I want a check of this proof because I'm not completely sure about the manipulation in some inequalities.

Let $f_n$ integrable in $[a,b]$ for all $n$. Show that if $(f_n)\to f$ uniformly in $[a,b]$ then $f$ is integrable in $[a,b]$

Using Darboux upper and lower sums if $f_n$ is integrable in $[a,b]$ this mean

$$\forall \varepsilon>0,\exists P\in\mathcal P: \sum_{j=1}^{H} (M_j-m_j)\Delta x_j<\varepsilon\tag{1}$$

where $M_j$ and $m_j$ are the supremum and infimum of $f_n(x)$ in $[x_j,x_{j+1}]$, and where $\mathcal P$ is the set of all partitions of $[a,b]$. And we have that if $f(x)\to f$ uniformly in $[a,b]$ then

$$\forall\varepsilon>0,\exists N\in\Bbb N,\forall x\in[a,b]:|f_n(x)-f(x)|<\varepsilon,\,\forall n\ge N\tag{2}$$

Then due to $(2)$ there exists $N$ such that $|f_n(x)-f(x)|<\frac{\varepsilon}{3(b-a)},\forall n\ge N$ and $\forall x\in[a,b]$. And due to $(1)$ for $\frac{\varepsilon}{3}$ and $f_n$ exists some $P_{\varepsilon,n}\in\mathcal P$ such that

$$\sum_{j=1}^{H} (M_{n,j}-m_{n,j})\Delta x_j <\frac{\varepsilon}{3}$$

If $\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j$ is the difference between upper and lower sum of $f$ using the partition $P_{\varepsilon,n}$ then

$$\sum_{j=1}^{H} (M_{n,j}-m_{n,j})\Delta x_j - \sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j+\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3}$$

$$\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3}-\sum_{j=1}^{H} (M_{n,j}-m_{n,j})\Delta x_j + \sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j$$

$$\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3} + \sum_{j=1}^{H} ((M_{j}-M_{n,j})+(m_{n,j}-m_{j}))\Delta x_j$$

$$\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3} + \sum_{j=1}^{H} (|M_{j}-M_{n,j}|+|m_{n,j}-m_{j}|)\Delta x_j$$

$$\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3} + \sum_{j=1}^{H} \frac{2\varepsilon}{3(b-a)}\Delta x_j=\frac{\varepsilon}{3} + \frac{2\varepsilon}{3(b-a)}\sum_{j=1}^{H}\Delta x_j$$

$$\sum_{j=1}^{H} (M_{j}-m_{j})\Delta x_j<\frac{\varepsilon}{3} + \frac{2\varepsilon}{3(b-a)}(b-a)=\varepsilon$$

Then for $f$ exists partitions where the difference between upper and lower sum is arbitrarily close to zero, so $f$ is integrable in $[a,b]$.


For a function $g:[a,b]\to\mathbb R$ and a partition $\mathcal P=\{x_0,\ldots, x_k\}\subset [a,b]$ set \begin{align} M_i(g) &:= \sup\{g(x):x_i\leqslant g\leqslant x_{i+1}\},\quad i=0,1,\ldots,k-1\\ m_i(g) &:= \inf\{g(x):x_i\leqslant g\leqslant x_{i+1}\},\quad i=0,1,\ldots,k-1.\\ \end{align} Since $f_n\to f$ uniformly, we have $$a_n := \sup\{|f_n(x)-f(x)|:x\in[a,b]\}\stackrel{n\to\infty}\longrightarrow0.$$ Let $\varepsilon>0$ and choose $N$ so that $a_N<\frac\varepsilon{3(b-a)}$. Choose $k$ so that $$\max\{M_i(f_N)-m_i(f_N):0\leqslant i\leqslant k-1\}<\frac\varepsilon{3(b-a)},$$ where $$\mathcal P_k=\bigcup_{i=0}^{k-1} \left\{a + \frac{i(b-a)}{k-1} \right\}. $$ Then \begin{align} |M_i(f)-M_i(f_N)|&\leqslant a_N<\frac\varepsilon{3(b-a)},\quad i=0,1,\ldots,k-1\\ |m_i(f)-m_i(f_N)|&\leqslant a_N<\frac\varepsilon{3(b-a)},\quad i=0,1,\ldots,k-1,\\ \end{align} and hence \begin{align} |U_f(\mathcal P_k)-L_f(\mathcal P_k)| &= \left|\sum_{i=0}^{k-1} (b-a)(k-1)^{-1}M_i(f)-\sum_{i=0}^{k-1} (b-a)(k-1)^{-1}m_i(f) \right|\\ &\leqslant (b-a)k^{-1}\sum_{i=0}^{k-1} |M_i(f)-m_i(f)|\\ &\leqslant (b-a)k^{-1}\sum_{i=0}^{k-1}\left(|M_i(f)-M_i(f_N)|+|M_i(f_N)-m_i(f_N)|+|m_i(f_N)-m_i(f)|\right)\\ &< (b-a)k^{-1}\sum_{i=0}^{k-1}\left(\frac\varepsilon{3(b-a)} + \frac\varepsilon{3(b-a)} + \frac\varepsilon{3(b-a)}\right)\\ &=(b-a)\cdot\frac\varepsilon{b-a}\\ &=\varepsilon, \end{align} so that $f$ is integrable on $[a,b]$.

  • $\begingroup$ @Masacroso My intention was to write a shorter proof than yours. Looks like I was not successful ;) $\endgroup$ – Math1000 Jun 19 '16 at 5:41
  • 1
    $\begingroup$ I appreciate it a lot anyway because I can see a different way to do it. $\endgroup$ – Masacroso Jun 19 '16 at 5:48

Some easy facts:

  1. If $f:E \to \mathbb R$ and $c$ is a constant, then $\sup_E (f+c) = (\sup_E f) + c.$ Same for $\inf_E (f+c).$

  2. If $c$ is a constant, and $f:[a,b]\to \mathbb R$ is bounded, then $U(f+c,P) = U(f,P) + c(b-a)$ for any partition $P$ of $[a,b].$ Same for $L(f+c,P).$

  3. If $f \le g$ on $[a,b],$ then $U(f,P)\le U(g,P)$ for any partition $P$ of $[a,b].$ Similarly, if $f \ge g$ on $[a,b],$ then $L(f,P)\ge L(g,P).$

Now suppose $f_n \to f $ uniformly on $[a,b]$ and that each $f_n$ is Riemann integrable on $[a,b].$ Let $\epsilon > 0.$ Then there exists $N$ such that $|f-f_N|<\epsilon$ on $[a,b].$ This implies $f< f_N + \epsilon$ and $f> f_N - \epsilon$ on $[a,b].$ Therefore, using the facts above,

$$U(f,P) \le U(f_N+\epsilon,P) = U(f_N,P) + \epsilon(b-a)$$


$$L(f,P) \ge L(f_N-\epsilon,P) = L(f_N,P) - \epsilon(b-a)$$

for any partition $P.$ Now $f_N$ is Riemann integrable, so we can choose a partition $P$ such that $U(f_N,P) - L(f_N,P)< \epsilon.$ For this $P$ we then have $U(f,P) - L(f,P) < \epsilon + 2\epsilon(b-a).$ This proves $f$ is Riemann integrable on $[a,b].$


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