LEC CAL2 3

Recall

Let $R\subseteq {\mathbb{R}}^n$ be a rectangle. Let $f:R\to \mathbb{R}$ be bounded. Partitions of $R$ are products of partitions. $U(P,f)$ and $L(P,f)$ are defined analogously to the one variable case. As expected, $f$ is said to be Riemann integrable on $R$ if

$$\underset{P}{\inf }U(P,f)=\underset{P}{\sup }L(P,f)\equiv {\int }_{R}f.$$

Example 1(Example).

Let $f:R\to \mathbb{R}$ be the constant function, $f(x)=c$ for all $x\in R$. Then,

$${\int }_{R}f=c\text{Vol}(R).$$

Example 2(Example).

Let $f:[0,1]\times [0,1]\to R$, be defined by

$$f(x,y)=\{\begin{array}{ll}0& x\in \mathbb{Q}\\ 1& x\notin \mathbb{Q}.\end{array}$$

Clearly, $L(P,f)=0$ and $U(P,f)=1$ for every partition $P$. It follows that $f$ is not integrable on $[0,1]\times [0,1]$.

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Integration on general subsets

Measure zero and content zero

Definition 3(Definition).

Let $A\subseteq {\mathbb{R}}^n$. We say that $A$ has measure zero if for all $ϵ>0$ there is a countable cover $\{U_1,U_2,\dots \}$ of $A$ by closed rectangles $U_i$ such that

$$\begin{array}{r}\sum _{i=1}^{\infty }\text{Vol}(U_i)<ϵ.\end{array}$$

Remarks:

1. If $A$ is finite, then $A$ has measure zero.
2. If $A$ is countable, then $A$ has measure zero (it is easy to construct a countable cover consisting of shrinking rectangles such that the sum of their areas is less than $ϵ$).
3. If $A$ has measure $0$ and $B\subseteq A$, $B$ has measure $0$.
4. $[0,1]$ does not have measure $0$.
5. Open rectangles can be used in place of closed rectangles in the definition of measure zero (we will be using this fact often).
6. A countable union of measure zero sets is measure zero (the proof is exactly what you would expect).

Definition 4(Definition).

A subset $A\subseteq {\mathbb{R}}^n$ has content zero if for all $ϵ>0$, there is a finite cover $\{U_1,\dots ,U_k\}$ of $A$ by closed rectangles such that

$$\sum _{i=1}^k\text{Vol}(U_i)<ϵ.$$

Clearly, $A$ has content zero $⟹$ $A$ has measure zero. The converse is true if $A$ is compact:

Lemma 5(Lemma).

If $A$ is compact and has measure zero, $A$ has content zero.

Proof.

Let$ϵ>0$. Since $A$ has measure $0$, there exists a cover of $A$ by open rectangles such that their cumulative volume is less than $ϵ$. Since $A$ is compact, a finite number $U_1,\dots ,U_n$ of the $U_i$ also cover $A$.□

Oscillations

Definition 6(Definition).

Let $f:A\to \mathbb{R}$ be a bounded function, $A\subseteq {\mathbb{R}}^n$. Let $x\in A$, $\delta >0$.

$$\begin{array}{rl}& M(x,f,\delta )\equiv \sup \{f(a):a\in A,|x-a|<\delta \}\\ & m(x,f,\delta )\equiv \inf \{f(a):a\in A,|x-a|<\delta \}\\ & \\ & o(f,x)\equiv \underset{\delta \to 0}{\lim }(M(x,f,\delta )-m(x,f,\delta )).\end{array}$$

$o(f,x)$ is called the oscillation of $f$ at $x$.

Theorem 7(Spivak (1965) 1-10).

The bounded function $f$ is continuous at $a$ iff $o(f,a)=0$.

[!Proof]-
$(⟹)$

Theorem 8(Spivak (1965) 1-11).

Let $A\subseteq {\mathbb{R}}^n$ be closed. If $f:A\to \mathbb{R}$ is any bounded function, and $ϵ>0$, then $\{x\in A:o(f,x)\ge ϵ\}$ is closed.

Characterizing integrable functions

Lemma 9(Lemma).

Let $R\subseteq {\mathbb{R}}^n$ be a closed rectangle, $f:R\to \mathbb{R}$ bounded function such that $o(f,x)\le ϵ$ for all $x\in R$. Then, there exists a partition $P$ such that $U(P,f)-L(P,f)<ϵ\text{Vol}(R)$.

Proof.

