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, $0\le f(x)\le 1$ for all $x\in \mathbb{ℝ}$, $f(x)=1$ if $\left|x\right|\le 1$ and $f(x)=0$ if $\left|x\right|\ge 2$.

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 $ϵ>0$ such that $g$ coincides with a polynomial on the interval $(c-ϵ,c+ϵ)$. 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 $x_0,x_1,\dots ,x_n$ 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 $[x_0,x_1],[x_1,x_2],\dots ,[x_{n-1},x_n]$.

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

$\Vert P\Vert =\underset{1\le j\le n}{\max }(x_j-x_{j-1})$

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 $\Vert Q\Vert \le \Vert P\Vert$.

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 $P=\{x_0,x_1,\dots ,x_n\}$ 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

$U(f,P)=\sum _{j=1}^n{M}_j(f){\mathrm{\Delta }}_j$

where ${\mathrm{\Delta }}_j=x_j-x_{j-1}$ and $M_j(f)=\sup f([x_{j-1}])$ for $j=1,2,\dots ,n$.

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

$L(f,P)=\sum _{j=1}^n{m}_j(f){\mathrm{\Delta }}_j$

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

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

Properties

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

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

$U(f,p)\le \sup f([a,b])\cdot (b-a)$

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

$\inf f([a,b])\cdot (b-a)\le L(f,P)$

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 $P_0=\{a,b\}$ is the trivial partition.

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

Every subinterval $J$ created by partition $P$ is the union of one or more subintervals $J_1,J_2,\dots ,J_k$ created by $Q$. Since $\sup f(J_i)\le \sup f(J)$ for $1\le i\le k$ ($J$ is a larger set, so its supremum can only increase compared with $J_i$), it follows that $\sum _{i=1}^k\sup f(J_i)\cdot \left|J_i\right|\le \sup f(J)\cdot \sum _{i=1}^k\left|J_i\right|=\sup f(J)\cdot \left|J\right|$. Summing up this inequality over all subintervals $J$, we obtain $U(f,Q)\le U(f,P)$. The inequality $L(f,P)\le L(f,Q)$ is proved similarly, considering supremum and flipping the inequalities.

$L(f,P)\le U(f,Q)$ 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 $L(f,P)\le L(f,P\cup Q)$ and $U(f,P\cup Q)\le U(f,Q)$. Besides, $L(f,P\cup Q)\le U(f,P\cup Q)$. 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

${\bar{\int }}_a^{b}f(x)\mathrm{d}x$ or $(U)\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$

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

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

${\underset{\_}{\int }}_a^{b}f(x)\mathrm{d}x$ or $(L)\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$

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

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$

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

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

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

$U(f,P)<\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x+\frac{ϵ}{2}$

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

$L(f,Q)>\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x-\frac{ϵ}{2}$

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

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 $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=c(b-a)$.

Proof. Indeed, for the trivial partition $P_0=\{a,b\}$, 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 $ϵ<0$.

quod erat demonstrandum

Step Function. $f(x)=\{\begin{array}{ll}1& x>0\\ 0& x\le 0\end{array}$ is integrable on $[-1,1]$ and $\underset{-1}{\overset{1}{\int }}f(x)\mathrm{d}x=1$.

Proof. For any $ϵ$, consider $P_{ϵ}=\{-1,-ϵ,ϵ,1\}$. Then

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

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

quod erat demonstrandum

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

Proof. For any $0<\delta \le 1$, the interval $[a,b]$ contains only finitely many points $y_1,y_2,\dots ,y_k$ such that $f(y_i)\ge \delta$. 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 $U(f,P_{\delta })\le 2\delta +\delta (b-a)$. 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 $ϵ>0$, there exists $\delta >0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\frac{ϵ}{b-a}$ for all $x,y\in [a,b]$. Obviously, there exists a partition $P=\{x_0,x_1,\dots ,x_n\}$ of $[a,b]$ that satisfies $\Vert P\Vert <\delta$.

Let $J=[x_{j-1},x_j]$ 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 $\Vert P\Vert <\delta$ the length of $J$ satisfies $\left|J\right|<\delta$. Then $|x_{+}-x_{-}|\le \left|J\right|<\delta$ so that $|f(x_{+})-f(x_{-})|<\frac{ϵ}{b-a}$. It follows that $\sup f(J)\cdot \left|J\right|-\inf f(J)\cdot \left|J\right|<\frac{ϵ\left|J\right|}{b-a}$. Summing up the latter inequality over all subintervals $J$, we obtain that $U(f,P)-L(f,P)<ϵ$.

quod erat demonstrandum