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

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.
• $\mu$ 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