I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I am interested to prove that if each $f_n$ is integrable and $f_n \to f$ uniformly on $[a,b]$, then the limiting function $f$ is also integrable. Do you have a clue/hint on how to proceed without giving away the entire solution/proof?
[Abbott 7.2.5] Assume that for each $n$, $f_n$ is an integrable function on $[a,b]$. If $(f_n) \to f$ uniformly on $[a,b]$, prove that $f$ is also integrable on this set.
Proof.
For starters, I wrote the definitions of both uniform convergence and integrability.
We are given that $\displaystyle ( f_{n})\rightarrow f$ uniformly. By definition of uniform convergence, we have:
\begin{equation*} ( \forall \epsilon >0)( \exists N\in \mathbf{N})( \forall x\in [ a,b])( \forall n\geq N)( |f_{n}( x) -f( x) |< \epsilon ) \end{equation*}
Since each $\displaystyle f_{n}$ is integrable, we can write:
\begin{equation*} ( \forall \epsilon >0)( \forall n\in \mathbf{N})( \exists P_{\epsilon } \in \mathcal{P})( U( f_{n} ,P_{\epsilon }) -L( f_{n} ,P_{\epsilon }) < \epsilon ) \end{equation*}
Edit. I have added my proof attempt as an answer to the question.
Proof.
Pick an arbitrary $\displaystyle \epsilon >0$.
Let $\displaystyle P$ be any arbitrary partition. We can write:
\begin{equation*} \begin{array}{ c l } 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) \end{array} \end{equation*}
Since each $\displaystyle f_{n}$ is integrable, we can write:
\begin{equation*} ( \exists P_{\epsilon } \in \mathcal{P})\left( U( f_{n} ,P_{\epsilon }) -L( f_{n} ,P_{\epsilon }) < \frac{\epsilon }{3}\right) \end{equation*}
We choose $\displaystyle P=P_{\epsilon }$. Assume that $\displaystyle P_{\epsilon } =\{x_{0} =a,x_{1} ,\dotsc ,x_{m} =b\}$.
Define $\displaystyle z_{k} \in [ x_{k-1} ,x_{k}]$ such that $\displaystyle f( z_{k}) =\sup \{f( x) :x\in [ x_{k-1} ,x_{k}]\}$. We have:
\begin{equation*} \sup \{f_{n}( x) :x\in [ x_{k-1} ,x_{k}]\} \geq f_{n}( z_{k}) \end{equation*}
Thus, we can write the expression $\displaystyle U( f,P_{\epsilon }) -U( f_{n} ,P_{\epsilon })$ as : \begin{equation*} \begin{array}{ c l } U( f,P) -U( f_{n} ,P) & \leq \sum ( f( z_{k}) -f_{n}( z_{k})) \Delta x_{k}\\ & \leq \sum |f( z_{k}) -f_{n}( z_{k}) |\Delta x_{k} \end{array} \end{equation*}
Define $\displaystyle y_{k} \in [ x_{k-1} ,x_{k}]$ such that $\displaystyle f( y_{k}) =\inf\{f( x) :x\in [ x_{k-1} ,x_{k}]\}$. We have:
\begin{equation*} \inf\{f_{n}( x) :x\in [ x_{k-1} ,x_{k}]\} \leq f_{n}( y_{k}) \end{equation*}
Thus, we can re-write the expression $\displaystyle L( f_{n} ,P_{\epsilon }) -L( f,P_{\epsilon })$ as:
\begin{equation*} \begin{array}{ c l } L( f_{n} ,P_{\epsilon }) -L( f,P_{\epsilon }) & \leq \sum ( f_{n}( y_{k}) -f( y_{k})) \Delta x_{k}\\ & \leq \sum |f_{n}( y_{k}) -f( y_{k}) \Delta x_{k} \end{array} \end{equation*}
Since $\displaystyle ( f_{n})\rightarrow f$ uniformly, there exists $\displaystyle N\in \mathbf{N}$, such that for all $\displaystyle n\geq N$, and for all $\displaystyle x\in [ a,b]$, we have:
\begin{equation*} |f_{n}( x) -f( x) |< \frac{\epsilon }{3m( b-a)} \end{equation*}
Pick $\displaystyle n\geq N$. Then, we have:
\begin{equation*} \begin{array}{ c l } 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) \end{array} \end{equation*}
\begin{equation*} \begin{array}{ c l } U( f,P_{\epsilon }) -L( f,P_{\epsilon }) & =U( f,P) -U( f_{n} ,P) +U( f_{n} ,P) -L( f_{n} ,P) +L( f_{n} ,P) -L( f,P)\\ & \leq \sum _{k=1}^{m} |f_{n}( z_{k}) -f( z_{k}) |\Delta x_{k} +\frac{\epsilon }{3} +\sum _{k=1}^{m} |f_{n}( y_{k}) -f( y_{k}) |\Delta x_{k}\\ & \leq \sum _{k=1}^{m} |f_{n}( z_{k}) -f( z_{k}) |( b-a) +\frac{\epsilon }{3} +\sum _{k=1}^{m} |f_{n}( y_{k}) -f( y_{k}) |( b-a)\\ & < \sum _{k=1}^{m}\frac{\epsilon }{3m( b-a)} \cdotp ( b-a) +\frac{\epsilon }{3} +\sum _{k=1}^{m}\frac{\epsilon }{3m( b-a)} \cdotp ( b-a)\\ & =\frac{\epsilon }{3} +\frac{\epsilon }{3} +\frac{\epsilon }{3} =\epsilon \end{array} \end{equation*}
This closes the proof.