MATH 409 Lecture 19

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Lecture Slides

Darboux vs. Riemann Sums

Darboux Sums

Riemann Sum

Riemann Sums

Recall upper and lower Darboux sums and upper and lower integrals

A Riemann sum of a function $f:[a,b]\to \mathbb{ℝ}$ with respect to a partition $P=\{x_0,x_1,\dots ,x_n\}$ of $[a,b]$ generated by samples $t_j\in [x_{j-1},x_j]$ is a sum

${𝒮}(f,P,t_j)=\sum _{j=1}^{n}f(t_j)(x_j-x_{j-1})$

Note: The function $f$ need not be bounded. If it is bounded, then $L(f,P)\le {𝒮}(f,P,t_j)\le U(f,P)$ for any choice of samples $t_j$.

The riemann sums ${𝒮}(f,P,t_j)$ converge to a limit $I(f)$ as the norm $\Vert P\Vert \to 0$ if for every $ϵ>0$ there exists $\delta >0$ such that $\Vert P\Vert <\delta$ implies $|{𝒮}(f,P,t_j)-I(f)|<ϵ$ for any partition $P$ and choice of samples $t_j$.

Theorem. The Riemann sums ${𝒮}(f,P,t_j)$ converge to a limit $I(f)$ as $\Vert P\Vert \to 0$ if and only if the function $f$ is integrable on $[a,b]$ and $I(f)=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$.

Proof of (⇒). Assume the Riemann sums ${𝒮}(f,P,t_j)$ converge to a limit $I(f)$ as $\Vert P\Vert \to 0$. Given $ϵ>0$, we choose $\delta >0$ so that for every partition $P$ with $\Vert P\Vert <\delta$, we have $|{𝒮}(f,P,t_j)-I(f)|<ϵ$ for any choice of samples $t_j$. Let ${\tilde{t}}_j$ be a different set of samples for the same partition $P$. Then $|{𝒮}(f,P,{\tilde{t}}_j)-I(f)|<ϵ$. We can choose the samples $t_j,{\tilde{t}}_j$ so that $f(t_j)$ is arbitrarily close to $\sup f([x_{j-1},x_j])$ while $f({\tilde{t}}_j)$ is arbitrarily close to Failed to parse (syntax error): {\displaystyle \inf{f([x_{j-1},x_])}} . That way ${𝒮}(f,P,t_j)$ gets arbitrarily close to $U(f,P)$, while ${𝒮}(f,P,{\tilde{t}}_j)$ gets arbitarily close to $L(f,P)$.

Hence it follows from the above inequalities that $|U(f,P)-I(f)|\le ϵ$ and $|L(f,P)-I(f)|\le ϵ$. As a consequence, the distance between $U(f,P)$ and $L(f,P)$ is $U(f,P)-L(f,P)\le 2ϵ$. In particular, the function $f$ is bounded. We conclude that $f$ is integrable.

Let $I=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$. The number $I$ lies between $L(f,P)$ and $U(f,P)$. The inequalities $U(f,P)-L(f,P)\le 2ϵ$ and $|U(f,P)-I(f)|\le ϵ$ imply that $|I-I(f)|\le 3ϵ$ as $ϵ$ can be arbitrarily small, $I=I(f)$.

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Properties of Integrals

Integration as a Linear Operation

Theorem. If $f,g$ are integrable on an interval $[a,b]$, then the sum $f+g$ is also integrable on $[a,b]$ and

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

Theorem. If $f$ is integrable on an interval $[a,b]$ and $\alpha$ is a real constant, then the scalar product $\alpha f$ is integrable on $[a,b]$ and

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

Proof of both. Let $I(f)$ denote the integral of $f$ and $I(g)$ denote the integral of $g$ over $[a,b]$. The key observation is that the Riemann sums depend linearly on a function. Namely, ${𝒮}(f+g,P,t_j)={𝒮}(f,P,t_j)+{𝒮}(g,P,t_j)$ and ${𝒮}(\alpha f,P,t_j)=\alpha \cdot {𝒮}(f,P,t_j)$ for any partition $P$ of $[a,b]$ and choice of samples $t_j$. It follows that

$\begin{array}{rl}|{𝒮}(f+g,P,t_j)-I(f)-I(g)|& \le |{𝒮}(f,P,t_j)-I(f)|+|{𝒮}(g,P,t_j)-I(g)|\\ |{𝒮}(\alpha f,P,t_j)-\alpha I(f)|& =\left|\alpha \right|\cdot |{𝒮}(f,P,t_j)-I(f)|\end{array}$

