Exercise 1.1.7 (Regions under the graphs are Jordan measurable)

We will often denote the graph of the function by

and the area under the graph of the function by

(i)

We have to demonstrate that

$$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}A\subset G\hfill \\ A\text{ is elementary}\hfill \end{array}\right\}=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m(B):\begin{array}{c}G\subset B\hfill \\ B\text{ is elementary}\hfill \end{array}\right\}$$

First, note that the set of inner covers of $G$ must be empty (since $G$ is a line). Assume, for the sake of contradiction, that we could find a box inside $G$, i.e., $B=[a_1,b_1)\times \dots \times [a_d,b_d)\times [a_{d+1},b_{d+1})$. By definition it must satisfy $f\left([a_1,b_1)\times \dots \times [a_d,b_d)\right)=y\in ℝ$ which contradicts the fact that the whole $[a_{d+1},b_{d+1})$ is included (one could also consider the intervals containing a single point, but for this we would need an infinite number of “point”-boxes to cover the image). Thus, the Jordan lower measure of $G$ is zero.

Next, consider the Jordan outer measure. To show that it is also zero, we demonstrate that it is less that any $𝜖>0$. Thus, fix $𝜖>0$. Since $f$ is a continuous function (w.r.t. $L^p$-metric) on a compact set, it must be furthermore absolutely continuous, i.e., for the $𝜖$ of our choice

$$\exists <!--\; FUNCTION\; APPLICATION\; -->{\delta }_{𝜖}\forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in B:\parallel x-y\parallel <{\delta }_{𝜖}⟹|f(x)-f(y)|<𝜖∕2$$

Following the usual procedure, we partition $B$ into boxes of “small enough” measure. Set the granularity $n\in \mathbb{N}$ such that in each dimension $\forall <!--\; FUNCTION\; APPLICATION\; -->i=1,\dots ,d$ the mesh size is small ${\mathrm{\Delta }}_i:=\frac{b_i-a_i}{n}<\frac{𝜖}{2{\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d(b_i-a_i)}$ and similarly for ${\mathrm{\Delta }}_{d+1}$ in the interval $f(B)$.

$$\begin{array}{cc}B\hfill & =\bigcup _{k\in \{(k_1,\dots ,k_d)|k_i\in \mathbb{Z}:0\le k_i<n\}}\left[a_1+k_1{\mathrm{\Delta }}_1,a_1+k_1{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_1\right)\times \dots \times \left[a_d+k_d{\mathrm{\Delta }}_d,a_d+k_d{\mathrm{\Delta }}_d+{\mathrm{\Delta }}_d\right)\hfill \\ \\ \hfill & =\bigcup _{k\in {\mathbb{Z}}^d:0\le k_i<n}\left(\prod _{i=1}^d\left[a_i+k_i{\mathrm{\Delta }}_i,a_i+k_i{\mathrm{\Delta }}_i+{\mathrm{\Delta }}_i\right)\right)\hfill \end{array}$$

We can cover $G$ by covering the box $B$ with itself plus somehow covering the line produced by $f$:

$$\overline{G}=\bigcup _{k\in {\mathbb{Z}}^d:0\le k_i<n}\left(\prod _{i=1}^d\left[a_i+k_i{\mathrm{\Delta }}_i,a_i+k_i{\mathrm{\Delta }}_i+{\mathrm{\Delta }}_i\right)\right)\times \left[f(a+k)-{\mathrm{\Delta }}_{d+1},f(a+k)+{\mathrm{\Delta }}_{d+1}\right)$$

(this is true, since $\parallel x-y\parallel =\sqrt[p]{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{(x_i-y_i)}^p}\le \sqrt[p]{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{\mathrm{\Delta }}_i^p}=\sqrt[p]{d}{\mathrm{\Delta }}_i\le {\delta }_{𝜖}⟹|f(x)-f(y)|\le 𝜖$. By additivity, the elementary measure of this cover is

