Lebesgue integral of $f$ is equal to the Lebesgue measure of the area under the curve of $f$

Question

I am self-studying measure theory, and I am trying to prove the following statement:

Definition 1.3.2 (Simple function) We call a function $f:\mathbb{R}^d\to\mathbb{C}$ simple iff $f = c_1 1_{E_1} + \ldots + c_k 1_{E_k}$ for some Lebesgue measurable sets $E_1,\ldots,E_k \subseteq \mathbb{R}^d$ and scalars $c_1,\ldots,c_k \in \mathbb{C}$. We have $\int f = c_1m(E_1) + \ldots + c_k m(E_k)$.

Definition 1.3.12 (Lebesgue integral) Let $f:\mathbb{R}^d\to[0,+\infty]$ be a Lebesgue measurable function. Then $\int_{\mathbb{R}^d} f = \sup_{g\leq f, \text{g simple}} \int_{\mathbb{R}^d} g = \inf_{f\leq h, \text{h simple}} \int_{\mathbb{R}^d} h$

Exercise 1.3.13 (Area interpretation of Lebesgue integral) Let $f:\mathbb{R}^d\to [0,+\infty]$ be a measurable function. Show that $$\int_{\mathbb{R}^d} f = m\left(\{(x,t)\in\mathbb{R}^d\times\mathbb{R}: 0 \leq t \leq f(x) \} \right) $$

I did not find the proof of this statement on the internet or on math.stackexchange, so I would like to ask you to look over my own proof and tell me whether this is ok. This would be really kind!

Proof: Part 1, $\geq$. We use the outer regularity condition to demonstrate that upper Lebesgue integral of $f$ is equal to the Lebesgue measure of $A(f)$: $$ \inf \left\{ \text{Simp} \int_{\mathbb{R}^d} h: \begin{array}{l} h \text{ is simple}\\ f \leq h \end{array} \right\} \geq \inf \left\{ m\left(U\right): \begin{array}{l} U \text{ is open}\\ A(f) \subseteq U\end{array} \right\}$$ Pick an arbitrary simple integral $\text{Simp} \int_{\mathbb{R}^d} h$ of a simple functions $h$ majorizing $f$ from the left-hand side set. We then have $h = c_1 1_{I_1} + \ldots + c_k 1_{I_k}$ for some measurable sets $I_1,\ldots,I_k$. Pick an arbitrary $\epsilon>0$. We then can find open sets $U_1',\ldots,U_k',U_{k+1}'$ containing $I_1,\ldots,I_k,\mathbb{R}^d/$ such that $m(U_i'/I_i) \leq \epsilon$. Define new sets in $\mathbb{R}^{d+1}$ by \begin{align*} U_1&:= U_1' \times (-\epsilon,c_1 + \epsilon)\\ &\vdots \\ U_k &:= U'_k \times (-\epsilon,c_k + \epsilon)\\ U_{k+1} &:= U'_{k+1} \times \{0\} \end{align*} 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 the Cartesian product property of the Lebesgue measure we obtain \begin{align*} m\left(\bigcup_{i=1}^{k+1} U_i\right) &= \sum_{i=1}^{k}m(U_i')\times m((-\epsilon,c_i + \epsilon)) \\ &\leq \sum_{i=1}^{k} \left(m(I_i) + {\epsilon}\right)\times \left(c_i + 2\epsilon\right)\\ &= \sum_{i=1}^{k} c_i m(I_i) + \epsilon \left[2m(I_i) + {c_i} + {2\epsilon} \right]\\ &= \text{Simp} \int_{\mathbb{R}^d} h + \epsilon' \end{align*} Since the measure of this element is in the right-hand side set, $\text{Simp} \int_{\mathbb{R}^d} h < \inf_{U\text{ open, } A(f) \subseteq U} m(U)$ would lead to a contradiction when taking $\epsilon' \to 0$. We hence conclude the opposite, and taking infimums yields $\inf_{\text{h simple},h\geq f} \text{Simp} \int_{\mathbb{R}^d} h \geq \inf_{U\text{ open,} A(f) \subseteq U} m(U) $

Part 2, $\leq$. \item[$\leq$)] This time we demonstrate the the lower Lebesgue integral is equal to the Lebesgue measure using the inner regularity criterion: $$ \sup \left\{ \text{Simp} \int_{\mathbb{R}^d} g: \begin{array}{l} g \text{ is simple}\\ g \leq f \end{array} \right\} \leq \sup \left\{ m(K): \begin{array}{l} K \text{ is compact}\\ K \subseteq A(f) \end{array} \right\}$$ Similarly to the previous part, pick an arbitrary simple integral $\text{Simp} \int_{\mathbb{R}^d} g$ from the left-hand side associated with a simple function $g = c_1 1_{I_1} + \ldots + c_k 1_{I_k}$ for some measurable sets $I_1,\ldots,I_k$. Pick an arbitrary $\epsilon>0$. We then can find compact sets $K_1',\ldots,K_k',K_{k+1}'$ which are contained in $I_1,\ldots,I_k,\mathbb{R}^d/$ such that $m(I_i/K_i') \leq \epsilon$. Define new compact sets in $\mathbb{R}^{d+1}$ by \begin{align*} K_1&:= K_1' \times [0,c_1]\\ &\vdots \\ K_k &:= K'_k \times [0,c_k]\\ K_{k+1} &:= K'_{k+1} \times \{0\} \end{align*} 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 $\epsilon' := \sum c_i \epsilon$ we give ourselves an epsilon of the room; by the finite additivity of Lebesgue measure + knowing how Lebesgue measure interacts with the Cartesian products we obtain \begin{align*} m\left(\bigcup_{i=1}^{k+1} K_i\right) +\epsilon' &= \sum_{i=1}^{k}m(K_i')\times m([0,c_i]) + \epsilon c_i\\ &= \sum_{i=1}^{k} \left(m(K_i') + \epsilon\right) \times m([0,c_i]) \\ &\geq \sum_{i=1}^{k} m(I_i')\times c_i= \text{Simp} \int_{\mathbb{R}^d} g \end{align*} Since the measure of this element is contained in the right-hand side set, $\text{Simp} \int_{\mathbb{R}^d} g > \sup_{K\text{ closed,} K \subseteq A(f)} m(K)$ would lead to a contradiction when taking $\epsilon' \to 0$. We hence conclude the opposite, and taking supremums w.r.t. $g$ yields $\sup_{\text{g simple},g\leq f} \text{Simp} \int_{\mathbb{R}^d} g \leq \sup_{K\text{ closed,} K \subseteq A(f)} m(K) $.