As $\Vert P\Vert \to 0$, the Riemann sums ${𝒮}(f,P,t_j)$ and ${𝒮}(g,P,t_j)$ get arbitrarily close to $I(f)$ and $I(g)$, respectively. Then ${𝒮}(f+g,P,t_j)$ will be getting arbitrarily close to $I(f)+I(g)$ while ${𝒮}(\alpha f,P,t_j)$ will be getting arbitrarily close to $\alpha I(f)$. Thus $I(f)+I(g)$ is the integral of $f+g$ and $\alpha I(f)$ is the integral of $\alpha f$ over $[a,b]$.

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Integrability of Subintervals

Theorem. If a function $f$ is integrable on $[a,b]$, then it is integrable on each subinterval $[c,d]\subset [a,b]$.

Proof. Since $f$ is integrable on $[a,b]$, for any $ϵ>0$ there is a partition $P_{ϵ}$ of $[a,b]$ such that $U(f,P_{ϵ})-L(f,P_{ϵ})<ϵ$. Given a subinterval $[c,d]\subset [a,b]$, let $P_{ϵ}\text{'}=P_{ϵ}\cup \{c,d\}$ and $Q_{ϵ}={P_{ϵ}}^{\prime }\cap [c,d]$. Then $P_{ϵ}\text{'}$ is a partition of $[a,b]$ that refines $P_{ϵ}$. Hence

$U(f,{P_{ϵ}}^{\prime })-L(f,{P_{ϵ}}^{\prime })\le U(f,P_{ϵ})-L(f,P_{ϵ})<ϵ$

Since $Q_{ϵ}$ is a partition of $[c,d]$ contained in $P_{ϵ}\text{'}$, it follows that

$U(f,Q_{ϵ})-L(f,Q_{ϵ})\le U(f,{P_{ϵ}}^{\prime })-L(f,{P_{ϵ}}^{\prime })<ϵ$

We conclude that $f$ is integrable on $[c,d]$.

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Integration over Subintervals

Theorem. If a function $f$ is integrable on $[a,b]$, then for any $c\in (a,b)$,

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

Proof. Since $f$ is integrable on $[a,b]$, it is also integrable on the subintervals $[a,c]$ and $[c,b]$. Let $P$ be a partition of $[a,c]$ and $\left\{t_j\right\}$ be some samples for that partition. Further, let $Q$ be a partition of $[c,b]$ and $\left\{{\tau }_i\right\}$ be some samples for that partition. Then $P\cup Q$ is a partition of $[a,b]$ and $\left\{t_j\right\}\cup \left\{{\tau }_i\right\}$ are samples for it. The key observation is that

${𝒮}(f,P\cup Q,\left\{t_j\right\}\cup \left\{{\tau }_i\right\})={𝒮}(f,P,t_j)+{𝒮}(f,Q,{\tau }_i)$

If $\Vert P\Vert \to 0$ and $\Vert Q\Vert \to 0$, then $\Vert P\cup Q\Vert =\max (\Vert P\Vert ,\Vert Q\Vert )\to 0$ as well. Therefore the Riemann sums in the latter equality will converge to the integrals $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$, $\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x$, and $\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x$, respectively.

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Theorem. If a function $f$ is integrable on $[a,b]$ and $f([a,b])\subset [A,B]$, then for each continuous function $g:[A,B]\to \mathbb{ℝ}$, the composition $g\circ f$ is also integrable on $[a,b]$.

(proof omitted)

Corollary If function $f$ and $g$ are integrable on $[a,b]$, then so is $fg$.

Proof. We have $(f+g{)}^2=f^2+g^2+2fg$. Since $f$ and $g$ are integrable on $[a,b]$, so is $f+g$. Since $h(x)=x^2$ is a continuous function on $\mathbb{ℝ}$, the compositions $h\circ f=f^2$, $h\circ g=g^2$, and $h\circ (f+g)=(f+g{)}^2$ are integrable on $[a,b]$. Then $fg=\frac{1}{2}{(f+g)}^2-\frac{1}{2}{f}^2-\frac{1}{2}{g}^2$ is integrable on $[a,b]$.

