[Measure Theory] Limit Theorems
The main reason the Lebesgue integral is so much easier to work with than the Riemann integral is that it behaves nicely when taking limits. In this chapter we first prove four main results:
- monotone convergence theorem
- linearity of the Lebesgue integral
- Fatou’s lemma
- the dominated convergence theorem
And from these results, we can obtain a variety of useful theorems, including generalized dominated convergence theorem, bounded convergence theorem, etc.
Convergence Theorems
Monotone Convergence Theorem
One of the most important results concerning Lebesgue integration is the monotone convergence theorem. It is one of the fundamental limit theorems that is useful to prove remaining theorems.
Suppose $f_n$ is a sequence of non-negative measurable functions that $f_n (x) \to f(x)$ for all $x$, i.e. with $f_1 (x) \leq f_2 (x) \leq \cdots$ for all $x$ and with $$ \begin{aligned} \lim_{n \to \infty} f_n (x) = f(x) \end{aligned} $$ for all $x$. Then, $$ \begin{aligned} \lim_{n \to \infty} \int f_n \; d\mu = \int \lim_{n \to \infty} f_n \; d\mu = \int f \; d\mu. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
From the property of integral, $\int f_n$ is an increasing sequence. Let $L$ be the limit, i.e. $\lim_{n \to \infty} \int f_n \; d\mu = L$. Since $f_n \leq f$ for all $x$, hence $L \leq \int f$.
For the reverse direction, let $s = \sum_{i=1}^m a_i \chi E_i$ be any non-negative simple function such that $s \leq f$ and let $c \in (0, 1)$. Define
\[\begin{aligned} A_n = \{ x \mid f_n (x) \geq c s (x) \}. \end{aligned}\]Since $f_n (x)$ increases to $f(x)$ for each $x$ and $c < 1$, then $A_n \uparrow X$. For each $n$,
\[\begin{aligned} \int f_n & \geq \int_{A_n} f_n \geq c \int_{A_n} s \\ & = \int_{A_n} \sum_{i=1}^m a_i \chi_{E_i} \\ & = \sum_{i=1}^m a_i \mu (E_i \cap A_n). \end{aligned}\]By letting $n \to \infty$, by continuity from below,
\[\begin{aligned} L \geq c \sum_{i=1}^m a_i \mu (E_i) = c \int s \end{aligned}\]Since $c$ is arbitrary in the interval $(0, 1)$, then $L \geq \int s$. By taking supremum over all $s \leq f$, we obtain $L \geq \int f$.
\[\tag*{$\blacksquare$}\]Here are some counterexamples when some of conditions are not satisfied.
$\color{green}{\mathbf{Examples.}}$
$\color{green}{\mathbf{Example\ 1.}}$ Let $X = [0, \infty)$ and $f_n (x) = - 1 / n$ for all $x$. Then $\int f_n = - \infty$, but $f_n \uparrow f$ where $f = 0$ and $\int f = 0$. The reason the monotone convergence theorem does not apply here is that the $f_n$ are not non-negative.
\[\tag*{$\blacksquare$}\]$\color{green}{\mathbf{Example\ 2.}}$ Suppose $f_n = n \chi_{(0, 1/n)}$. Then $f_n \geq 0$, $f_n \to 0$ for each $x$, but $\int f_n = 1$ doesn’t converge to $\int 0 = 0$. The reason the monotone convergence theorem does not apply here is that the $f_n (x)$ does not increase to $0$ for all $x$.
\[\tag*{$\blacksquare$}\]The monotone convergence theorem is an essential tool in many situations, but its immediate significance for us is as follows. The definition of $\int f$ involves the supremum over a huge (usually uncountable) family of simple functions, so it may be difficult to evaluate directly from the definition. The monotone convergence theorem, however, assures us that to compute $\int f$ it is enough to compute $\lim_{n \to \infty} \int s_n$ where \(\{ s_n \}\) is any sequence of simple functions that increase to $f$.
Linearity of the integral
Once we have the monotone convergence theorem, we can prove that the Lebesgue integral is linear.
