Exercise 1.3.13 (Lebesgue integral and Lebesgue measure)

Let $A(f):=\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f(x)\}$ denote the area under the graph of $f$. We have demonstrated in Exercise 1.3.6 that $A(f)$ is a Lebesgue measurable set, and we have demonstrated in Exercise 1.3.11 that the lower and upper integrals of Lebesgue measurable functions coincide.

• We use the outer regularity condition from Lemma 1.2.12 to demonstrate that upper Lebesgue integral of $f$ is equal to the Lebesgue measure of $A(f)$:

  $$\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h:\begin{array}{c}h\text{ is simple}\hfill \\ f\le h\hfill \end{array}\right\}\ge \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m\left(U\right):\begin{array}{c}U\text{ is open}\hfill \\ A(f)\subseteq U\hfill \end{array}\right\}$$

  Pick an arbitrary simple integral $\mathrm{Simp}{\int }_{{ℝ}^d}h$ of a simple functions $h$ majorizing $f$ from the left-hand side set. We then have $h=c_1{1}_{I_1}+\dots +c_k{1}_{I_k}$ for some measurable sets $I_1,\dots ,I_k$. Pick an arbitrary $𝜖>0$. We then can find open sets $U_1^{\prime },\dots ,U_k^{\prime },U_{k+1}^{\prime }$ containing $I_1,\dots ,I_k,{ℝ}^d∕$ such that $m(U_i^{\prime }∕I_i)\le 𝜖$. Define new sets in ${ℝ}^{d+1}$ by

  $$\begin{array}{llll}\hfill U_1& :=U_1^{\prime }\times (-𝜖,c_1+𝜖)& \hfill & \\ \hfill & ⋮& \hfill & \\ \hfill U_k& :=U_k^{\prime }\times (-𝜖,c_k+𝜖)& \hfill & \\ \hfill U_{k+1}& :=U_{k+1}^{\prime }\times \{0\}& \hfill & \end{array}$$

  The Cartesian product of two open sets is again open, and it is then easy to verify that

  $$A(f)\subseteq \bigcup _{i=1}^{k+1}{U}_i$$

  Now we look at the measure of the above approximation. By the finite additivity of Lebesgue measure combined with Exercise 1.2.22 we obtain

  $$\begin{array}{llll}\hfill m\left(\bigcup _{i=1}^{k+1}{U}_i\right)& =\sum _{i=1}^{k}m(U_i^{\prime })\times m((-𝜖,c_i+𝜖))& \hfill & \\ \hfill & \le \sum _{i=1}^k\left(m(I_i)+𝜖\right)\times \left(c_i+2𝜖\right)& \hfill & \\ \hfill & =\sum _{i=1}^k{c}_{i}m(I_i)+𝜖\left[2m(I_i)+c_i+2𝜖\right]& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h+{𝜖}^{\prime }& \hfill & \end{array}$$

  Since the measure of this element is in the right-hand side set, $\mathrm{Simp}{\int }_{{ℝ}^d}h<\underset{U\text{ open, }A(f)\subseteq U}{\inf <!--\; FUNCTION\; APPLICATION\; -->}m(U)$ would lead to a contradiction when taking ${𝜖}^{\prime }\to 0$. We hence conclude the opposite, and taking infimums yields $\underset{\text{h simple},h\ge f}{\inf <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{Simp}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_{{ℝ}^d}h\ge \underset{U\text{ open,}A(f)\subseteq U}{\inf <!--\; FUNCTION\; APPLICATION\; -->}m(U)$

• This time we demonstrate the the lower Lebesgue integral is equal to the Lebesgue measure using the criterion from Exercise 1.2.15:

  $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g:\begin{array}{c}g\text{ is simple}\hfill \\ g\le f\hfill \end{array}\right\}\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(K):\begin{array}{c}K\text{ is compact}\hfill \\ K\subseteq A(f)\hfill \end{array}\right\}$$

  Similarly to the previous part, pick an arbitrary simple integral $\mathrm{Simp}{\int }_{{ℝ}^d}g$ from the left-hand side associated with a simple function $g=c_1{1}_{I_1}+\dots +c_k{1}_{I_k}$ for some measurable sets $I_1,\dots ,I_k$. Pick an arbitrary $𝜖>0$. We then can find compact sets $K_1^{\prime },\dots ,K_k^{\prime },K_{k+1}^{\prime }$ which are contained in $I_1,\dots ,I_k,{ℝ}^d∕$ such that $m(I_i∕K_i^{\prime })\le 𝜖$. Define new compact sets in ${ℝ}^{d+1}$ by

  $$\begin{array}{llll}\hfill K_1& :=K_1^{\prime }\times [0,c_1]& \hfill & \\ \hfill & ⋮& \hfill & \\ \hfill K_k& :=K_k^{\prime }\times [0,c_k]& \hfill & \\ \hfill K_{k+1}& :=K_{k+1}^{\prime }\times \{0\}& \hfill & \end{array}$$

