# MATH 409 Lecture 18

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Lecture Slides

## Challenges

### Challenge 14

Find a function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ such that ${\displaystyle f}$ is infinitiely differentiable, ${\displaystyle 0\leq f(x)\leq 1}$ for all ${\displaystyle x\in \mathbb {R} }$, ${\displaystyle f(x)=1}$ if ${\displaystyle \left|x\right|\leq 1}$ and ${\displaystyle f(x)=0}$ if ${\displaystyle \left|x\right|\geq 2}$.

### Challenge 15

Suppose that a function ${\displaystyle g:\mathbb {R} \to \mathbb {R} }$ is locally a polynomial, which means that for every ${\displaystyle c\in \mathbb {R} }$ there exists ${\displaystyle \epsilon >0}$ such that ${\displaystyle g}$ coincides with a polynomial on the interval ${\displaystyle \left(c-\epsilon ,c+\epsilon \right)}$. Prove that ${\displaystyle g}$ is a polynomial.

## The Riemann Integral

### Partitions of an Interval

A partition of a closed bounded interval ${\displaystyle [a,b]}$ is a finite subset ${\displaystyle P\subset [a,b]}$ that includes the endpoints ${\displaystyle a}$ and ${\displaystyle b}$.

Let ${\displaystyle x_{0},x_{1},\ldots ,x_{n}}$ be the list of elements of ${\displaystyle P}$ arranged in ascending order so that ${\displaystyle x_{0}. By definition, ${\displaystyle x_{0}=a}$ and ${\displaystyle x_{n}=b}$. These points split the interval ${\displaystyle [a,b]}$ into finitely many subintervals ${\displaystyle [x_{0},x_{1}],[x_{1},x_{2}],\ldots ,[x_{n-1},x_{n}]}$.

The norm of the partition ${\displaystyle P}$, denoted ${\displaystyle \|P\|}$ is the maximum of lengths of those subintervals, hence

${\displaystyle \|P\|=\max _{1\leq j\leq n}(x_{j}-x_{j-1})}$

Given two partitions ${\displaystyle P}$ and ${\displaystyle Q}$ of the same interval, we say that ${\displaystyle Q}$ is a refinement of ${\displaystyle P}$ (or that ${\displaystyle Q}$ is finer than ${\displaystyle P}$) if ${\displaystyle P\subset Q}$. Observe that ${\displaystyle P\subset Q}$ implies ${\displaystyle \|Q\|\leq \|P\|}$.

For any two partitions ${\displaystyle P}$ and ${\displaystyle Q}$ of the interval ${\displaystyle [a,b]}$, the union ${\displaystyle P\cup Q}$ is also a partition that refines both ${\displaystyle P}$ and ${\displaystyle Q}$.

### Darboux Sums

Let ${\displaystyle P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}}$ be a partition of an interval ${\displaystyle [a,b]}$, where ${\displaystyle x_{0}=a. Let ${\displaystyle f:[a,b]\to \mathbb {R} }$ be a bounded function.

The upper Darboux sum (or the upper Riemann sum) of the function ${\displaystyle f}$ over the partition ${\displaystyle P}$ is the number

${\displaystyle U(f,P)=\sum _{j=1}^{n}M_{j}(f)\,\Delta _{j}\,\!}$

where ${\displaystyle \Delta _{j}=x_{j}-x_{j-1}}$ and ${\displaystyle M_{j}(f)=\sup {f([x_{j-1}])}}$ for ${\displaystyle j=1,2,\ldots ,n}$.

Likewise, the lower Darboux sum (or the lower Riemann sum) of ${\displaystyle f}$ over ${\displaystyle P}$ is the number

${\displaystyle L(f,P)=\sum _{j=1}^{n}m_{j}(f)\,\Delta _{j}\,\!}$

Where ${\displaystyle m_{j}(f)=\inf {f([x_{j-1}])}}$ for ${\displaystyle j=1,2,\ldots ,n}$.