If $f$ and $g$ are non-negative and measurable, or if $f$ and $g$ are integrable, then $$ \begin{aligned} \int (f + g) \; d\mu = \int f \; d\mu + \int g \; d\mu. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Case 1: non-negative simple functions
First, assume that $f$ and $g$ are non-negative and simple:
\[\begin{aligned} f = \sum_{i=1}^m a_i \chi_{A_i} \text{ and } g = \sum_{j=1}^n b_j \chi_{B_j}. \end{aligned}\]Let’s express $f + g$ by a simple function. For $i \leq m$, let $c_i = a_i$ and $C_i = A_i$. For $m + 1 \leq i \leq m + n$, let $c_i = b_{i - m}$ and $C_i = B_{i - m}$. Then
\[\begin{aligned} f + g = \sum_{i=1}^{m + n} c_i \chi_{C_i} \end{aligned}\]and we see that
\[\begin{aligned} \int (f + g) & = \sum_{i=1}^{m + n} c_i \mu (C_i) \\ & = \sum_{i=1}^m c_i \mu (C_i) + \sum_{j=1}^n c_{j + m} \mu (C_{j + m}) \\ & = \sum_{i=1}^m a_i \mu (A_i) + \sum_{j=1}^n b_{j} \mu (B_{j}) \\ & = \int f + \int g. \end{aligned}\]Thus the theorem holds in this case.
Case 2: non-negative functions
Next, let $f$ and $g$ be non-negative. By simple function approximation, take $s_n$ and $t_n$ non-negative simple functions increasing to $f$ and $g$, respectively.
Then, $s_n + t_n$ are also simple functions increasing to $f + g$, so the result follows from the monotone convergence theorem:
\[\begin{aligned} \int (f+g) \; d\mu = \lim_{n \to \infty} \int (s_n + t_n) = \lim_{n \to \infty} \left(\int s_n + \int t_n \right) = \lim_{n \to \infty} \int s_n + \lim_{n \to \infty} \int t_n = \int f + \int g \end{aligned}\]Case 3: integrable functions
Suppose now that $f$ and $g$ are real-valued and integrable but take both positive and negative values. Since
\[\begin{aligned} \int | f + g | \leq \int ( |f| + |g| ) = \int |f| + \int |g| < \infty, \end{aligned}\]therefore $f + g$ is also integrable. Write
\[\begin{aligned} (f + g)^+ - (f + g)^- = f+g = f^+ - f^- + g^+ - g^- \end{aligned}\]so that
\[\begin{aligned} (f + g)^+ + f^- + g^- = f^+ + g^+ + (f + g)^-. \end{aligned}\]Using the result for non-negative functions
\[\begin{aligned} \int (f + g)^+ + \int f^- + \int g^- = \int f^+ + \int g^+ + \int (f + g)^-. \end{aligned}\]By rearranging,
\[\begin{aligned} \int (f + g) & = \int (f+g)^+ - (f + g)^- = \int (f+g)^+ - \int (f + g)^- \\ & = \int f^+ - \int f^- + \int g^+ - \int g^- \\ & = \int f + \int g. \end{aligned}\]If $f$ and $g$ are complex-valued, apply the above to the real and imaginary parts. $\tag*{$\blacksquare$}$
Then, from the linearity of integral, we now extend the properties of integral of non-negative functions to integrable functions:
- If $a \leq f(x) \leq b$ for all $x$ and $\mu (X) < \infty$, then $$ \begin{aligned} a \mu(X) \leq \int f \; d\mu \leq b\mu(X); \end{aligned} $$
- If $f$ and $g$ are integrable, and $f(x) \leq g(x)$ for all $x$, then $$ \begin{aligned} \int f \; d\mu \leq \int g \; d\mu; \end{aligned} $$
- If $f$ is complex-valued and integrable, and $c$ is a complex number, then $$ \begin{aligned} \int cf \; d\mu = c \int f \; d\mu; \end{aligned} $$
- If $\mu (A) = 0$ and $f$ is measurable, then $$ \begin{aligned} \int f \chi_A \; d\mu = 0. \end{aligned} $$
$\color{#bf1100}{\mathbf{Proof.}}$
These follow from the definition of the Lebesgue integral and linearity.
(2)
Write $f = f^+ - f^-$ and $g = g^+ - g^-$.
Then
implies
\[\begin{aligned} 0 \leq f^+ (x) + g^- (x) \leq g^+ (x) + f^- (x) \end{aligned}\]for all $x$. Then it implies $\int (f^+ + g^-) \leq \int (g^+ + f^-)$ and by linearity $\int f \leq \int g$.
\[\tag*{$\blacksquare$}\]$\color{#bf1100}{\mathbf{Proof.}}$
Let $F_N = \sum_{n=1}^N f_n$. Then, since $f_n$ is non-negative for each $n$, $F_N (x) \uparrow \sum_{n=1}^\infty f_n (x)$ for all $x$. Then, from the monotone convergence theorem and linearity:
\[\begin{aligned} \int \sum_{n=1}^\infty f_n & = \lim_{N \to \infty} \int F_N \\ & = \lim_{N \to \infty} \int \sum_{n=1}^N f_n \\ & = \lim_{N \to \infty} \sum_{n=1}^N \int f_n \\ & = \sum_{n=1}^\infty \int f_n. \end{aligned}\] \[\tag*{$\blacksquare$}\]$\color{#bf1100}{\mathbf{Proof.}}$
For the real case, it is easy: $f \leq | f |$, so $\int f \leq \int | f |$. Also, $-f \leq | f |$, so $- \int f \leq \int | f |$.
