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:
- $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
Proof
Lemma
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:
- $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
$\Box$
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:
- $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$
$\blacksquare$
Source of Name
This entry was named for Henri Léon Lebesgue.
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $11.2$
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $2.4$: Limit Theorems: $2.4.5$