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

# The Pointwise Bounded Convergence Theorem for Uniformly 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 a sequence of bounded functions.2) $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ uniformly on $E$.Then $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$. |

*In the theorem above, since $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $E$ we have that $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x)}$. So the conclusion of theorem above is equivalence to saying that we can interchange the limit and integral symbol, that is, $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E \lim_{n \to \infty} f_n}$.*

**Proof:**Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of bounded Lebesgue measurable functions defined on a Lebesgue measurable set $E$ with $m(E) < \infty$.

- We first acknowledge that since the sequence of functions $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ uniformly on $E$ and each functions $f_n$ is bounded on $E$, we have that the limit function $f$ is also bounded $E$. Furthermore, $f$ is also measurable on $E$.

- We break the proof of this theorem into two cases.

**Case 1:**If $m(E) = 0$ then $\displaystyle{\int_E f_n = 0}$ for all $n \in \mathbb{N}$ and furthermore, $\displaystyle{\int_E f = 0}$ and clearly the conclusion of the theorem holds.

**Case 2:**Let $m(E) > 0$. Let $\epsilon > 0$ be given. Since $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $E$, for $\displaystyle{\epsilon_0 = \frac{\epsilon}{m(E)} > 0}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then for all $x \in E$ we have that:

\begin{align} \quad | f_n(x) - f(x) | < \epsilon_0 = \frac{\epsilon}{m(E)} \end{align}

- By the additivity and monotonicity properties of the Lebesgue integral for bounded Lebesgue measurable functions we have that:

\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| < \int_E \epsilon_0 = \epsilon_0 m(E) = \frac{\epsilon}{m(E)} m(E) = \epsilon \end{align}

- Therefore $\displaystyle{\lim_{n \to \infty} \int_E f_n = \int_E f}$. $\blacksquare$