For the complex case, $\int f$ is a complex number. If it is $0$, the inequality is simple. If it is not, then $\int f = r e^{i \theta}$ for some $r$ and $\theta$. Then
\[\begin{aligned} \left| \int f \right| = r = e^{-i \theta} \int f = \int e^{-i \theta} f. \end{aligned}\]From the definition of $\int f$ when $f$ is complex, it follows that $\text{Re}(\int f) = \int \text{Re} (f)$. Since $| \int f |$ is real, we have
\[\begin{aligned} \left| \int f \right| = \text{Re} \left( \int e^{-i\theta} f \right) = \int \text{Re}(e^{-i \theta} f) \leq \int | f | \end{aligned}\]as desired.
\[\tag*{$\blacksquare$}\]Fatou’s Lemma
Recall that increasing condition is essential for the monotone convergence theorem. Let $X = \mathbb{R}$, and $f_n = n \chi_{0, 1/n}$. Although $f_n \to 0$ pointwisely, the monotone convergence theorem does not hold: $\int f_n = 1$ for each $n$.
Upon sketching the graphs, it becomes evident that the issue in this example lies in the fact that the area under the graph diverges towards infinity as $n \to \infty$. Consequently, the area in the limit turns out to be less than one might anticipate. And Fatou’s lemma generalizes this intuition:
Suppose the $f_n$ are non-negative and measurable. Then $$ \begin{aligned} \int \underset{n \to \infty}{\text{lim inf}} \; f_n \leq \underset{n \to \infty}{\text{lim inf}} \; \int f_n. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Let \(g_n = \text{inf}_{i \geq n} \; f_i\). Then the $g_n$ are non-negative, measurable and $g_n$ increases to \(\text{sup}_n \; \text{inf}_{i \geq n} \; f_i = \underset{n \to \infty}{\text{lim inf}} \; f_n\). Clearly $g_n \leq f_i$ for each $i \geq n$, so $\int g_n \leq \int f_i$. Therefore
\[\begin{aligned} \int g_n \leq \underset{i \geq n}{\text{ inf }} \int f_i. \end{aligned}\]By taking the limit $n \to \infty$,
\[\begin{aligned} \int \underset{n \to \infty}{\text{lim inf}} \; f_n & \leq \underset{n}{\text{sup }} \underset{i \geq n}{\text{ inf }} \int f_i \\ & = \underset{n \to \infty}{\text{lim inf}} \; \int f_n. \end{aligned}\] \[\tag*{$\blacksquare$}\]Dominated Convergence Theorem
In the context of our intuitions in Fatou’s lemma, if the graph of $f_n$ is confined to a region of the plane with finite area, then now $\int f_n \; d\mu \to \int f \; d\mu$ might be ensured. And the dominated convergence theorem is saying about this statement:
Suppose the $f_n$ are measurable real-valued functions and $f_n (x) \to f(x)$ for each $x$. Suppose there exists a non-negative integrable function $g$ such that $| f_n (x) | \leq g(x)$ for all $x$. Then $$ \begin{aligned} \lim_{n \to \infty} \int f_n \; d\mu = \int f \; d\mu. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Since $f_n + g \geq 0$, by Fatou’s lemma,
\[\begin{aligned} \int f + \int g = \int (f + g) \leq \underset{n \to \infty}{\text{lim inf}} \int (f_n + g) = \underset{n \to \infty}{\text{lim inf}} \int f_n + \int g. \end{aligned}\]Since $g$ is integrable,
\[\begin{aligned} \int f \leq \underset{n \to \infty}{\text{lim inf}} \int f_n. \end{aligned}\]Similarly, $g - f_n \geq 0$, so
\[\begin{aligned} \int g - \int f = \int (g - f) \leq \underset{n \to \infty}{\text{lim inf}} \int (g - f_n) = \underset{n \to \infty}{\text{lim inf}} \int (-f_n) + \int g. \end{aligned}\]and hence
\[\begin{aligned} - \int f \leq \underset{n \to \infty}{\text{lim inf}} \int (-f_n) = - \underset{n \to \infty}{\text{lim sup}} \int f_n \end{aligned}\]Therefore
\[\begin{aligned} \int f \geq \underset{n \to \infty}{\text{lim sup}} \int f_n \end{aligned}\]which proves the theorem.
