Lebesgue's Dominated Convergence Theorem

# Lebesgue's Dominated Convergence Theorem

On the Levi's Monotone Convergence Theorems page we looked at a bunch of very useful theorems collectively known as Levi's Monotone Convergence Theorems. We now introduce another convergence theorem known as Lebesgue's Dominated Convergence Theorem.

Theorem (Lebesgue's Dominated Convergence): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue integrable functions that converge to a limit function $f$ almost everywhere on $I$. Suppose that there exists a Lebesgue integrable function $g$ such that $\mid f_n \mid \leq g$ almost everywhere on $I$ and for all $n \in \mathbb{N}$. Then $f$ is Lebesgue integrable on $I$ and $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I f(x) \: dx}$. |

*Recall that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I f(x) \: dx}$ can also be rewritten as $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I \lim_{n \to \infty} f_n(x) \: dx}$.*

**Proof:**Suppose that $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions that converge to a limit function $f$ almost everywhere on $I$. Then the following equality holds almost everywhere on $I$:

\begin{align} \quad \limsup_{n \to \infty} f_n(x) = f(x) = \liminf_{n \to \infty} f_n(x) \end{align}

- Define two new sequences of functions $(G_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ whose general terms are defined by:

\begin{align} \quad G_n(x) = \sup_{k \geq n} \{ f_k(x) \} \quad \mathrm{and} \quad g_n(x) = \inf_{k \geq n} \{ f_k(x) \} \end{align}

- Then $(G_n(x))_{n=1}^{\infty}$ is a decreasing sequence of functions that converges to $f(x)$ almost everywhere on $I$. Also, $(g_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions that converges to $f(x)$ almost everywhere on $I$. Both $(G_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ are sequences of Lebesgue integrable functions.

- So, the following inequality holds almost everywhere on $I$:

\begin{align} \quad g_n(x) \leq f(x) \leq G_n(x) \end{align}

- We are given that $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$, for all $n \in \mathbb{N}$, and for some Lebesgue integrable function $g$ on $I$. So, the following chain of inequalities holds almost everywhere on $I$:

\begin{align} \quad -g(x) \leq g_n(x) \leq f_n(x) \leq G_n(x) \leq g(x) \end{align}

- Now consider the sequence of functions $(g - G_n)_{n=1}^{\infty}$. Then $g - G_n$ is Lebesgue integrable on $I$, and moreover, is an increasing sequence almost everywhere on $I$. Furthermore, the sequence $\displaystyle{\left ( \int_I [g(x) - G_n(x)] \: dx \right )_{n=1}^{\infty}}$ converges since:

\begin{align} \quad \int_I [g(x) - G_n(x)] \: dx \leq \int_I 2g(x) \: dx = 2 \int_I g(x) \: dx < \infty \end{align}

- So by Levi's Monotonic Convergence theorems for Lebesgue integrable functions, we see that the function $g(x) - f(x) = \lim_{n \to \infty} [g(x) - G_n(x)]$ is Lebesgue integrable on $I$ and:

\begin{align} \quad \int_I [g(x) - f(x)] = \lim_{n \to \infty} \int_I [g(x) - G_n(x)] \: dx = \int_I g(x) \: dx - \lim_{n \to \infty} \int_I G_n(x) \: dx = \int_I g(x) \: dx - \int_I f(x) \: dx \end{align}

- Therefore $f$ is Lebesgue integrable on $I$, and $\displaystyle{\int_I f(x) \: dx = \lim_{n \to \infty} \int_I G_n(x) \: dx}$. $(*)$

- Now consider the sequence of functions $(g_n(x) + g(x))_{n=1}^{\infty}$. This is an increasing sequence of Lebesgue integrable functions on $I$. Furthermore, the sequence $\left ( \int_I [g_n(x) + g(x)] \: dx \right )_{n=1}^{\infty}$ converges since:

\begin{align} \quad \int_I [g_n(x) + g(x)] \: dx \leq \int_I 2g(x) \: dx = 2 \int_I g(x) \: dx < \infty \end{align}

- So by Levi's Convergence Theorem for Lebesgue integrable functions we have that $\displaystyle{g(x) + f(x) = \lim_{n \to \infty} [g(x) + g_n(x)]}$ is a Lebesgue integrable function and that:

\begin{align} \quad \int_I [g(x) + f(x)] \: dx = \lim_{n \to \infty} \int_I [g(x) + g_n(x)] \: dx = \int_I g(x) \: dx + \lim_{n \to \infty} \int_I g_n(x) \: dx = \int_I g(x) \: dx + \int_I f(x) \: dx \end{align}

- Once again, this shows that $f$ is Lebesgue integrable on $I$ and that $\displaystyle{\int_I f(x) \: dx = \lim_{n \to \infty} \int_I g_n(x) \: dx}$. $(**)$.

- Since $g_n(x) \leq f(x) \leq G_n(x)$ almost everywhere on $I$, then the following chain of inequalities hold almost everywhere on $I$:

\begin{align} \quad \int_I g_n(x) \:dx \leq \int_I f(x) \: dx \leq \int_I G_n(x) \: dx \\ \quad \lim_{n \to \infty} \int_I g_n(x) \: dx \leq \lim_{n \to \infty} \int_I f(x) \: dx \leq \lim_{n \to \infty} \int_I G_n(x) \: dx \\ \quad \int_I f(x) \: dx \leq \lim_{n \to \infty} \int_I f(x) \: dx \leq \int_I f(x) \: dx \end{align}

- Hence $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I f(x) \: dx}$. $\blacksquare$