MATH 409 Lecture 19

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Lecture Slides

Darboux vs. Riemann Sums

 Darboux Sums Riemann Sum

Riemann Sums

A Riemann sum of a function ${\displaystyle f:[a,b]\to \mathbb {R} }$ with respect to a partition ${\displaystyle P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}}$ of ${\displaystyle [a,b]}$ generated by samples ${\displaystyle t_{j}\in \left[x_{j-1},x_{j}\right]}$ is a sum

${\displaystyle {\mathcal {S}}(f,P,t_{j})=\sum _{j=1}^{n}f(t_{j})\,(x_{j}-x_{j-1})}$
Note: The function ${\displaystyle f}$ need not be bounded. If it is bounded, then ${\displaystyle L(f,P)\leq {\mathcal {S}}(f,P,t_{j})\leq U(f,P)}$ for any choice of samples ${\displaystyle t_{j}}$.

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

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

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

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

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

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Properties of Integrals

Integration as a Linear Operation

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

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

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

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

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

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

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Integrability of Subintervals

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

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

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

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

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

We conclude that ${\displaystyle f}$ is integrable on ${\displaystyle [c,d]}$.

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Integration over Subintervals

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

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

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

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

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

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

(proof omitted)

Corollary If function ${\displaystyle f}$ and ${\displaystyle g}$ are integrable on ${\displaystyle [a,b]}$, then so is ${\displaystyle f\,g}$.

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

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Comparison Theorem for Integrals

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

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

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

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

Proof. Comparison theorem with ${\displaystyle g(x)=0}$.

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

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

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

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

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

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

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

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

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

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

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

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

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Sets of Measure Zero

A subset ${\displaystyle E}$ of the real line ${\displaystyle \mathbb {R} }$ is said to have measure zero if for any ${\displaystyle \epsilon >0}$, the set ${\displaystyle E}$ can be covered by countably many open intervals ${\displaystyle J_{1},J_{2},\ldots }$ such that ${\displaystyle \sum _{n=1}^{\infty }\left|J_{n}\right|<\epsilon }$.

Examples

Theorem. Any countable set has measure zero.

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

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

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

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

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

Lebesgue's Criterion for Riemann Integrability

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

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