\[\tag*{$\blacksquare$}\]Generalized Dominated Convergence Theorem
Suppose the $f_n$, $g_n$, $f$, and $g$ are integrable, and $f_n \to f$ a.e., $g_n \to g$ a.e. Let $| f_n | \leq g_n$ for each $n$, and $\int g_n \to \int g$. Then $$ \begin{aligned} \int f_n \to \int f \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Since $f_n + g_n \geq 0$, by Fatou’s lemma,
\[\begin{aligned} \int f + \int g = \int (f + g) \leq \underset{n \to \infty}{\text{lim inf}} \int (f_n + g_n) = \underset{n \to \infty}{\text{lim inf}} \int f_n + \int g. \end{aligned}\]Since $g$ is integrable,
\[\begin{aligned} \int f \leq \underset{n \to \infty}{\text{lim inf}} \int f_n. \end{aligned}\]Similarly, $g_n - f_n \geq 0$, so
\[\begin{aligned} \int g - \int f = \int (g - f) & \leq \underset{n \to \infty}{\text{lim inf}} \int (g_n - f_n) \leq \underset{n \to \infty}{\text{lim inf}} \int g - f_n \\ & = \underset{n \to \infty}{\text{lim inf}} \int (-f_n) + \int g. \end{aligned}\]and hence
\[\begin{aligned} - \int f \leq \underset{n \to \infty}{\text{lim inf}} \int (-f_n) = - \underset{n \to \infty}{\text{lim sup}} \int f_n \end{aligned}\]Therefore
\[\begin{aligned} \int f \geq \underset{n \to \infty}{\text{lim sup}} \int f_n \end{aligned}\]which proves the theorem.
\[\tag*{$\blacksquare$}\]Bounded Convergence Theorem
Suppose $\mu(X) < \infty$ and $f_n$ is a sequence of uniformly bounded measurable real-valued functions that converge to $f$ pointwisely. Then $$ \begin{aligned} \int f_n \to \int f \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Let $| f_n (x) | \leq M$ for all $x \in X$ and $n \in \mathbb{N}$. Let $g = M \cdot \chi_X$. Then, $\int g = M \cdot \mu(X) < \infty$, therefore $g$ is integrable. Done by the dominated convergence theorem.
\[\tag*{$\blacksquare$}\]Uniformly Integrability
Uniform integrability is one of the most important concepts, which emerges a lot in the probability theory.
Let $(X, \mathcal{A}, \mu)$ be a measure space. Then, a family of measurable functions $\{ f_n \}$ is uniformly integrable if given $\varepsilon$ there exists $M$ such that $$ \begin{aligned} \int_{\{ x: | f_n (x) | > M \}} | f_n (x) | \; d\mu < \varepsilon \end{aligned} $$ for all $n$.
And there is a connection between the notion of continuity called uniformly absolutely continuous when $\mu$ is a finite measure.
Let $(X, \mathcal{A}, \mu)$ be a measure space. Then, a family of measurable functions $\{ f_n \}$ is uniformly absolutely continuous if given $\varepsilon$ there exists $\delta$ such that $$ \begin{aligned} \left| \int_{A} f_n (x) \; d\mu \right| < \varepsilon \end{aligned} $$ for all $n$ if $\mu(A) < \delta$.
Suppose $\mu$ is a finite measure. Let $\{ f_n \}$ be a family of measurable functions. Then the followings are equivalent:
- $\{ f_n \}$ is uniformly integrable
- $\text{sup}_n \int | f_n | \; d\mu < \infty$ and $\{ f_n \}$ is uniformly absolutely continuous
$\color{red}{\mathbf{Proof.}}$
1 $\to$ 2
Let any $\varepsilon > 0$ be given.