$$\begin{array}{cc}m(\overline{G})\hfill & \stackrel{\mathit{ad}}{\text{=}}\sum _{k\in {\mathbb{Z}}^d:0\le k_i<n}m\left(\left(\prod _{i=1}^d\left[a_i+k_i{\mathrm{\Delta }}_i,a_i+k_i{\mathrm{\Delta }}_i+{\mathrm{\Delta }}_i\right)\right)\times \left[f(a+k)-{\mathrm{\Delta }}_{d+1},f(a+k)+{\mathrm{\Delta }}_{d+1}\right)\right)\hfill \\ \\ \hfill & \stackrel{1.1.4}{\text{=}}\sum _{k\in {\mathbb{Z}}^d:0\le k_i<n}{\mathrm{\Delta }}_1\cdot \dots \cdot {\mathrm{\Delta }}_d\cdot 2{\mathrm{\Delta }}_{d+1}\hfill \\ \\ \hfill & =n^d\cdot {\mathrm{\Delta }}_1\cdot \dots \cdot {\mathrm{\Delta }}_d\cdot 2{\mathrm{\Delta }}_{d+1}=n{\mathrm{\Delta }}_1\cdot \dots \cdot n{\mathrm{\Delta }}_d\cdot 2{\mathrm{\Delta }}_{d+1}\hfill \\ \\ \hfill & =\prod _{i=1}^d(b_i-a_i)\cdot 2{\mathrm{\Delta }}_i\le 𝜖\hfill \end{array}$$

(ii)

To show that the area under the graph is Jordan measurable, we use the criterion (2) from Exercise 1.1.5. Fix an $𝜖>0$. Using the same logic as before, define

$$\overline{A}=\bigcup _{k\in {\mathbb{Z}}^d:0\le k_i<n}\left[a_1+k_1{\mathrm{\Delta }}_1,a_1+k_1{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_1\right)\times \dots \times \left[a_d+k_d{\mathrm{\Delta }}_d,a_d+k_d{\mathrm{\Delta }}_d+{\mathrm{\Delta }}_d\right)\times \left[0,f(a+k)+{\mathrm{\Delta }}_{d+1}\right)$$

$$\underline{A}=\bigcup _{k\in {\mathbb{Z}}^d:0\le k_i<n}\left[a_1+k_1{\mathrm{\Delta }}_1,a_1+k_1{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_1\right)\times \dots \times \left[a_d+k_d{\mathrm{\Delta }}_d,a_d+k_d{\mathrm{\Delta }}_d+{\mathrm{\Delta }}_d\right)\times \left[0,f(a+k)-{\mathrm{\Delta }}_{d+1}\right)$$

We then have

$$\begin{array}{cc}m(\overline{U})-m(\underline{U})\hfill & =\sum _{k\in {\mathbb{Z}}^d:0\le k_i<n}{\mathrm{\Delta }}_1\cdot \dots \cdot {\mathrm{\Delta }}_d\cdot (f(a+k)+{\mathrm{\Delta }}_i)-\sum _{k\in {\mathbb{Z}}^d:0\le k_i<n}{\mathrm{\Delta }}_1\cdot \dots \cdot {\mathrm{\Delta }}_d\cdot (f(a+k)-{\mathrm{\Delta }}_i)\hfill \\ \\ \hfill & =\sum _{k\in {\mathbb{Z}}^d:0\le k_i<n}{\mathrm{\Delta }}_1\cdot \dots \cdot {\mathrm{\Delta }}_d\cdot 2{\mathrm{\Delta }}_{d+1}\le 𝜖\hfill \end{array}$$

as desired.

$f$ is uniformly continuous in $B$; that is, for every $𝜀>0$ there exists $\delta >0$ such that for any ||$x-y\left|\right|<\delta$ we have $|f(x)-f(y)|<𝜀$.

Write $B={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_{\delta }\left(x_n\right)$ where ${\left(B_{\delta }\left(x_n\right)\right)}_n$ are disjoint boxes of diameter less than $\delta$, with $x_n$ the center of $B_{\delta }$. Uniform continuity of $f$ gives that for any $x\in B$ there exists (a unique $)n$ such that $x\in B_{\delta }\left(x_n\right)$, and $\mathit{so}(x,f(x))\in B_{\delta }\left(x_n\right)\times \left(f\left(x_n\right)-𝜀,f\left(x_n\right)+𝜀\right)=:Q_{\delta ,n}$. Thus, $\{(x,f(x)):x\in B\}\subset \bigcup <!--\; FUNCTION\; APPLICATION\; -->Q_{\delta ,n}$. Since $m\left(Q_{\delta ,n}\right)=m\left(B_{\delta }\left(x_n\right)\right)\cdot 2𝜀$, we have that

Taking $𝜀\to 0$ gives the first assertion.

