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:

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:

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:


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