(1) $\text{sup}_n \int | f_n | \; d\mu < \infty$
\[\begin{aligned} \int | f_n | \; d\mu & = \int_{\{ x \; : \; | f_n (x) | > M \}} | f_n | \; d\mu + \int_{\{ x \; : \; | f_n (x) | \leq M \}} | f_n | \; d\mu \\ & = \frac{\varepsilon}{2} + M \cdot \mu (X) < \infty \end{aligned}\](2) uniformly absolutely continous
Let $\delta = \frac{\varepsilon}{2M}$. Let any $A \in \mathcal{A}$ with $\mu(A) < \delta$ be given/
\[\begin{aligned} \int_A | f_n | \; d\mu & = \int_{A \cap \{ x \; : \; | f_n (x) | > M \}} | f_n | \; d\mu + \int_{A \cap \{ x \; : \; | f_n (x) | \leq M \}} | f_n | \; d\mu \\ & = \frac{\varepsilon}{2} + M \cdot \mu (A) < \varepsilon \end{aligned}\]Thus,
\[\begin{aligned} \left| \int_A f_n \; d\mu \right| \leq \int_A | f_n | \; d\mu < \varepsilon. \end{aligned}\]2 $\to$ 1
Let any $\varepsilon > 0$ be given. Let $L = \text{sup}_n \int | f_n | \; d\mu < \infty$.
Then, we have
\[\begin{aligned} \mu \left( \{ x \; : \; | f_n (x) | > M \} \right) \leq \int_{\{ x \; : \; | f_n (x) | > M \}} \frac{| f_n (x) |}{M} \; d \mu \leq \frac{L}{M} < \delta \end{aligned}\]for sufficiently large $M > 0$.
Then
\[\begin{aligned} \int_{\{ x \; : \; | f_n (x) | > M \}} | f_n | \; d\mu & = \int_{\{ x \; : \; f_n (x) > M \}} f_n \; d\mu + \int_{\{ x \; : \; f_n (x) < - M \}} - f_n \; d\mu \\ & \leq \left| \int_{\{ x \; : \; f_n (x) > M \}} f_n \; d\mu \right| + \left| \int_{\{ x \; : \; f_n (x) < - M \}} f_n \; d\mu \right| < \varepsilon \end{aligned}\] \[\tag*{$\blacksquare$}\]Vitali Convergence Theorem
When $\int f_n \to f$, it is not always imply that $f_n$ converges to $f$ in $L^1$-metric, in other words, $\int | f_n - f | \to 0$ does not hold in general. But there are some additional conditions that makes the statement hold.
$\color{red}{\mathbf{Proof.}}$
Let any $\varepsilon > 0$ be given.
Define
\[\begin{aligned} A_n = \{ x: | f_n (x) - f (x) | \leq \frac{\varepsilon}{2 \mu(X)} \} \end{aligned}\]Then, from $f_n \to f$ a.e., $\mu (A_n^c) \to 0$ as $n \to \infty$. In other words, there exists $N \in \mathbb{N}$ such that
\[\begin{aligned} \mu (A_n^c) \leq \frac{\varepsilon}{2 \mu (X)} \quad \text{ for all } n \geq N. \end{aligned}\]Note that it is possible as $\mu (X) < \infty$. Therefore, for any $n \geq N$,
\[\begin{aligned} \int | f_n - f_n | \; d\mu & = \int_{A_n} | f_n - f_n | \; d\mu + \int_{A_n^c} | f_n - f_n | \; d\mu \\ & \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \end{aligned}\] \[\tag*{$\blacksquare$}\]The following is known as the Vitali convergence theorem.
Suppose $(X, \mathcal{A}, \mu)$ is a measure space and $\mu$ is a finite measure. If $f_n$ is a sequence of uniformly integrable real-valued functions and $f_n \to f$ a.e., then $$ \begin{aligned} \int | f_n - f | \; d\mu \to 0. \end{aligned} $$
$\color{red}{\mathbf{Proof.}}$
Let any $\varepsilon > 0$ be given.
Step 1. $f$ is integrable
From Fatou’s lemma,
\[\begin{aligned} \int \text{ lim inf}_n | f_n | \; d\mu = \int | f | \; d\mu \leq \text{ lim inf}_n \int | f_n | \; d\mu \end{aligned}\]and for any $n \in \mathbb{N}$, there exists a uniform bound $L$ due to uniformly integrability and finite measure:
\[\begin{aligned} \int | f_n | \; d\mu = \int_{| f_n | \geq M} | f_n | \; d\mu + \int_{| f_n | < M} | f_n | \; d\mu \leq \varepsilon + M \cdot \mu (X) \end{aligned}\]Thus
\[\begin{aligned} \text{ lim inf}_n \int | f_n | \; d\mu < \infty \end{aligned}\]It implies that $| f | < \infty$ a.e., which can be proved easily. By the dominated convergence theorem,
\[\begin{aligned} \int_{\{ x: \; | f (x) | > M \}} | f | \to 0 \end{aligned}\]as $M \to \infty$.
