Exercise 1.3.6 (Measurability of a function and the area under its graph)

Notice that we can represent the area under the graph as the union of the two following sets:

We argue that the former is Lebesgue measurable, while the latter is a null set.

• $\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t<f(x)\right\}$
  Since $f$ is an unsigned Lebesgue measurable function, by Lemma 1.3.9 (iv) we can find an increasing sequence ${(f_n)}_{n\in \mathbb{N}}$ of unsigned simple functions $f_n\le f_{n+1}<f$ such that $f(x)=\sup <!--\; FUNCTION\; APPLICATION\; -->\{f_n(x):n\in \mathbb{N}\}$ for all $x\in {ℝ}^d$. Consider the areas under the curves of these simple functions:

  $$\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f_n(x)\right\}.$$

  This set is Lebesgue measurable for any $n\in \mathbb{N}$ since it can be represented as a union of measurable sets

  $$\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f_n(x)\right\}=\bigcup _{i=1}^m\left\{(x,t)\in E_i\times ℝ:0\le t\le c_i\right\}=\bigcup _{i=1}^m{E}_i\times [0,c_i]$$

  where $c_1,\dots ,c_m$ are the finite number of values that $f_n$ takes on the corresponding Lebesgue measurable sets $E_1,\dots ,E_m$. In other words, the theorem is at least true for simple functions. The trick is to represent the area under $f$ as the countable union of the Lebesgue measurable areas under the simple functions:

  $$\bigcup _{n\in \mathbb{N}}\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f_n(x)\right\}=\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t<f(x)\right\}.$$

  It is easy to verify that this is indeed the case:

  • From $f_n<f$ we trivially have

    $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:\left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f_n(x)\right\}\subseteq \left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f(x)\right\}$$

  • Pick $(x,t)$ from the right-hand side, i.e., $t\le f(x)$. Set $𝜖:=f(x)-t$, and find an $n\in \mathbb{N}$ such that $f(x)-f_n(x)\le 𝜖$. Then, $(x,t)\in \left\{(x,t)\in {ℝ}^d\times ℝ:0\le t\le f_n(x)\right\}$ since $f_n(x)\ge f(x)-𝜖=t$.

  Thus, the strict area under the graph is Lebesgue measurable by Lemma 1.2.13 (vi).

• $\left\{(x,f(x)):x\in {ℝ}^d\right\}$
  Assume that $f:{ℝ}^d\to ℝ$ is measurable. We prove that $\left\{(x,f(x)):x\in {ℝ}^d\right\}$ is a Lebesgue null set.It is enough to prove the claim when $f:B\to ℝ$ is measurable and $B\subset {ℝ}^d$ is a box. Then, since ${ℝ}^d$ is a countable union of boxes, the claim will follow. Let $G:=\left\{(x,f(x)):x\in B\right\}$, and for $n\in \mathbb{Z}$, let $F_n:=G\bigcap <!--\; FUNCTION\; APPLICATION\; -->(B\times [n,n+1])$. It is enough to prove that $F_n$ is a null set for every $n\in \mathbb{Z}$, so without loss of generality, we show it for the unit interval product $F:=F_0$.Fix some $k\in \mathbb{N}$, and let $I_1,\dots ,I_{k-1}$ be a sequence of intervals $[0,\frac{1}{k}],[\frac{1}{k},\frac{2}{k}],\dots ,[\frac{k-1}{k},1]$. Then, for any $j=0,\dots ,k-1$ we have

  $$m^{\text{∗}}(G\cap (B\times I_j))\le m^{\text{∗}}(f^{-1}(I_j)\times I_j)\le m^{\text{∗}}(f^{-1}(I_j))\times \frac{1}{k}.$$

  which implies, by Exercise 1.2.22 and some basic set theory

  $$\begin{array}{llll}\hfill m^{\text{∗}}(F)& =m^{\text{∗}}\left(\bigcup _{j=0}^{k-1}G\cap (B\times I_j)\right)=\sum _{j=0}^{k-1}{m}^{\text{∗}}(G\cap (B\times I_j))\le \sum _{j=0}^{k-1}{m}^{\text{∗}}(f^{-1}(I_j)\times I_j)& \hfill & \end{array}$$

  At this point we use the assumption that $f$ is Lebesgue measurable, and apply Lemma 1.3.9 (ix) which states that the inverse must remain Lebesgue measurable

  $$\begin{array}{llll}\hfill & \le \sum _{j=0}^{k-1}{m}^{\text{∗}}(f^{-1}(I_j))\times \frac{1}{k}=\frac{1}{k}{m}^{\text{∗}}\left(\bigcup _{j=0}^{k-1}{f}^{-1}(I_j)\right)=\frac{1}{k}\times m^{\text{∗}}(f^{-1}([0,1]))\le \frac{1}{k}\times m^{\text{∗}}(B)& \hfill & \end{array}$$

  Since $k$ was arbitrary, we can take $k\to \infty$ and we are done.