Theorem: Lebesgue's Dominated Convergence Theorem

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 a sequence of $\Sigma$-measurable functions $f_n : X \to \overline \R$ such that: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
and:
:$\size {\map {f_n} x} \le \map g x$
hold for $\mu$-almost all $x \in X$.
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:
=== Lemma ===
{{:Lebesgue's Dominated Convergence Theorem/Lemma}}{{qed|lemma}}
Since: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
and:
:$\size {\map {f_n} x} \le \map g x$
hold for $\mu$-almost all $x \in X$, there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever: 
:either $\ds \lim_{n \mathop \to \infty} \map {u_n} x$ does not exist or $\ds \map u x \ne \lim_{n \mathop \to \infty} \map {u_n} x$
or: 
:$\size {\map {f_n} x} > \map g x$
then $x \in N_1$.
From Integrable Function is A.E. Real-Valued, there exists a $\mu$-null set $N_2 \subseteq X$ such that whenever: 
:$\map g x = \infty$ 
we have $x \in N_2$.
Let: 
:$N = N_1 \cap N_2$
From Sigma-Algebra Closed under Countable Intersection, we have that $N$ is $\Sigma$-measurable. 
From Intersection is Subset, we have: 
:$N \subseteq N_1$ 
From Null Sets Closed under Subset, we have: 
:$N$ is a $\mu$-null set. 
Now define a function $u : X \to \overline \R$ by: 
:$\map u x = \map f x \map {\chi_{X \setminus N} } x$
for each $x \in X$. 
Define a function $v : X \to \overline \R_{\ge 0}$ by: 
:$\map v x = \map g x \map {\chi_{X \setminus N} } x$
for each $x \in X$. 
For each $n \in \N$ define a function $u_n : X \to \overline \R$ by: 
:$\map {u_n} x = \map {f_n} x \map {\chi_{X \setminus N} } x$
for each $x \in X$. 
We now show that: 
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
and:
:$\size {\map {u_n} x} \le \map v x$
for each $x \in X$, and: 
:$v$ is $\mu$-integrable with $\map v x < \infty$ for each $x \in X$
so that we can apply the {{Lemma|Lebesgue's Dominated Convergence Theorem}}.
Let $x \in N$.
Then: 
:$\map {u_n} x = 0$ for each $n$
and:
:$\map u x = 0$ 
So we have: 
:$\ds \lim_{n \mathop \to \infty} \map {u_n} x = \map u x$ 
We also trivially have: 
:$\size {\map {u_n} x} \le \map v x$
Now, let $x \in X \setminus N$. 
We have: 
:$\map {u_n} x = \map {f_n} x$ for each $n \in \N$
and: 
:$\map u x = \map f x$
So, we have: 
:$\ds \lim_{n \mathop \to \infty} \map {u_n} x = \map u x$
We also have: 
:$\map g x = \map v x$ 
From the definition of $N$, we also have: 
:$\size {\map {f_n} x} \le \map g x$ 
So that: 
:$\size {\map {u_n} x} \le \map v x$ 
It remains to verify that: 
:$v$ is $\mu$-integrable with $\map v x < \infty$ for each $x \in X$
By the definition of $N$, we have: 
:$\map g x < \infty$ for all $x \in X \setminus N$
So, for $x \in X \setminus N$, we have: 
:$\map v x < \infty$
From Integral of Integrable Function over Measurable Set is Well-Defined, we also have: 
:$v$ is $\mu$-integrable.
So, from the {{Lemma|Lebesgue's Dominated Convergence Theorem}}, we obtain: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$ 
and: 
:$u_n$ is $\mu$-integrable for each $n$, and $u$ is  $\mu$-integrable.
That is: 
:$\ds \int \paren {f \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int \paren {f_n \times \chi_{X \setminus N} } \rd \mu$
From {{Corollary|Characteristic Function of Null Set is A.E. Equal to Zero}}, we have: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere. 
Then, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$f_n \times \chi_{X \setminus N} = f_n$ $\mu$-almost everywhere for each $n \in \N$
and:
:$f \times \chi_{X \setminus N} = f$ $\mu$-almost everywhere.
So, from {{Corollary|A.E. Equal Positive Measurable Functions have Equal Integrals|1}}, we have: 
:$f_n$ is $\mu$-integrable for each $n \in \N$
with: 
:$\ds \int f_n \rd \mu = \int \paren {f_n \times \chi_{X \setminus N} } \rd \mu$ for each $n \in \N$.
We also obtain: 
:$f$ is $\mu$-integrable
with:
:$\ds \int f \rd \mu = \int \paren {f \times \chi_{X \setminus N} } \rd \mu$ 
So, we obtain: 
:$\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
{{qed}}