Step 2. $f_n \to f$ in $L^1$ metric.
A sequence of $f_n$ is uniformly integrable, hence there exists $M > 0$ such that
\[\begin{aligned} \int_{\{x : \; | f_n (x) | > M \}} | f_n | < \frac{\varepsilon}{5} \end{aligned}\]So we can choose $M > 0$ such that for all $n \in \mathbb{N}$
\[\begin{aligned} & \int_{\{x : \; | f_n (x) | > M \}} | f_n | < \frac{\varepsilon}{5} \\ & \int_{\{x : \; | f (x) | > M \}} | f | < \frac{\varepsilon}{5} \end{aligned}\]Furthermore, for each $n$ construct
\[\begin{aligned} A_n = \{ x : \; |f_n (x) - f(x)| < \frac{\varepsilon}{5 \mu (X)} \} \end{aligned}\]and since $f_n \to f$ a.e. so that $\mu (A_n^c) \to 0$ as $n \to \infty$, there exists $N \in \mathbb{N}$ such that
\[\begin{aligned} \mu (A_n^c) \leq \frac{\varepsilon}{5M} \quad \text{ for all } n \geq N. \end{aligned}\]Then, we obtain the following:
\[\begin{aligned} \int | f_n - f | & = \int_{A_n} | f_n - f | + \int_{A_n^c} | f_n - f | \\ & \leq \int_{A_n} | f_n - f | + \int_{A_n^c} | f_n |+ \int_{A_n^c} | f | \\ & = \int_{A_n} | f_n - f | + \int_{A_n^c \cap \{ x : \; |f_n (x)| > M \}} | f_n | + \int_{A_n^c \cap \{ x : \; |f_n (x)| \leq M \}} | f_n | \\ & \quad + \int_{A_n^c \cap \{ x: \; |f (x)| > M \}} | f | + \int_{A_n^c \cap \{ x : \; |f (x)| \leq M \}} | f | \\ & \leq \frac{\varepsilon}{5} + \frac{\varepsilon}{5} + \frac{\varepsilon}{5} + \frac{\varepsilon}{5} + \frac{\varepsilon}{5} = \varepsilon. \end{aligned}\]for all $n \geq N$.
\[\tag*{$\blacksquare$}\]Furthermore, if $\mu$ is finite measure the converse of theorem also holds:
$\color{red}{\mathbf{Proof.}}$
Let any $\varepsilon > 0$ be given.
As $| f | \wedge M \uparrow | f |$, by monotone convergence theorem there exists a sufficiently large $M > 0$ such that $\int | f | - \int | f | \wedge M = \int_{| f | \geq M} < \varepsilon/3$.
Also, from $f_n \to f$ a.e. and $\mu$ is finite for sufficiently large $N$ we have $\int | f_n - f | < \varepsilon/3$ for each $n \geq N$.
Finally, for each $n \geq M$, define
\[\begin{aligned} A_n = \{ x: \; |f_n (x)| \geq 2M, |f(x)| < M \} \end{aligned}\]Thus, for $x \in A_n$ $| f_n (x) - f(x) | > M$,
\[\begin{aligned} M \cdot \mu(A_n) \leq \int_{A_n} | f_n (x) - f(x) | \leq \int | f_n (x) - f(x) | < \frac{\varepsilon}{3}. \end{aligned}\]By combining these facts, we obtain
\[\begin{aligned} \int_{\{ | f_n | \geq 2M \}} |f_n| & \leq \int_{\{ | f_n | \geq 2M \}} |f_n - f| + | f | \\ & = \int_{\{ | f_n | \geq 2M \}} |f_n - f| + \int_{\{ | f_n | \geq 2M \}} | f | \\ & = \int_{\{ | f_n | \geq 2M \}} |f_n - f| + \int_{\{ | f_n | \geq 2M, | f | \geq M \}} | f | + \int_{A_n} | f | \\ & = \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ & = \varepsilon \end{aligned}\]for all $n \geq N$. In case of $n = 1, 2, \cdots, N - 1$, by monotone convergence theorem there exists $M_n$ such that $\int_{| f_n | \geq M_n} | f_n | < \varepsilon$.
Done by setting $M = \text{max } (M_1, \cdots M_{n-1}, 2M)$.
\[\tag*{$\blacksquare$}\]Reference
[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley & Sons, 1999.
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