For the second assertion, let $V=\{(x,t):x\in B,0\le t\le f(x)\}.$ Fix $𝜀,\delta >0$ and $B={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{B}_{\delta }(x)$ as above. For every $n$ set

Note that $\left|M_n-m_n\right|\le 𝜀$ by uniform continuity. Now, set $K_n=\lfloor \frac{m_n}{𝜀}\rfloor$. For $0\le k\le K_n-1$ set $Q_{n,k}=B_{\delta }\left(x_n\right)\times 𝜀[k,k+1)$ and ${\tilde{Q}}_n=B_{\delta }\left(x_n\right)\times \left[𝜀K_n,M_n\right]$.

For any $n$, if $0\le k<K_n$ and $(x,t)\in Q_{n,k}$ then $x\in B_{\delta }\left(x_n\right)$ and $t<𝜀K_n\le m_n\le f(x)$. So

On the other hand, if $x\in B$ and $0\le t\le f(x)$ then there exists $n$ such that $x\in B_{\delta }\left(x_n\right)$ and so $f(x)\in \left[m_n,M_n\right]$. Thus, either $t<𝜀K_n$ in which case $(x,t)\in Q_{n,k}$ for some $0\le k<K_n$ or $𝜀K_n\le t\le f(x)\le M_n$ in which case $(x,t)\in {\tilde{Q}}_n$. Thus,

Now, let $\Lambda :=m\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{\bigcup }_{k=0}^{K_n-1}{Q}_{n,k}\right)$. Then,

Taking $𝜀\to 0$ completes the proof.

1.

1. Step (measure zero): First, let us suppose that the set is measurable and show that if this was the case it must have measure zero. Since $f$ is continuous it is Riemann integrable over $B$ which gives an immediate cover of $G$ (the graph of $f$): By definition, for every $𝜖>0$ there exists some partition of $B$ into cuboids $Q_1,\dots ,Q_n$ s.t. $U(f,P)-L(f,P)<𝜖$. This gives us the cover (writing $l(f,Q)=\underset{x\in Q}{\inf <!--\; FUNCTION\; APPLICATION\; -->}f(x)$ and $u(f,Q)=\underset{x\in Q}{\sup <!--\; FUNCTION\; APPLICATION\; -->}f(x)$)

$$\begin{array}{c}\hfill G\subset \underset{C}{\underset{⏟}{\underset{k=1}{\overset{n}{\bigcup }}{Q}_k\times \left[l(f,Q_k),u(f,Q_k)\right]}}\end{array}$$

whereby

$$\begin{array}{c}\hfill m(G)\le m\left(C\right)=\underset{k=1}{\overset{n}{\sum }}m(Q_k)(u(f,Q_k)-l(f,Q_k))=U(f,P)-L(f,P)<𝜖.\end{array}$$

2. Step (measurable): Let $𝜖>0$ be as above. Since for every elementary set $A\subset G$ we have $m(A)\le m\left(C\right)$ it follows that $0\le m\left(A\right)<𝜖$ so in particular $m\left(C\setminus A\right)=m\left(C\right)-m\left(A\right)<𝜖$ as desired.

2.

For (ii) we can follow a similar approach: Riemann integrability gives us a partition $P=\{Q_1,\dots ,Q_n\}$ of $B$ s.t. $U(f,P)-L(f,P)<𝜖$. We can now choose our outer (resp. inner) set to be the area under the upper (resp. lower) step function, i.e.

$$\begin{array}{c}\hfill O=\underset{k=1}{\overset{n}{\bigcup }}{Q}_k\times [0,u(f,Q_k)],I=\underset{k=1}{\overset{n}{\bigcup }}{Q}_k\times [0,l(f,Q_k)]\end{array}$$

yielding (as above)

$$\begin{array}{c}\hfill m\left(O\setminus I\right)=m\left(O\right)-m\left(I\right)=U(f,P)-L(f,P)<𝜖.\end{array}$$