For all$x\in R$, there is a closed rectangle $U_x$ such that $x\in U_x^{\circ }$ and $M_{U_x}(f)-m_{U_x}(f)<ϵ$. Since $R$ is compact, a finite subcollection $\alpha =\{U_{x_1},\dots ,U_{x_r}\}$ covers $R$. Choose a partition $P$ such that every rectangle is inside some $U_{x_i}\in \alpha$. So, if $S\in P$, then $M_S(f)-m_S(f)<ϵ$. Thus, $U(P,f)-L(P,f)<ϵ\text{Vol}(R)$.□

Theorem 10(Theorem).

Let $R\subseteq {\mathbb{R}}^n$ be a closed rectangle. Let $f:R\to \mathbb{R}$ be a bounded function. Then, $f$ is integrable iff the set of discontinuities of $f$ has measure zero.

Proof.

Let$B$ be the set of discontinuities of $f$.

$(⟸)$ Assume $B$ has measure zero. Let $ϵ>0$. We will use the Cauchy criterion.

$$\begin{array}{r}{B}_{ϵ}\equiv \{x\in R|o(f,x)\ge ϵ\}\subseteq B.\end{array}$$

Since $B$ has measure zero, $B_{ϵ}$ has measure zero. Since $B_{ϵ}$ is closed an bounded, $B_{ϵ}$ is compact, so $B_{ϵ}$ has content zero. Thus, there exist closed rectangles $U_1,\dots ,U_r$ such that $B_{ϵ}\subseteq {\bigcup }_{i=1}^r{U}_i^{\circ }$ (working with interiors requires some work) and ${\sum }_{i=1}^r\text{Vol}(U_i)<ϵ$.

Choose a partition $P$ of $R$ such that every subrectangle $S$ of $P$ is one of two types:

1. $S\subseteq U_i$ for some $i$, or
2. $S\cap B_{ϵ}=\varnothing$,
   (Show that this can be done!). Let $P={\mathcal{S}}_1\bigsqcup {\mathcal{S}}_2$, where ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$ represent rectangles of type $1$ and $2$ respectively.

Next, let $|f(x)|\le M$ for all $x\in R$. Then, $M_S(f)-m_S(f)\le 2M$ for all $S\in P$.

$$\begin{array}{r}\sum _{S\in {\mathcal{S}}_1}(M_S(f)-m_S(f))\text{Vol}(S)<2Mϵ.\end{array}$$

Let $S\in {\mathcal{S}}_2$. Then, $o(f,x)<ϵ$ for all $x\in S$. From Lemma 9, there exists a partition of $S$ such that $U-L<ϵ\text{Vol}(S)$. Thus, there exists a refinement $P^{\prime }$ of $P$ such that for all $S\in {\mathcal{S}}_2$, we have

$$\begin{array}{r}\sum _{S^{\prime }\subseteq S\in {\mathcal{S}}_2}(M_{S^{\prime }}(f)-m_{S^{\prime }}(f))\text{Vol}(S^{\prime }).<ϵ\text{Vol}(S)\end{array}$$

Finally,

$$\begin{array}{rl}U(P^{\prime },f)-L(P^{\prime },f)& =\sum _{S^{\prime }\subseteq S\in {\mathcal{S}}_1}\left(\ast \right)+\sum _{S^{\prime }\subseteq S\in {\mathcal{S}}_2}(\ast )\\ & <2Mϵ+\sum _{S\in {\mathcal{S}}_2}ϵ\text{Vol}(S)\\ & \le 2Mϵ+ϵ\text{Vol}(R).\end{array}$$

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$(⟹)$ Let $B_{ϵ}$ be as defined previously. Note that $B=B_1\cup B_2\cup \dots$. We will show that each $B_{\frac{1}{n}}$ has measure zero. Let $ϵ>0$. By the Cauchy criterion, there exists a partition $P$ of $R$ such that $U(f,P)-L(f,P)<ϵ/n$. Let $\mathcal{S}\equiv \left\{S\in P|S\cap B_{\frac{1}{n}}\ne \varnothing \right\}$. $\mathcal{S}$ is a cover of $B_{\frac{1}{n}}$. $S\in \mathcal{S}⟹\exists x\in So(f,x)\ge \frac{1}{n}⟹M_S(f)-m_S(f)\ge \frac{1}{n}$. So,

$$\begin{array}{r}\frac{1}{n}\sum _{S\in \mathcal{S}}\text{Vol}(S)\le \sum _{S\in \mathcal{S}}(M_S(f)-m_S(f))\text{Vol}(S)\le U(P,f)-L(P,f)<\frac{ϵ}{n}.\end{array}$$ □

[!Definition]
$f:{\mathbb{R}}^n\to \mathbb{R}$ is integrable if there is a closed rectangle $R\subseteq {\mathbb{R}}^n$ such that $f(x)=0$ for all $x\notin R$ and $f$ is integrable on $R$.

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References

Spivak, M. (1965). Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus. Addison-Wesley publ.