The Dominated Convergence Theorem for Meas. Functs.

The Dominated Convergence Theorem for Measurable Functions

Recall from The Lebesgue Dominated Convergence Theorem that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E$ such that:

  • 1) There exists a nonnegative Lebesgue measurable function $g$ such that $|f_n(x)| \leq g(x)$ on $E$ for all $n \in \mathbb{N}$.
  • 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.

Then $f$ is Lebesgue integrable on $E$ and $\displaystyle{\lim_{n \to \infty} \int_E f_n(x) \: dx = \int_E f(x) \: dx}$.

We now state a similar theorem for general complete measure spaces.

Theorem 1 (The Dominated Convergence Theorem for Measurable Functions): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $(f_n(x))_{n=1}^{\infty}$ be a sequence of measurable functions defined on a Lebesgue measurable set $E$ such that:
1) There exists a nonnegative measurable function $g$ such that $|f_n(x)| \leq g(x)$ on $E$.
2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.
Then $f$ is integrable on $E$ and $\displaystyle{\lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu}$.
  • Proof: For each $n \in \mathbb{N}$ we have that the measurable function $f_n$ is such that $|f_n(x)| \leq g(x)$ where $g$ is an integrable function. By The Comparison Test for Integrability, each $f_n$ is integrable. Furthermore, $|f(x)| \leq g(x)$ so $f$ is integrable by the comparison test.
  • Consider the sequences $(g(x) - f_n(x))_{n=1}^{\infty}$ and $(g(x) + f_n(x))_{n=1}^{\infty}$ of Lebesgue nonnegative measurable functions (these are also sequences of integrable functions) defined on $E $], that converge to [[$ g(x) - f(x)$ and $g(x) + f(x)$ respectively.
(1)
\begin{align} \quad \int_E g(x) \: d \mu - \int_E f(x) \: d \mu \overset{\mathrm{monotonicity}} = \int_E [g(x) - f(x)] \: d \mu \overset{\mathrm{Fatou's \: Lemma}} \leq \liminf_{n \to \infty} \int_E [g(x) - f_n(x)] \: d \mu = \int_E g(x) \: d \mu - \limsup_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
  • Therefore $\displaystyle{\limsup_{n \to \infty} \int_E f_n(x) \: d \mu \leq \int_E f(x) \: d \mu}$. $(*)$
  • Similarly, by Fatou's Lemma on the sequence of functions $(g(x) + f_n(x))_{n=1}^{\infty}$ we have that:
(2)
\begin{align} \quad \int_E g(x) \: d \mu + \int_E f(x) \: d \mu \overset{\mathrm{monotonicity}} = \int_E [g(x) + f(x)] \: d \mu \overset{\mathrm{Fatou's \: Lemma}} \leq \liminf_{n \to \infty} \int_E [g(x) + f_n(x)] \: d \mu = \int_E g(x) \: d \mu + \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
  • Therefore $\displaystyle{\int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu}$. $(**)$
  • So from $(*)$ and $(**)$ we have:
(3)
\begin{align} \quad \limsup_{n \to \infty} \int_E f_n(x) \: d \mu \leq \int_E f(x) \: d \mu \leq \liminf_{n \to \infty} \int_E f_n(x) \: d \mu \end{align}
  • Therefore:
(4)
\begin{align} \quad \lim_{n \to \infty} \int_E f_n(x) \: d \mu = \int_E f(x) \: d \mu \quad \blacksquare \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License