Lebesgue's Dominated Convergence Theorem/Lemma
Lemma
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $g : X \to \overline \R_{\ge 0}$ be a $\mu$-integrable function.
Let $\sequence {f_n}_{n \mathop \in \N}$ be an sequence of $\Sigma$-measurable function $f_n : X \to \overline \R$ such that:
- $\ds (1): \quad \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
- $\ds (2): \quad \size {\map {f_n} x} \le \map g x$
- $\ds (3): \quad \map {f_n} x < \infty$
- $\ds (4): \quad \map g x < \infty$
hold for each $x \in X$.
Then:
- $f$ is $\mu$-integrable and $f_n$ is $\mu$-integrable for each $n \in \N$
and:
- $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
Proof
We first show that:
- $f$ is $\mu$-integrable and $f_n$ is $\mu$-integrable for each $n \in \N$
From Integral of Positive Measurable Function is Monotone, we have:
- $\ds \int \size {f_n} \rd \mu \le \int g \rd \mu < \infty$
so, from Characterization of Integrable Functions:
- $f_n$ is $\mu$-integrable for each $n \in \N$.
Note that we also have:
- $\size {\map f x} \le \map g x$
from Modulus of Limit.
From Integral of Positive Measurable Function is Monotone, we then obtain:
- $\ds \int \size f \rd \mu \le \int g \rd \mu < \infty$
so, from Characterization of Integrable Functions:
- $f$ is $\mu$-integrable.
From Convergence of Limsup and Liminf, we have:
- $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
- $\ds \int f \rd \mu = \limsup_{n \mathop \to \infty} \int f_n \rd \mu = \liminf_{n \mathop \to \infty} \int f_n \rd \mu$
We have:
- $\ds \liminf_{n \mathop \to \infty} \int f_n \rd \mu \le \limsup_{n \mathop \to \infty} \int f_n \rd \mu$
We will show that:
- $\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int f \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$
so that:
- $\ds \liminf_{n \mathop \to \infty} \int f_n \rd \mu = \limsup_{n \mathop \to \infty} \int f_n \rd \mu$
then we will have:
- $\ds \int f \rd \mu = \limsup_{n \mathop \to \infty} \int f_n \rd \mu = \liminf_{n \mathop \to \infty} \int f_n \rd \mu$
We first show that:
- $\ds \int f \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$
From Pointwise Sum of Measurable Functions is Measurable, we have:
- $g + f_n$ is $\Sigma$-measurable for each $n \in \N$.
Since we also have:
- $\map {f_n} x \ge -\map g x$
we have:
- $\map {f_n} x + \map g x \ge 0$
for each $n \in \N$.
We also have:
- $\ds \map f x + \map g x = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map g x}$
from Sum Rule for Real Sequences.
From Convergence of Limsup and Liminf, this gives:
- $\ds \map f x + \map g x = \liminf_{n \mathop \to \infty} \paren {\map {f_n} x + \map g x}$
From Fatou's Lemma for Integrals: Positive Measurable Functions, we have:
- $\ds \int \paren {f + g} \rd \mu \le \liminf_{n \mathop \to \infty} \int \paren {f_n + g} \rd \mu$
We then have:
- $\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
from Integral of Integrable Function is Additive.
We also have:
\(\ds \liminf_{n \mathop \to \infty} \int \paren {f_n + g} \rd \mu\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \int \paren {f_k + g} \rd \mu}\) | Definition of Limit Inferior | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \paren {\int f_k \rd \mu + \int g \rd \mu} }\) | Integral of Integrable Function is Additive | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \int f_k \rd \mu + \int g \rd \mu}\) | Infimum Plus Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \int f_k \rd \mu} + \int g \rd \mu\) | Sum Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \liminf_{n \mathop \to \infty} \int f_n \rd \mu + \int g \rd \mu\) | Definition of Limit Inferior |
So we have:
- $\ds \int f \rd \mu + \int g \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu + \int g \rd \mu$
so:
- $\ds \int f \rd \mu \le \liminf_{n \mathop \to \infty} \int f_n \rd \mu$
We now show that:
- $\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int f \rd \mu$
From Pointwise Difference of Measurable Functions is Measurable, we have:
- $g - f_n$ is $\Sigma$-measurable for each $n \in \N$.
Since we also have:
- $-\map g x \le \map {f_n} x \le \map g x$
so:
- $-\map g x \le -\map {f_n} x \le \map g x$
giving:
- $0 \le \map g x - \map {f_n} x \le 2 \map g x$
We also have:
- $\ds \map g x - \map f x = \lim_{n \mathop \to \infty} \paren {\map g x - \map {f_n} x}$
from Difference Rule for Sequences.
From Convergence of Limsup and Liminf, we have:
- $\ds \map g x - \map f x = \limsup_{n \mathop \to \infty} \paren {\map g x - \map {f_n} x}$
From Fatou's Lemma for Integrals: Positive Measurable Functions, we have:
- $\ds \int \paren {g - f} \rd \mu \le \liminf_{n \mathop \to \infty} \int \paren {g - f_n} \rd \mu$
We then have:
- $\ds \int \paren {g - f} \rd \mu = \int g \rd \mu - \int f \rd \mu$
from Integral of Integrable Function is Additive: Corollary $2$.
We also have:
\(\ds \liminf_{n \mathop \to \infty} \int \paren {g - f_n} \rd \mu\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \int \paren {g - f_n} \rd \mu}\) | Definition of Limit Inferior | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\inf_{n \ge k} \paren {\int g \rd \mu - \int f_n \rd \mu} }\) | Integral of Integrable Function is Additive | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\int g \rd \mu + \inf_{n \ge k} \paren {-\int f_n \rd \mu} }\) | Infimum Plus Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\int g \rd \mu - \sup_{n \ge k} \paren {\int f_n \rd \mu} }\) | Negative of Supremum is Infimum of Negatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \int g \rd \mu - \lim_{n \mathop \to \infty} \paren {\sup_{n \ge k} \int f_k \rd \mu}\) | Difference Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \int g \rd \mu - \limsup_{n \mathop \to \infty} \int f_n \rd \mu\) | Definition of Limit Superior |
So we have:
- $\ds \int g \rd \mu - \int f \rd \mu \le \int g \rd \mu - \limsup_{n \mathop \to \infty} \int f_n \rd \mu$
so:
- $\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int f \rd \mu$
$\blacksquare$