  The Cartesian product of two compact sets is again compact, and it is then easy to verify that

  $$\bigcup _{i=1}^{k+1}{K}_i\subseteq A(f).$$

  Now we look at the measure of the above approximation. Defining ${𝜖}^{\prime }:=\sum <!--\; FUNCTION\; APPLICATION\; -->c_i𝜖$ we give ourselves an epsilon of the room; by the finite additivity of Lebesgue measure combined with Exercise 1.2.22 we obtain

  $$\begin{array}{llll}\hfill m\left(\bigcup _{i=1}^{k+1}{K}_i\right)+{𝜖}^{\prime }& =\sum _{i=1}^{k}m(K_i^{\prime })\times m([0,c_i])+𝜖c_i& \hfill & \\ \hfill & =\sum _{i=1}^k\left(m(K_i^{\prime })+𝜖\right)\times m([0,c_i])& \hfill & \\ \hfill & \ge \sum _{i=1}^{k}m(I_i^{\prime })\times c_i=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g& \hfill & \end{array}$$

  Since the measure of this element is contained in the right-hand side set, $\mathrm{Simp}{\int }_{{ℝ}^d}g>\underset{K\text{ closed,}K\subseteq A(f)}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)$ would lead to a contradiction when taking ${𝜖}^{\prime }\to 0$. We hence conclude the opposite, and taking supremums w.r.t. $g$ yields $\underset{\text{g simple},g\le f}{\sup <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{Simp}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_{{ℝ}^d}g\le \underset{K\text{ closed,}K\subseteq A(f)}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)$.

Simple $f$

Assume $f$ is a simple function with $f={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{c}_i{1}_{E_i}$. We can assume the $E_i$ are disjoint by MTP3.2a. Define $E_{k+1}={ℝ}^d\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{E}_i$. Then for $x\in E_{k+1}$ we have $f(x)=0$. That is,

Then,

(by MTE2.22)

(by definition of $f$)

(by above observations for $E_{k+1}$)

(by finite additivity)

So the claim is proven for simple functions.

$f$is infinite valued

Let $F$ be the set where $f(x)=+\infty$ for $x\in F$. If $m(F)>0$, then

and by monotonicity also

Also, $+\infty 1_F$ is a simple function minorizing $f$. We have

and so by monotonicity also

If $m(F)=0$, define $f^{\prime }=f$ if $x\notin F$, and $f^{\prime }=0$ if $x\in F$. Then

On the other hand,

(by MTE2.22)

Also,

(by finite additivity)

Thus, we only need to prove the equality for the set of equivalent finite-valued functions and will assume in the following that $f$ is finite-valued.

$f$is bounded and has finite support

Next, we assume $f$ is bounded and has finite support. Then by MTE3.11, the upper and lower Lebesgue integrals coincide. Let $𝜀>0$. By the definitions of the intervals, there are simple functions $g,h:{ℝ}^d\to [0,+\infty ]$ with $g\le f\le h$.

Let $𝜀>0$. By the definitions of the intervals, there are simple functions $g,h:{ℝ}^d\to [0,+\infty )$, with $g\le f\le h$.

By monotonicity, we have

and if we set in the claim for simple functions:

By $g\le f\le h$, we also have

By monotonicity, this gives

This gives us

Taking $𝜀\to 0$ gives the claim.

Case: $f$ has finite support

Now assume $f$ has finite support. Then the functions $\min <!--\; FUNCTION\; APPLICATION\; -->(f,n)$ are bounded, so that we can apply the previous result:

Now, by vertical truncation:

Since $f$ is real-valued, for each $x$, there is an $n_x$ such that $f(x)\le n_x$, and

The claim follows.

Case: general real-valued $f$

Let $E_n={[-n,n]}^d$. Then for each $f{1}_{E_n}$, the previous result applies. First, note that for $x\notin E_n$, we have $f{1}_{E_n}(x)=0$, and so

By horizontal truncation:

By finite additivity:

By the above consideration:

Since $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{E}_n={ℝ}^d$ and $1_{{ℝ}^d}=1$, we have:

This proves the claim for general $f$, and we are done.