# Math Help - Today's integration proof #1

1. ## Today's integration proof #1

1. Let $f:[a,b]\to\mathbb R$ be a monotone function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$
2. Let $f:[a,b]\to\mathbb R$ be a continuous function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$

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Please, be adviced, don't post URL where the proof is, if you have a proof, post it, otherwise, don't post.

2. Three things

1. I am only currently finishing up Riemann-Stieltjes integration...I am still a novice. I would not even be posting if this were not for "fun"

2.I assume that a knowledge of the defintion of the Riemann integral is knownw

3. I assume you mean integrable in respect to x

Originally Posted by Krizalid
1. Let $f:[a,b]\to\mathbb R$ be a monotone function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$
Let me prove a stronger result.

Suppose that $\left|\sum_i \left\{M_i-m_i\right\}\right|\leqslant M$ then $f\in\mathfrak{R}$

Proof: We must merely prove that given $\varepsilon>0$ we may find a partion $P$ of $[a,b]$ such that $U\left(P,f\right)-L\left(P,f\right)<\varepsilon$.

Now for each partition $P=\left\{a=x_0,x_1,\cdots,x_n=b\right\}$ of $[a,b]$ choose $\Delta x_i=\frac{b-a}{n}$ (note this can be done because x is continuous).

So

\begin{aligned}U-L&=\sum_{i=1}^{n}\left[M_i-m_i\right]\Delta x_i\\
&=\frac{b-a}{n}\sum_{i=1}^{n}\left[M_i-m_i\right]\\

The last step is true for sufficiently large n.

Now I will show that if $f$ is monotone then $\sum\left[M_i-m_i\right]$ is bounded:

Without loss of generality assume that $f$ is montonically increasing. Then we can see that

$M_i=\sup_{[x_{i-1},x_i]}f=f(x_i)$ and

$m_i=\inf_{[x_{i-1},x_i]}f=f\left(x_{i-1}\right)$

So

\begin{aligned}\sum_{i=1}^{n}\left[M_i-m_i\right]&=\sum_{i=1}^{n}\left[f(x_i)-f\left(x_{i-1}\right)\right]\\
&=\left(f(x_1)-f(x_0)\right)+\left(f(x_2)-f(x_1)\right)+\cdots+\left(f(x_n)-f(x_n-1)\right)\\
&=f(x_n)-f(x_0)\\
&=f(b)-f(a)\end{aligned}

And because Riemann integration is only applicable to bounded functions we can see that if $f$ is montonically increasing that $\sum_{i}\left[M_i-m_i\right]$ is bounded. The proof for montonically decreasing functions is almost identical.

Let $f:[a,b]\to\mathbb R$ be a continuous function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$
Similarly we must just show that for any $\varepsilon>0$ there exists a corresponding partition $P$ of $[a,b]$ such that $U\left(P,f\right)-L\left(P,f\right)<\varepsilon$.

So for every $\varepsilon>0$ choose $\eta>0$ such that $\eta[b-a]<\varepsilon$. Now since continuity and uniform continuity are equivalent on compact spaces we may choose a $\delta>0$ such that $|x-y|<\delta\implies|f(x)-f(y)|<\eta {\color{blue}(*)}$. Now choose a partition $P$ of $[a,b]$ such that $\max\left\{\Delta x_i\right\}<\delta$, so then

\begin{aligned}U-L&=\sum_{i=1}^{n}\left[M_i-m_i\right]\Delta x_i\\
&\leqslant \eta\sum_{i=1}^{n}\Delta x_i{\color{red}(*)}\\

$\color{red}(*)$ is justified by $\color{blue}(*)$

3. Originally Posted by Krizalid
1. Let $f:[a,b]\to\mathbb R$ be a monotone function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$
2. Let $f:[a,b]\to\mathbb R$ be a continuous function on the compact interval $[a,b].$ Prove that $f$ is integrable on $[a,b].$
These are well-known and nice results from analysis.

1)Without lose of generality say $f$ is non-decreasing. Notice that $f(a) \leq f(x) \leq f(b)$ for $x\in [a,b]$ therefore $|f| \leq M$ for some $M>0$ hence the function is bounded. Now for a partition $P = \{ x_0,x_1,...,x_n\}$ with $x_j < x_{j+1}$ where $0\leq j\leq n-1$ and $x_0=a,x_n=b$ define $\Delta_j = x_j - x_{j-1}$ and $M_j = \sup \{ f(x) : x\in [x_{j-1},x_j]\}$ and $m_j = \inf \{ f(x) : x\in [x_{j-1},x_j] \}$. Let $\Delta = \max \{ \Delta_j : 1\leq j \leq n\}$. Let $U_P$ be the upper sum and $L_P$ be the lower sum. This means $U_P - L_P = \sum_{j=1}^n (M_j - m_j) \Delta_j$. Now $M_j = f(x_j)$ and $m_j = f(x_{j-1})$. Therefore, $U_P - L_P = \sum_{j=1}^n (f(x_j) - f(x_{j-1}) ) \Delta_j$. Therefore, $U_P - L_P \leq \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \Delta = \Delta (f(x_n) - f(x_0)) = \Delta(f(b)-f(a))$. This means for $\epsilon > 0$ pick $\Delta < \frac{\epsilon}{f(b)-f(a)}$ (assuming that $f(b)\not = f(a)$, but if $f(b) = f(a)$ then the function is constant!) Therefore, $U_P - L_P < \epsilon$ which forces $f$ be to integrable by Cauchy's criterion.

2)Here we use the fact that $f$ is uniformly continous. Thus, for any $\epsilon > 0$ there is $\delta > 0$ so that if $|x-y| < \delta \text{ and }x,y\in [a,b]\implies |f(x)-f(y)|<\epsilon$. Let $P$ be a partition $\Delta < \delta$. Now $M_j = \max \{ f(x) : x\in [x_{j-1},x_j] \}$ and $m_j = \min \{ f(x) : x\in [x_{j-1},x_j]\}$ by extreme-value theorem. Consequently, $M_j - m_j < \epsilon$ since $\Delta_j \leq \Delta < \delta$. Thus, $U_P - L_P = \sum_{j=1}^n (M_j - m_j) \Delta_j \leq \sum_{j=1}^n \epsilon \Delta_j = \epsilon (b-a)$. And since this is true for any $\epsilon > 0$ it means we can make $U_P - L_P$ arbitrary small and it is integrable by Cauchy's criterion.