Note: These sums were originally introduced by Darboux, not Riemann

#### Properties

${\displaystyle L(f,P)\leq U(f,P)}$

indeed ${\displaystyle \inf {f(J)}\leq \sup {f(J)}}$ for any subinterval ${\displaystyle J\in [a,b]}$.

${\displaystyle U(f,p)\leq \sup {f([a,b])}\cdot (b-a)}$

We have ${\displaystyle \sup {f(J)}\leq \sup {f([a,b])}}$ for any subinterval ${\displaystyle J\subset [a,b]}$. Then ${\displaystyle \sup {f(J)}\cdot \left|J\right|\leq \sup {f([a,b])}\cdot \left|J\right|}$, where ${\displaystyle \left|J\right|}$ is the length of ${\displaystyle J}$. Summing up over all subintervals ${\displaystyle J}$ created by the partition ${\displaystyle P}$, we obtain ${\displaystyle U(f,P)\leq \sup {f([a,b])}\cdot (b-a)}$.

${\displaystyle \inf {f([a,b])}\cdot (b-a)\leq L(f,P)}$

Analogous to previous proof

Remark. Observe that ${\displaystyle \sup {f([a,b])}\cdot (b-a)=U(f,P_{0})}$ and ${\displaystyle \inf {f([a,b])}\cdot (b-a)=L(f,P_{0})}$, where ${\displaystyle P_{0}=\left\{a,b\right\}}$ is the trivial partition.

${\displaystyle L(f,P)\leq L(f,Q)\leq U(f,Q)\leq U(f,P)}$ for any partition ${\displaystyle Q}$ that refines ${\displaystyle P}$.

Every subinterval ${\displaystyle J}$ created by partition ${\displaystyle P}$ is the union of one or more subintervals ${\displaystyle J_{1},J_{2},\ldots ,J_{k}}$ created by ${\displaystyle Q}$. Since ${\displaystyle \sup {f(J_{i})}\leq \sup {f(J)}}$ for ${\displaystyle 1\leq i\leq k}$ (${\displaystyle J}$ is a larger set, so its supremum can only increase compared with ${\displaystyle J_{i}}$), it follows that ${\displaystyle \sum _{i=1}^{k}\sup {f(J_{i})}\cdot \left|J_{i}\right|\leq \sup {f(J)}\cdot \sum _{i=1}^{k}\left|J_{i}\right|=\sup {f(J)}\cdot \left|J\right|}$. Summing up this inequality over all subintervals ${\displaystyle J}$, we obtain ${\displaystyle U(f,Q)\leq U(f,P)}$. The inequality ${\displaystyle L(f,P)\leq L(f,Q)}$ is proved similarly, considering supremum and flipping the inequalities.

${\displaystyle L(f,P)\leq U(f,Q)}$ for any partitions ${\displaystyle P}$ and ${\displaystyle Q}$ of the same interval ${\displaystyle [a,b]}$.

Since ${\displaystyle P\cup Q}$ refines both ${\displaystyle P}$ and ${\displaystyle Q}$, it follows from above that ${\displaystyle L(f,P)\leq L(f,P\cup Q)}$ and ${\displaystyle U(f,P\cup Q)\leq U(f,Q)}$. Besides, ${\displaystyle L(f,P\cup Q)\leq U(f,P\cup Q)}$. Thus the property follows by transitivity.

### Upper and Lower Integrals

Suppose ${\displaystyle f:[a,b]\to \mathbb {R} }$ is a bounded function.

The upper integral of ${\displaystyle f}$ on ${\displaystyle [a,b]}$, denoted

${\displaystyle {\overline {\int }}_{a}^{b}f(x)\,\mathrm {d} x}$ or ${\displaystyle (U)\int _{a}^{b}f(x)\,\mathrm {d} x}$

is the number ${\displaystyle \inf {\left\{U(f,P)~\mid ~P{\mbox{ is a partition of }}[a,b]\right\}}}$.

