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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.

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    $\begingroup$ Given any partition, try to bound $U(P,f)-L(P,f)$. Then work with what assumptions you are given. $\endgroup$
    – Andrew
    Jan 21, 2023 at 12:10
  • $\begingroup$ So, I am interested to show that, given an arbitrary $\epsilon > 0$, there exists a partition $P$, such that $\sum_{k=1}^{n} (M_k - m_k)\Delta x_k$ can be made arbitrarily small. Here, $M_k=\sup \{f(x):x \in [x_k,x_{k-1}]\}$ and $m_k=\inf \{f(x):x \in [x_k,x_{k-1}]\}$ where $[x_k,x_{k-1}]$ is the $k$th sub-interval of $P$. $\endgroup$
    – Quasar
    Jan 21, 2023 at 12:18
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    $\begingroup$ Yes, you can try to use the “triple triangle inequality” $\endgroup$
    – Andrew
    Jan 21, 2023 at 12:24
  • $\begingroup$ @AndrewZhang, I tried to use the hints you provided, to bound $U(f,P) - L(f,P)$. Does the proof check out? $\endgroup$
    – Quasar
    Jan 21, 2023 at 13:15
  • $\begingroup$ @SineoftheTime, if each $f_n$ is continuous and $f_n \to f$ uniformly, then $f$ is continuous. However, we make no assumption about the continuity of $f_n$'s. $\endgroup$
    – Quasar
    Jan 21, 2023 at 13:41

1 Answer 1

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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.

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