Exercise 7.2.5 Assume that for each $n, f_{n}$ is an integrable function on $[a,b]$. If $\left(f_{n}\right) \rightarrow f$ uniformly on $[a,b]$ then $f$ is also integrable on $[a,b]$.
let $\epsilon > 0$. We want to use the fact that
\begin{equation*}
\begin{split}
U(f,P) - L(f,P) &= U(f,P) - U(f_{N}, P) + U(f_{N}, P) - L(f_{N}, P) + L(f_{N}, P) - L(f, P) \\
&\leq |U(f,P) - U(f_{N},P)| + \left(U(f_{N},P) - L(f_{N},P)\right) + |L(f,P) - L(f_{N},P)|. \\
\end{split}
\end{equation*}
Since $\left(f_{n}\right) \rightarrow f$ uniformly we can certainly pick $N \in\mathbb{N} \ni$ $$ \left|f_{N}(x) - f(x)\right| \leq \frac{\epsilon}{3(b - a)}.$$ Since $f_{N}$ is integrable $\exists$ $P$, a partition for which $$U(f_{N}, P) - L(f_{N}, P) < \frac{\epsilon}{3}.$$ Consider $[x_{k - 1}, x_{k}]$ for $P$. Let $$M_{k} = \sup\{f(x) : x \in [x_{k - 1}, x_{k}]\} \textrm{ and } \Xi_{k} =\sup\{f_{N}(x) : x \in [x_{k - 1}, x_{k}]\}.$$ Then our choice of $f_{N}$ guarantees that $$\left|M_{k} - \Xi_{k}\right| < \frac{\epsilon}{3(b - a)}.$$ Hence $$\left|U(f, P) - U(f_{N}, P)\right| = \left|\sum_{k = 1}^{n}\left(M_{k} - \Xi_{k}\right)(x_{k} - x_{k - 1}) \right| \leq \sum_{k = 1}^{n}\left(\frac{\epsilon}{3(b -a)}\right)(b - a) = \frac{\epsilon}{3}.$$ Now let $$m_{k} = \inf\{f(x): x\in[x_{k - 1}, x_{k}]\} \textrm{ and } \xi_{k} = \inf\{f_{N}(x): x\in[x_{k - 1}, x_{k}]\}. $$ Then our choice of $f_{N}$ ensures that $$\left|m_{k} - \xi_{k}\right| < \frac{\epsilon}{3(b-a)}.$$ So, $$\left|L(f,P) - L(f_{N}, P)\right| = \left|\sum_{k = 1}^{n}\left(m_{k} - \xi_{k}\right)(x_{k} - x_{k -1})\right| \leq \sum_{k = 1}^{n}\left(\frac{\epsilon}{3(b -a)}\right)(b - a) = \frac{\epsilon}{3}.$$ Therefore
\begin{equation*}
\begin{split}
U(f,P) - L(f,P) &= U(f,P) - U(f_{N}, P) + U(f_{N}, P) - L(f_{N}, P) + L(f_{N}, P) - L(f, P) \\
&\leq |U(f,P) - U(f_{N},P)| + \left(U(f_{N},P) - L(f_{N},P)\right) + |L(f,P) - L(f_{N},P)|. \\
&< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}\\
&= \epsilon \\
\end{split}
\end{equation*}
So by Theorem 7.2.8 $f$ is integrable on $[a,b].$
