Let $f:[a,b] \rightarrow \mathbb{R}$ a integrable function sucht that $f(x)\geq 0$. Define :

$$A = \{ (x,y) \in \mathbb{R}^2 \vert x\in [a,b], y \in [0,f(x)] \} $$

Prove that $A$ is Jordan measurable with Jordan measure equal to $\int_{a}^{b} f(x) dx$.

$\textbf{My attempt (with help of "RRL")}$

Consider the block $E = [a,b] \times [0,M]$ where $M$ is a bound of $f$. Let $\epsilon>0$, how $f$ is integrable exists a partition $P'=\{t_o<t_1<\cdots < t_{n-1} <t_n \}$ of $[a,b]$ such that :

$$S(f,P')-s(f,P') < \epsilon $$

Define the blocks : $R_j = [t_{j-1},t_j]\times [0,m_j]$ and $R'_j = [t_{j-1},t_j]\times [0,M_j]$. Let $P$ the partition of $E$ such that ( intuitively obtained by prolonging the faces of the block $R_j$ and $R'_j$) :

Each $R_j$ is union of blocks of $P$ such that are contained in $A$.

Each $R'_j$ is the union of blocks of $P$ such that intersect $A$.

Then :

$$ S(\chi_{A}, P) - s(\chi_{A},P) \leq \sum_{i=1}^n (M_i-m_i)(t_i-t_{i-1})=S(f,P')-s(f,P')< \epsilon $$

So $\chi_{A}$ is integrable and so $A$ is Jordan measurable.


Consider a partition $P: a = x_0 < x_1 < \ldots < x_n = b$ of $[a,b]$ and let $M_j = \sup_{x \in [x_{J-1},x_j]}f(x)$ and $m_j = \inf_{x \in [x_{J-1},x_j]}f(x)$.

Define rectangles $R_j = [x_{j-1},x_j] \times [0,m_j]$ with volume $|R_j| = m_j(x_j - x_{j-1})$ and $R_j' =[x_{j-1},x_j] \times [0,M_j] $ with volume $|R_j'| = M_j(x_j - x_{j-1})$ and notice that

$$L(P,f) = \sum_{j=1}^n |R_j| = \left|\bigcup_{j=1}^n R_j \right|, \quad U(P,f) = \sum_{j=1}^n |R_j'|= \left|\bigcup_{j=1}^n R_j' \right|,$$

where $L(P,f)$ and $U(P,f)$ are lower and upper Darboux sums, respectively.

Assuming that $f$ is Riemann integrable (although not necessarily continuous) we have

$$\tag{*}\sup_P \left|\bigcup_{j=1}^n R_j \right| = \int_a^b f(x) \, dx = \inf_P\left|\bigcup_{j=1}^n R_j' \right|$$

The Jordan measure of a Jordan measurable set $A$ is given by

$$|A| = \sup_{E \subset A} |E| = \inf_{A \subset E'} |E'|,$$

where $E$ and $E'$ denote elementary sets (finite unions of non-overlapping rectangles) contained in and containing, respectively, $A$.

Take $A = \{ (x,y) \in \mathbb{R}^2 \vert x\in [a,b], y \in [0,f(x)] \}$ which you have already shown is Jordan measurable. Since $E= \bigcup_{j=1}^n R_j $ and $E' = \bigcup_{j=1}^n R_j'$ are examples of elementary sets contained in and containing $A$, we have

$$\sup_P \left|\bigcup_{j=1}^n R_j \right| \leqslant \sup_{E \subset A} |E| = |A| =\inf_{A \subset E'} |E'| \leqslant \inf_P \left|\bigcup_{j=1}^n R_j' \right|,$$

and from (*) it follows that $|A| = \int_a^b f(x) \, dx$.

  • $\begingroup$ Thanks very much, any hint for prove the same exercise but consider $[a,b]^n$ a block in $\mathbb{R}^n$ $\endgroup$
    – user411479
    Nov 14 '19 at 21:34
  • $\begingroup$ @Orested: The proof would be exactly the same. The lower and upper sums for the integral of a function $f :[a,b]^n \to \mathbb{R}$ are volumes of elementary sets in $\mathbb{R}^{n+1}$. In other words $L(P,f) = vol(\cup_{j=1}^N R_j)$ where $R_j$ is a block in $\mathbb{R}^{n+1}$.. $\endgroup$
    – RRL
    Nov 14 '19 at 21:44
  • $\begingroup$ How you prove that $A$ is Jordan measurable? in your proof. $\endgroup$
    – user411479
    Nov 14 '19 at 22:12
  • $\begingroup$ You can argue that the boundary has Jordan measure zero (which you did). The other way to prove a set $A$ is Jordan measurable is to show that the supremum of volumes of elementary sets contained in $A$ equals the infimum of volumes of elementary sets containing $A$. This is what I did -- by squeezing the "inner" measure and "outer" measure between lower and upper integrals that turn out to be equal. So I both proved measurability and found the value of the measure $\endgroup$
    – RRL
    Nov 14 '19 at 22:41

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