Answer 1

I think your proof is all right. I would, though suggest though to use Tonelli's Theorem.

First of all some notation; $m_{n}$ is the Lebesgue measure on $\mathbb{R}^n$, $n\geq 1$, $\chi_B$ is the characteristic function of a set $B \subset \mathbb{R}^n$, $n\geq 1$ and $A^f:= \{ (x,t)\in \mathbb{R}^n\times [0, +\infty) : \quad t<|f(x)|=f(x) \}$. Also,

• for a fixed $t\in [0,+\infty)$, $A_t^f:= \{ x\in \mathbb{R}^n : \quad t<|f(x)|=f(x) \}$

and

• for a fixed $x\in \mathbb{R}^n$, $A_x^f:= \{ t\in [0, +\infty) : \quad t<|f(x)|=f(x) \}$.

By using Tonelli's Theorem (and since $m_{n+1} = m_n \otimes m_1$) we have that, $$\int\limits_{\mathbb{R}^n \times [0,+\infty)} \chi_{A^f}\text{ d}m_{n+1} = \int_{\mathbb{R}^n} \int_{0}^{+\infty} \chi_{A_x^f}(t) \text{ d}m_1 \text{ d}m_n = \int_{0}^{+\infty} \int_{\mathbb{R}^n} \chi_{A_t^f}(x) \text{ d}m_n \text{ d}m_1$$

Now $$\int_{\mathbb{R}^n} f(x) \text{ d}m_{n} = \int_{\mathbb{R}^n} \left( \int_{0}^{f(x)} 1 \text{ d}m_{1}(t) \right) \text{ d}m_n(x) = \int_{\mathbb{R}^n} \left( \int_{0}^{+\infty} \chi_{[0, f(x))} \text{ d}m_{1}(t) \right) \text{ d}m_n(x)$$

Now you just have to observe that for every (fixed) $x\in \mathbb{R}^n$, $\chi_{A_x^f} = \chi_{[0, f(x))}$ everywhere in $[0, +\infty)$. Hence, $$\int_{\mathbb{R}^n} \int_{0}^{+\infty} \chi_{A_x^f}(t) \text{ d}m_1 \text{ d}m_n = \int_{\mathbb{R}^n} \int_{0}^{+\infty} \chi_{[0, f(x))} \text{ d}m_{1}(t) \text{ d}m_n(x)$$ and that completes the argument.

Answer 2

I'll give one solution that only uses knowledge till this point in Tao's text.

Denote the set $\{(x,t) \in \mathbb{R}^d \times \mathbb{R}: 0 \leq t \leq f(x) \}$ by $A(f)$. Let $h = \sum_{i=1}^{n}c_i\chi_{E_i}$ be a simple unsigned function defined using disjoint measurable sets $(E_i)_{1 \leq i\leq n}$. By definition, and Exercise $1.2.22$, $\text{Simp} \int_{\mathbb{R}^d} h = \int_{\mathbb{R}^d} h := \sum_{i=1}^{n}c_im(E_i) = m(\bigcup_{i=1}^{n}E_i \times [0, c_i]) = m(A(h)).$ So the statement holds for simple functions.

Suppose first that $f:\mathbb{R}^d\to[0,+\infty]$ is bounded measurable. By Exercise $1.3.4$, $f$ is the uniform limit of bounded simple functions $g_n$. In particular $\forall \epsilon > 0$, there is an $n$ large enough s.t $m(A(f)) - m(A(g_n)) \leq \epsilon/2$. By definition of the Lebesgue integral, we can also find $0 \leq g \leq f$, $g$ simple s.t $\int_{\mathbb{R}^d} f - \int_{\mathbb{R}^d} g \leq \epsilon/2$. By symmetry we assume that $\int_{\mathbb{R}^d} g \geq \int_{\mathbb{R}^d} g_n$. Then we have: $|\int_{\mathbb{R}^d} f - m(A(f))| \leq |\int_{\mathbb{R}^d} f - \int_{\mathbb{R}^d} g| + |\int_{\mathbb{R}^d} g -m(A(f))| \leq \epsilon/2 + \epsilon/2 = \epsilon$. Since $\epsilon > 0$ is arbitrary, this proves the statement for the case when $f$ is bounded.

The general case then follows from the horizontal truncation property: As $n \to \infty$, $\int_{\mathbb{R}^d} min(f, n) = m(A(min(f,n))) \to m(A(f))$. Also, $\int_{\mathbb{R}^d} min(f, n) \to \int_{\mathbb{R}^d} f$. Hence it must be that $\int_{\mathbb{R}^d} f = m(A(f))$.