Similarly, the lower integral of ${\displaystyle f}$ on ${\displaystyle [a,b]}$, denoted

${\displaystyle {\underline {\int }}_{a}^{b}f(x)\,\mathrm {d} x}$ or ${\displaystyle (L)\int _{a}^{b}f(x)\,\mathrm {d} x}$

### Integratability

A bounded function ${\displaystyle f:[a,b]\to \mathbb {R} }$ is called integrable (or Riemann integrable) on the interval ${\displaystyle [a,b]}$ if the upper and lower integrals of ${\displaystyle f}$ on ${\displaystyle [a,b]}$ coincide. The common value is called the integral of ${\displaystyle f}$ on ${\displaystyle [a,b]}$ (or over ${\displaystyle [a,b]}$) and is denoted

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x\,\!}$

Theorem. A bounded function ${\displaystyle f:[a,b]\to \mathbb {R} }$ is integrable on ${\displaystyle [a,b]}$ if and only if for every ${\displaystyle \epsilon >0}$, there is a partition ${\displaystyle P_{\epsilon }}$ of ${\displaystyle [a,b]}$ such that ${\displaystyle U(f,P_{\epsilon })-L(f,P_{\epsilon })<\epsilon }$

Proof. (⇒) ${\displaystyle 0\leq (U)\int _{a}^{b}f(x)\,\mathrm {d} x-(L)\int _{a}^{b}f(x)\,\mathrm {d} x\leq U(f,P)-L(f,P)}$ for any partition ${\displaystyle P}$.

(⇐) Conversely, assume ${\displaystyle f}$ is integrable on ${\displaystyle [a,b]}$ Given ${\displaystyle \epsilon >0}$, there exists a partition ${\displaystyle P}$ of ${\displaystyle [a,b]}$ such that

${\displaystyle U(f,P)<\int _{a}^{b}f(x)\,\mathrm {d} x+{\frac {\epsilon }{2}}}$

Also, there exists a partition ${\displaystyle Q}$ of ${\displaystyle [a,b]}$ such that

${\displaystyle L(f,Q)>\int _{a}^{b}f(x)\,\mathrm {d} x-{\frac {\epsilon }{2}}}$

Then ${\displaystyle U(f,P)-L(f,Q)<\epsilon }$. Now ${\displaystyle P\cup Q}$ is a partition of ${\displaystyle [a,b]}$ that refines both ${\displaystyle P}$ and ${\displaystyle Q}$. It follows that ${\displaystyle U(f,P\cup Q)\leq U(f,P)}$ and ${\displaystyle L(f,P\cup Q)\geq L(f,Q)}$. Hence ${\displaystyle U(f,P\cup Q)-L(f,P\cup Q)\leq U(f,P)-L(f,Q)<\epsilon }$.

quod erat demonstrandum
Note: Standard trick to form inequality with same partitions from an inequality with different partitions, take union and apply the refinement property

### Examples

Constant Function. ${\displaystyle f(x)=c}$ is integrable on any interval ${\displaystyle [a,b]}$ and ${\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=c(b-a)}$.

Proof. Indeed, for the trivial partition ${\displaystyle P_{0}=\left\{a,b\right\}}$, we obtain ${\displaystyle U(f,P_{0})=c(b-a)}$ and ${\displaystyle L(f,P_{0})=c(b-a)}$. Thus the lower and upper darboux sums are equivalent, and their difference is equivalent to 0, which is smaller than any ${\displaystyle \epsilon <0}$.

quod erat demonstrandum

Step Function. ${\displaystyle f(x)={\begin{cases}1&x>0\\0&x\leq 0\end{cases}}}$ is integrable on ${\displaystyle [-1,1]}$ and ${\displaystyle \int _{-1}^{1}f(x)\,\mathrm {d} x=1}$.

Proof. For any ${\displaystyle \epsilon }$, consider ${\displaystyle P_{\epsilon }=\left\{-1,-\epsilon ,\epsilon ,1\right\}}$. Then

{\displaystyle {\begin{aligned}U(f,P_{\epsilon })&=0+2\epsilon +(1-\epsilon )=1+\epsilon \\L(f,P_{\epsilon })&=0+0+(1-\epsilon )=1-\epsilon \end{aligned}}}

The difference between the upper and lower sums is ${\displaystyle 2\epsilon }$.

quod erat demonstrandum

Dirichlet Function. ${\displaystyle f(x)={\begin{cases}1&x\in \mathbb {Q} \\0&x\in \mathbb {R} \setminus \mathbb {Q} \end{cases}}}$ is not integrable on any interval ${\displaystyle [a,b]}$.

