Lebesgue's Dominated Convergence Theorem

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 an sequence of $\Sigma$-measurable function $f_n : X \to \overline \R$ such that:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

and:


 * $\ds \size {\map {f_n} x} \le \map g x$

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:


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

Lemma
Since:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

and:


 * $\ds \size {\map {f_n} x} \le \map g x$

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:


 * $\ds \size {\map {f_n} x} > \map g x$

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.

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, 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 Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, 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 A.E. Equal Positive Measurable Functions have Equal Integrals, 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:


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