}} g(x) dx \le \int_{E} g(x) dx < +\infty.$$ Then $m(\bigcup_{k=1}^{\infty} E_k(\epsilon)) \le m(\{x\in E: g(x) > \frac{\epsilon}{2}\}) \le \frac{2}{\epsilon} \int_{\{x \in E: g(x) > \frac{\epsilon}{2}\}}g(x) dx < +\infty.$

I have two questions:

Why for $\forall k \in \mathbb N$, $\{ x \in E: |f_k(x) - f(x)| > \epsilon \}$ is a subset of $\{x \in E: g(x) > \frac{\epsilon}{2}\}?$

How does the solution imply $\lim_{j \to \infty}m(\bigcup_{k \ge j}^{\infty} \{x \in E: |f_k(x) - f(x)| > \epsilon\})=0?$

• Is $E_k(\epsilon)$ the set $\{x\in E: |f_k(x) - f(x)| >\epsilon\}$? – Andrew Aug 7 '15 at 4:27
• @Andrew: Yes. Sorry about that. I will add it. – Bear and bunny Aug 7 '15 at 4:28

## 1 Answer

First question: By the triangle inequality, $$|f_k(x) - f(x)| \leq |f_k(x)| + |f(x)| \leq 2 g(x),$$ since $|f_k(x)| \leq g(x)$ implies $|f(x)| \leq g(x)$ by taking a limit. So, if the left hand side is greater than $\epsilon$, we get $\epsilon < 2g(x)$, so $g(x) > \epsilon/2$.

Second question: Define $A_j = \bigcup_{k\geq j} E_k(\epsilon)$. We know that the set $A_1=\bigcup_{n=1}^\infty E_k(\epsilon)$ has finite measure. We note that $A_j \supset A_{j+1}$, and $m(A_1) <\infty$. Thus, the continuity from above property of measures implies that $\lim_{j\to \infty} m(A_j) = m\left(\bigcap_j A_j\right).$ Now, if $x\in A_j$ for all $j$, then for all $j$ there is $k\geq j$ such that $|f_j(x)-f(x)|>\epsilon$, which is precisely the statement that $f_n(x) \not \to f(x)$. Since $f_n$ is assumed to converge pointwise to $f$ almost everywhere, we see that $\bigcap_j A_j$ must have measure $0$. Therefore, $\lim_j m(A_j) =0$, which is the desired result.

• "the continuity from above property of measures" what does this mean? – Bear and bunny Aug 7 '15 at 4:43
• See property 6 here: planetmath.org/propertiesformeasure – Andrew Aug 7 '15 at 4:44
• $\mu$ is a general measure that satisfy $\sigma$-additivity, right? – Bear and bunny Aug 7 '15 at 4:46
• Yes, by definition measures are $\sigma$-additive. Your case for Lebesgue measure is a special case. – Andrew Aug 7 '15 at 4:47
• Thank you! Please vote up as well :) I'm a grad student and I've taken a year of real analysis/measure theory, plus I've had a lot of experience with analysis in the past. It may take a while, but don't get discouraged. I've definitely found it helpful to talk to people in person about these ideas and ask them how they "intuit" these notions. Having intuition for what you're symbolically working with is very helpful. – Andrew Aug 7 '15 at 5:07