Proof. Indeed, any subinterval ${\displaystyle [a,b]}$ contains both rational and irrational points. Therefore ${\displaystyle U(f,P)=b-a}$ and ${\displaystyle L(f,P)=0}$ for all partitions ${\displaystyle [a,b]}$.

quod erat demonstrandum

Riemann Function. ${\displaystyle f(x)={\begin{cases}{\frac {1}{q}}&x={\frac {p}{q}}\in \mathbb {Q} \\0&x\in \mathbb {R} \setminus \mathbb {Q} \end{cases}}}$ is integrable on any interval ${\displaystyle [a,b]}$.

Proof. For any ${\displaystyle 0<\delta \leq 1}$, the interval ${\displaystyle [a,b]}$ contains only finitely many points ${\displaystyle y_{1},y_{2},\ldots ,y_{k}}$ such that ${\displaystyle f(y_{i})\geq \delta }$. Let ${\displaystyle P_{\delta }}$ be a partition of ${\displaystyle [a,b]}$ that includes points ${\displaystyle y_{i}\pm {\frac {\delta }{k}}}$. Then ${\displaystyle L(f,P_{\delta })=0}$ and ${\displaystyle U(f,P_{\delta })\leq 2\delta +\delta (b-a)}$. Thus we isolate ${\displaystyle k}$ points with intervals, each of which has a supremum of ${\displaystyle 1}$ and so the sum of all these contributions is ${\displaystyle k\cdot {\frac {2\delta }{k}}=2\delta }$.

quod erat demonstrandum

### Continuity and Integratability

Theorem. If a function ${\displaystyle f:[a,b]\to \mathbb {R} }$ is continuous on the interval ${\displaystyle [a,b]}$, then it is integrable on ${\displaystyle [a,b]}$.

Proof. Since the function ${\displaystyle f}$ is continuous, it is bounded on ${\displaystyle [a,b]}$. Furthermore, ${\displaystyle f}$ is uniformly continuous on ${\displaystyle [a,b]}$. Therefore, for every ${\displaystyle \epsilon >0}$, there exists ${\displaystyle \delta >0}$ such that ${\displaystyle \left|x-y\right|<\delta }$ implies ${\displaystyle \left|f(x)-f(y)\right|<{\frac {\epsilon }{b-a}}}$ for all ${\displaystyle x,y\in [a,b]}$. Obviously, there exists a partition ${\displaystyle P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}}$ of ${\displaystyle [a,b]}$ that satisfies ${\displaystyle \left\|P\right\|<\delta }$.

Let ${\displaystyle J=\left[x_{j-1},x_{j}\right]}$ be an arbitrary subinterval of ${\displaystyle [a,b]}$ created by ${\displaystyle P}$. By the extreme value theorem, there are ponits ${\displaystyle x_{-},x_{+}\in J}$ such that ${\displaystyle f(x_{+})=\sup {f(J)}}$ and ${\displaystyle f(x_{-})=\inf {f(J)}}$. Since ${\displaystyle \left\|P\right\|<\delta }$ the length of ${\displaystyle J}$ satisfies ${\displaystyle \left|J\right|<\delta }$. Then ${\displaystyle \left|x_{+}-x_{-}\right|\leq \left|J\right|<\delta }$ so that ${\displaystyle \left|f(x_{+})-f(x_{-})\right|<{\frac {\epsilon }{b-a}}}$. It follows that ${\displaystyle \sup {f(J)}\cdot \left|J\right|-\inf {f(J)}\cdot \left|J\right|<{\frac {\epsilon \,\left|J\right|}{b-a}}}$. Summing up the latter inequality over all subintervals ${\displaystyle J}$, we obtain that ${\displaystyle U(f,P)-L(f,P)<\epsilon }$.

quod erat demonstrandum