The Uni. Bound. Con. Theorem for Point. Con. Seqs. of Functs.

# The Uniform Bounded Convergence Theorem for Pointwise Convergent Sequences of Functions

 Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Suppose that: 1) $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded. 2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$. Then $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$.

We say that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded on $E$ if there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ and for all $x \in E$ we have that $|f_n(x)| \leq M$.

• Proof: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of uniformly bounded Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$.
• We have already noted that the limit function $f(x)$ is bounded and Lebesgue measurable. So we break this proof up into two cases.
• Case 1: If $m(E) = 0$ then once again for each $n \in \mathbb{N}$ we have that $\displaystyle{\int_E f_n = 0}$ and $\displaystyle{\int_E f = 0}$ and the conclusion to the theorem holds.
• Case 2: Let $m(E) > 0$. Let $\epsilon > 0$ be given. Since $(f_n(x))_{n=1}^{\infty}$ is uniformly bounded on $E$ we have that there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ and for all $x \in E$:
(1)
\begin{align} \quad | f_n(x) | \leq M \end{align}
• Furthermore, since $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$ we also have that $|f(x)| \leq M$ as well. So for all $n \in \mathbb{N}$ and for all $x \in E$:
(2)
\begin{align} \quad |f_n(x) - f(x)| \leq |f_n(x)| + |f(x)| \leq M + M = 2M \end{align}
• Now, by Egoroff's Theorem, since $E$ is Lebesegue measurable, $m(E) < \infty$, and $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$, for $\displaystyle{\epsilon_0 = \frac{\epsilon}{4M} > 0}$ there exists a closed set $F \subseteq E$ such that:
(3)
\begin{align} \quad m(E \setminus F) < \epsilon_0 = \frac{\epsilon}{4M} \quad \mathrm{and} \quad f_n \to f \: \mathrm{uniformly \: on} \: F \end{align}
• Since $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $F$, for $\displaystyle{\epsilon_1 = \frac{\epsilon}{2 m(E)} > 0}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that then for all $x \in F$:
(4)
\begin{align} \quad | f_n(x) - f(x) | < \epsilon_1 = \frac{\epsilon}{2m(E)} \quad (**) \end{align}
• So from the additivity and monotonicity properties of the Lebesgue integral for bounded Lebesgue measurable functions we have that for all $n \geq N$:
(5)
\begin{align} \quad \biggr \lvert \int_E f_n - \int_E f \biggr \rvert = \biggr \lvert \int_E (f_n - f) \biggr \rvert \leq \int_E |f_n - f| & \leq \int_F |f_n - f| + \int_{E \setminus F} |f_n - f| \\ & \leq \int_F \epsilon_1 + \int_{E \setminus F} 2M \\ & \leq \int_F \frac{\epsilon}{2m(E)} + \int_{E \setminus F} 2M \\ & \leq \frac{\epsilon}{2m(E)} m(E) + 2M \cdot m(E \setminus F) \\ & < \frac{\epsilon}{2} + 2M \frac{\epsilon}{4M} \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}
• Therefore $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$. $\blacksquare$