MATH 409 Lecture 18

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Lecture Slides

Challenges

Challenge 14

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

Challenge 15

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

The Riemann Integral

Partitions of an Interval

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

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

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

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

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

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

Darboux Sums

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

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

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

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

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

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

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

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

Properties

$L(f,P)\leq U(f,P)$

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

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

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

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

Analogous to previous proof

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

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

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

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

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

Upper and Lower Integrals

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

The upper integral of $f$ on $[a,b]$, denoted

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

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

Similarly, the lower integral of $f$ on $[a,b]$, denoted

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

Integratability

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

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

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

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

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

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

Also, there exists a partition $Q$ of $[a,b]$ such that

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

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

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

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

quod erat demonstrandum

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

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

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

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

quod erat demonstrandum

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

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

quod erat demonstrandum

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

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

quod erat demonstrandum

Continuity and Integratability

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

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

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

quod erat demonstrandum