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 $P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}$ of $[a,b]$ generated by samples $t_{j}\in \left[x_{j-1},x_{j}\right]$ is a sum

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

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

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

Theorem. The Riemann sums ${\mathcal {S}}(f,P,t_{j})$ converge to a limit $I(f)$ as $\left\|P\right\|\to 0$ if and only if the function $f$ is integrable on $[a,b]$ and $I(f)=\int _{a}^{b}f(x)\,\mathrm {d} x$.

Proof of (⇒). Assume the Riemann sums ${\mathcal {S}}(f,P,t_{j})$ converge to a limit $I(f)$ as $\left\|P\right\|\to 0$. Given $\epsilon >0$, we choose $\delta >0$ so that for every partition $P$ with $\left\|P\right\|<\delta$, we have $\left|{\mathcal {S}}(f,P,t_{j})-I(f)\right|<\epsilon$ for any choice of samples $t_{j}$. Let ${\tilde {t}}_{j}$ be a different set of samples for the same partition $P$. Then $\left|{\mathcal {S}}(f,P,{\tilde {t}}_{j})-I(f)\right|<\epsilon$. 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 ${\mathcal {S}}(f,P,t_{j})$ gets arbitrarily close to $U(f,P)$, while ${\mathcal {S}}(f,P,{\tilde {t}}_{j})$ gets arbitarily close to $L(f,P)$.

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

Let $I=\int _{a}^{b}f(x)\,\mathrm {d} x$. The number $I$ lies between $L(f,P)$ and $U(f,P)$. The inequalities $U(f,P)-L(f,P)\leq 2\epsilon$ and $\left|U(f,P)-I(f)\right|\leq \epsilon$ imply that $\left|I-I(f)\right|\leq 3\epsilon$ as $\epsilon$ 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

$\int _{a}^{b}\left(f(x)+g(x)\right)\,\mathrm {d} x=\int _{a}^{b}f(x)\,\mathrm {d} x+\int _{a}^{b}g(x)\,\mathrm {d} x$

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

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

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, ${\mathcal {S}}(f+g,P,t_{j})={\mathcal {S}}(f,P,t_{j})+{\mathcal {S}}(g,P,t_{j})$ and ${\mathcal {S}}(\alpha \,f,P,t_{j})=\alpha \cdot {\mathcal {S}}(f,P,t_{j})$ for any partition $P$ of $[a,b]$ and choice of samples $t_{j}$. It follows that

${\begin{aligned}\left|{\mathcal {S}}(f+g,P,t_{j})-I(f)-I(g)\right|&\leq \left|{\mathcal {S}}(f,P,t_{j})-I(f)\right|+\left|{\mathcal {S}}(g,P,t_{j})-I(g)\right|\\\left|{\mathcal {S}}(\alpha \,f,P,t_{j})-\alpha \,I(f)\right|&=\left|\alpha \right|\cdot \left|{\mathcal {S}}(f,P,t_{j})-I(f)\right|\end{aligned}}$

As $\left\|P\right\|\to 0$, the Riemann sums ${\mathcal {S}}(f,P,t_{j})$ and ${\mathcal {S}}(g,P,t_{j})$ get arbitrarily close to $I(f)$ and $I(g)$, respectively. Then ${\mathcal {S}}(f+g,P,t_{j})$ will be getting arbitrarily close to $I(f)+I(g)$ while ${\mathcal {S}}(\alpha \,f,P,t_{j})$ 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 $\left[a,b\right]$.

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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 $\epsilon >0$ there is a partition $P_{\epsilon }$ of $[a,b]$ such that $U(f,P_{\epsilon })-L(f,P_{\epsilon })<\epsilon$. Given a subinterval $[c,d]\subset [a,b]$, let ${P_{\epsilon }}'=P_{\epsilon }\cup \left\{c,d\right\}$ and $Q_{\epsilon }={P_{\epsilon }}'\cap [c,d]$. Then ${P_{\epsilon }}'$ is a partition of $[a,b]$ that refines $P_{\epsilon }$. Hence

$U(f,{P_{\epsilon }}')-L(f,{P_{\epsilon }}')\leq U(f,P_{\epsilon })-L(f,P_{\epsilon })<\epsilon$

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

$U(f,Q_{\epsilon })-L(f,Q_{\epsilon })\leq U(f,{P_{\epsilon }}')-L(f,{P_{\epsilon }}')<\epsilon$

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 $c\in (a,b)$,

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

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

${\mathcal {S}}(f,P\cup Q,\left\{t_{j}\right\}\cup \left\{\tau _{i}\right\})={\mathcal {S}}(f,P,t_{j})+{\mathcal {S}}(f,Q,\tau _{i})$

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

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Theorem. If a function $f$ is integrable on $[a,b]$ and $f([a,b])\subset \left[A,B\right]$, 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 $f\,g$.

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 $fg={\frac {1}{2}}\,\left(f+g\right)^{2}-{\frac {1}{2}}\,f^{2}-{\frac {1}{2}}g^{2}$ 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 $f(x)\leq g(x)$ for all $x\in [a,b]$, then

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

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

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

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

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

$m(b-a)\leq \int _{a}^{b}f(x)\,\mathrm {d} x\leq M(b-a)$

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

$\left|\int _{a}^{b}f(x)\,\mathrm {d} x\right|\leq \int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x$

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 $-\left|f(x)\right|\leq f(x)\leq \left|f(x)\right|$ for $x\in [a,b]$, the Comparison Theorem for integrals implies that

$-\int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x\leq \int _{a}^{b}f(x)\,\mathrm {d} x\leq \int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x$

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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 $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$ (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: $\left|f(t)\right|\leq M$ for some $M>0$ and all $t\in [a,b]$.

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

$\left|F(y)-F(x)\right|=\left|\int _{x}^{y}f(t)\,\mathrm {d} t\right|\leq \int _{x}^{y}\left|f(t)\right|\,\mathrm {d} t\leq M\left|y-x\right|$

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 $\epsilon >0$, the set $E$ can be covered by countably many open intervals $J_{1},J_{2},\ldots$ such that $\sum _{n=1}^{\infty }\left|J_{n}\right|<\epsilon$.

Examples

Theorem. Any countable set has measure zero.

Proof. Indeed, suppose $E$ is a countable set and let $x_{1},x_{2},\ldots$ be a list of all elements of $E$. Given $\epsilon >0$, let

$J_{n}=\left(x_{n}-{\frac {\epsilon }{2^{n+1}}},x_{n}+{\frac {\epsilon }{2^{n+1}}}\right)$ for $n=1,2,\ldots$

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

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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 $\left\{x\in S~\mid ~P(x)\ {\mbox{does not hold}}\right\}$ 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 $[a,b]$.