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Comparison Theorem for Integrals

Theorem. [Comparison theorem]. If functions $f,g$ are integrable on $[a,b]$ and $f(x)\le g(x)$ for all $x\in [a,b]$, then

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

Proof. Since $f\le g$ on the interval $[a,b]$ it follows that ${𝒮}(f,P,t_j)\le {𝒮}(g,P,t_j)$ for any partition $P$ of $[a,b]$ and choice of samples $t_j$. As $\Vert P\Vert \to 0$, the sum ${𝒮}(g,P,t_j)$ gets arbitrarily close to the integral of $g$. The theorem follows.

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Corollary. If $f$ is integrable on $[a,b]$ and $f(x)\ge 0$, for $x\in [a,b]$, then $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\ge 0$.

Proof. Comparison theorem with $g(x)=0$.

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Corollary. If $f$ is integrable on $[a,b]$ and $m\le f(x)\le M$ for $x\in [a,b]$, then

$m(b-a)\le \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le M(b-a)$

Corollary. If $f$ is integrable on $[a,b]$, then the function $\left|f\right|$ is also integrable on $[a,b]$ and

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

Proof. The function $\left|f\right|$ is the composition of $f$ with a continuous function $g(x)=\left|x\right|$. Therefore $\left|f\right|$ is integrable on $[a,b]$. Since $-|f(x)|\le f(x)\le |f(x)|$ for $x\in [a,b]$, the Comparison Theorem for integrals implies that

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

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Integral with Variable Limit

Suppose $f:[a,b]\to \mathbb{ℝ}$ is an integrable function. For any $x\in [a,b]$, let $F(x)=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t$ (we assume $F(a)=0$).

Theorem. The function $F$ is well-defined and continuous on $[a,b]$.

Proof. Since $f$ is integrable on $[a,b]$, it is also integrable on each subinterval of $[a,b]$. Hence the function $F$ is well-defined on $[a,b]$. Besides, $f$ is bounded: $|f(t)|\le M$ for some $M>0$ and all $t\in [a,b]$.

For any $x,y\in [a,b]$, where $x\le y$, we have $\underset{a}{\overset{y}{\int }}f(t)\mathrm{d}t=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t+\underset{x}{\overset{y}{\int }}f(t)\mathrm{d}t$. Equivalently, $F(y)=F(x)+\underset{x}{\overset{y}{\int }}f(t)\mathrm{d}x$. It follows that

$|F(y)-F(x)|=|\underset{x}{\overset{y}{\int }}f(t)\mathrm{d}t|\le \underset{x}{\overset{y}{\int }}|f(t)|\mathrm{d}t\le M|y-x|$

Thus $F$ is a Lipschitz function on $[a,b]$, which implies $F$ is uniformly continuous on $[a,b]$.

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Sets of Measure Zero

A subset $E$ of the real line $\mathbb{ℝ}$ is said to have measure zero if for any $ϵ>0$, the set $E$ can be covered by countably many open intervals $J_1,J_2,\dots$ such that $\sum _{n=1}^{\mathrm{\infty }}\left|J_n\right|<ϵ$.

Examples

Theorem. Any countable set has measure zero.

Proof. Indeed, suppose $E$ is a countable set and let $x_1,x_2,\dots$ be a list of all elements of $E$. Given $ϵ>0$, let

$J_n=(x_n-\frac{ϵ}{2^{n+1}},x_n+\frac{ϵ}{2^{n+1}})$ for $n=1,2,\dots$

Then $E\subset J_1\cup J_2\cup \dots$ and $\left|J_n\right|=\frac{ϵ}{2^n}$ for all $n\in \mathbb{\mathbb{N}}$, so that $\sum _{n=1}^{\mathrm{\infty }}\left|J_n\right|=ϵ$.

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Theorem. A nondegenerate interval $[a,b]$ is not a set of measure zero.

Theorem. There exist sets of measure zero that are of the same cardinality as $\mathbb{ℝ}$.

Lebesgue's Criterion for Riemann Integrability

Suppose $P(x)$ is a property depending on $x\in S$, where $S\subset \mathbb{ℝ}$. We say that $P(x)$ holds for almost all $x\in S$ (or almost everywhere on $S$) if the set $\{x\in S\mid P(x)\text{does not hold}\}$ has measure zero.

Theorem. A function $f:[a,b]\to \mathbb{ℝ}$ is Riemann integrable on the interval $[a,b]$ if and only if $f$ is bounded on $[a,b]$ and continuous almost everywhere on $[a,b]$.