Corollaries to Lebesgue's Dominated Convergence Theorem
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Corollaries to Lebesgue's Dominated Convergence Theorem

Recall from the Lebesgue's Dominated Convergence Theorem page that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions that converges to a limit function $f$ almost everywhere on $I$ and if there exists a Lebesgue integrable function $g$ on $I$ such that $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $n \in \mathbb{N}$ then $f$ is Lebesgue integrable on $I$ and moreover:

(1)
\begin{align} \quad \lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I f(x) \: dx \end{align}

We will now look at some nice corollaries to the Lebesgue's dominated convergence theorem, the first of which is called Lebesgue's bounded convergence theorem.

 Corollary 1 (Lebesgue's Bounded Interval Convergence Theorem): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue integrable functions on $I$ that converges almost everywhere on a BOUNDED INTERVAL $I$ to a limit function $f$. Furthermore, suppose that there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid f_n(x) \mid \leq M$ almost everywhere on $I$ for all $n \in \mathbb{N}$. Then $f$ is Lebesgue integrable on $I$ and $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I \lim_{n \to \infty} f_n(x) \: dx = \int_I f(x) \: dx}$.
• Proof: Suppose that there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid f_n(x) \mid \leq M$] almost everywhere on $I$ and for all $n \in \mathbb{N}$. Let:
(2)
\begin{align} \quad g(x) = M \end{align}
• Then $g$ is a Lebesgue integrable function since $I$ is a bounded interval, and by Lebesgue's dominated convergence theorem we must have that $f$ is Lebesgue integrable on $I$ and that:
(3)
\begin{align} \quad \lim_{n \to \infty} \int_I f_n(x) \: dx = \int_I \lim_{n \to \infty} f_n(x) \: dx = \int_I f(x) \: dx \quad \blacksquare \end{align}
 Corollary 2 (Test for Lebesgue Integrability): Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue integrable functions on $I$ that converges almost everywhere on $I$ to a limit function $f$. Suppose that there exists a Lebesgue integrable function $g$ on $I$ such that $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$. Then $f$ is Lebesgue integrable on $I$.
• Proof: Suppose that there exists a Lebesgue integrable function $g$ on $I$ such that $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$. Define a sequence of functions $(g_n(x))_{n=1}^{\infty}$ where for each $n \in \mathbb{N}$ we have that:
(4)
\begin{align} \quad g_n(x) = \max \{ \min \{ f_n(x), g(x) \}, -g(x) \} \end{align}
• Graphically, $g_n(x)$ is the function obtained from $f_n(x)$ and is contained between $-g$ and $g$. For example, if we had the following graphs for $g$, $-g$ (in red), and $f_n$ (in blue):
• Then the graph for $g_n$ (in yellow) would be:
• Notice that the sequence $(g_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions since the maximum/minimum of Lebesgue integrable functions is Lebesgue integrable.
• Also notice that since $(f_n(x))_{n=1}^{\infty}$ converges to $f$ and $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$ that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$. So for all $n \geq N$ we would have that $\mid g_n(x) \mid \leq g(x)$ almost everywhere on $I$, and moreover, $(g_n(x))_{n=1}^{\infty}$ converges almost everywhere on $I$ to $f$ since for $n \geq N$ we have that $-g(x) \leq g_n(x) \leq g(x)$ almost everywhere on $I$ which means that for $n \geq N$, $g_n(x) = f_n(x)$ almost everywhere on $I$.
• So, $(g_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions that converges to a limit function $f$ almost everywhere on $I$ and such that $\mid g_n(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $n \geq N$. By Lebesgue's dominating convergence theorem we must have that the limit function, $f$, is Lebesgue integrable on $I$. $\blacksquare$

Moreover, we also have that $\displaystyle{\lim_{n \to \infty} \int_I g_n(x) \: dx = \int_I \lim_{n \to \infty} g_n(x) \: dx = \int_I f(x) \: dx}$.

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