Region under its graph is Jordan-measurable

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 $$P'=\{t_o of $$[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 $$|A| = \int_a^b f(x) \, dx$$.

• Thanks very much, any hint for prove the same exercise but consider $[a,b]^n$ a block in $\mathbb{R}^n$
– user411479
Nov 14 '19 at 21:34
• @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 $\mathbb{R}^{n+1}$..
– RRL
Nov 14 '19 at 21:44
• How you prove that $A$ is Jordan measurable? in your proof.
– user411479
Nov 14 '19 at 22:12
• 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
– RRL
Nov 14 '19 at 22:41