Riemann Integrability of the Limit Function of a Uni. Con. Seq. of Functs.

# Riemann Integrability of the Limit Function of a Uniformly Convergent Sequence of Functions

Recall from the Uniform Convergence of Sequences of Functions page that a sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be uniformly convergent to the limit function $f(x)$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that for all $x \in X$ that:

(1)\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}

Now consider a sequence of real-valued Riemann integrable functions $(f_n(x))_{n=1}^{\infty}$ with common domain $[a, b]$ that uniformly converges to the limit function $f(x)$. As we will see in the following theorem, the limit function $f(x)$ will be Riemann integrable on $[a, b]$ and the Riemann integral of $f$ on $[a, b]$ will equal to the limit of the Riemann integrals of $f_n$ on $[a, b]$.

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions with common domain $[a, b]$ and suppose that $(f_n(x))_{n=1}^{\infty}$ uniformly converges to a limit function $f(x)$. Then $\displaystyle{\int_a^b f(x) \: dx = \int_a^b \lim_{n \to \infty} f_n(x) \: dx = \lim_{n \to \infty} \int_a^b f_n(x) \: dx}$. |

**Proof:**Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of real-valued functions with common domain $[a, b]$. We want to show that the numerical sequence $\displaystyle{\left ( \int_a^b f_n(x) \: dx \right )_{n=1}^{\infty}}$ converges to $\displaystyle{\int_a^b f(x) \: dx}$, i.e., show that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad \biggr \lvert \int_a^b f_n(x) \: dx - \int_a^b f(x) \: dx \biggr \rvert < \epsilon \end{align}

- Since $(f_n(x))_{n=1}^{\infty}$ is uniformly converge to $f(x)$ we have that for $\epsilon_1 = \frac{\epsilon}{b - a} > 0$ there exists an $N^* \in \mathbb{N}$ such that if $n \geq N^*$ then for all $x \in [a, b]$ we have that:

\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon_1 = \frac{\epsilon}{b - a} \quad (*) \end{align}

- Let $N = N^*$. Then for all $n \geq N$ we have that $(*)$ holds and so:

\begin{align} \quad \biggr \lvert \int_a^b f_n(x) \: dx - \int_a^b f(x) \: dx \biggr \rvert = \biggr \lvert \int_a^b [f_n(x) - f(x)] \: dx \biggr \rvert < \biggr \lvert \int_a^b \epsilon_1 \: dx = (b - a) \epsilon_1 = (b - a) \cdot \frac{\epsilon}{b - a} = \epsilon \end{align}

- Therefore the sequence $\displaystyle{\left ( \int_a^b f_n(x) \: dx \right )}$ converges to $\displaystyle{\int_a^b f(x) \: dx}$, i.e:

\begin{align} \quad \int_a^b f(x) \: dx = \int_a^b \lim_{n \to \infty} f_n(x) \: dx = \lim_{n \to \infty} \int_a^b f_n(x) \: dx \quad \blacksquare \end{align}