First, for any
\(a' < b', a \le a' < b' < b\text{,}\) we can apply
Theorem 2.2.7 on
\([a', b']\) to conclude that
\begin{equation*}
\int_{a'}^{b'} f(x)\, d\alpha=\lim_{n\to \infty}\int_{a'}^{b'} f_{n}(x)\, d\alpha.
\end{equation*}
Next, let
\(c\) be chosen according to the uniform integrability of the family
\(\left\{f_{n}\right\}\) in
(2.3.2) for a given
\(\epsilon>0\text{.}\) If we set
\(a'=c\text{,}\) then for any
\(c', c < c' < b\text{,}\) we would get
\begin{equation*}
\left|\int_{c}^{c'} f_{n}(x)\, d\alpha \right| < \epsilon \text{ for all } n.
\end{equation*}
This then implies that
\begin{equation*}
\left|\int_{c}^{c'} f(x)\, d\alpha \right| =\lim_{n\to \infty } \left|\int_{c}^{c'} f_{n}(x)\, d\alpha \right|
\le \epsilon,
\end{equation*}
which implies that the integral \(\int_{a}^{b} f(x)\, d\alpha\) is convergent at \(x=b\text{,}\) and
\begin{equation*}
\left|\int_{c}^{b} f(x)\, d\alpha \right| \le \epsilon.
\end{equation*}
Finally, using
\begin{align*}
\amp \left| \int_{a}^{b} f(x)\, d\alpha - \int_{a}^{b} f_{n}(x)\, d\alpha \right| \\
\le \amp \left|\int_{c}^{b} f(x)\, d\alpha - \int_{c}^{b} f_{n}(x)\, d\alpha \right|
+ \left| \int_{a}^{c} f(x)\, d\alpha - \int_{a}^{c} f_{n}(x)\, d\alpha \right|\\
\le \amp 2\epsilon + \left| \int_{a}^{c} f(x)\, d\alpha - \int_{a}^{c} f_{n}(x)\, d\alpha \right|
\end{align*}
and applying
Theorem 2.2.7 on
\([a, c]\text{,}\) we find some
\(N\) such that
\(\left| \int_{a}^{c} f(x)\, d\alpha - \int_{a}^{c} f_{n}(x)\, d\alpha \right| < \epsilon\) for all
\(n\ge N\text{,}\) which implies
\begin{equation*}
\left| \int_{a}^{b} f(x)\, d\alpha - \int_{a}^{b} f_{n}(x)\, d\alpha \right| < 3\epsilon
\end{equation*}
for all \(n\ge N\text{,}\) therefore proving the claimed conclusion.
