Lebesgue's Dominated Convergence Theorem/Lemma

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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:

$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$

if and only if:

$\ds \int f \rd \mu = \limsup_{n \mathop \to \infty} \int f_n \rd \mu = \liminf_{n \mathop \to \infty} \int f_n \rd \mu$

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:

$\ds \limsup_{n \mathop \to \infty} \int f_n \rd \mu \le \int f \rd \mu$

$\blacksquare$