I'm working with the next exercise and I don't know how to solve it.
Let $f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}$ an integrable function over $A$ where $A$ is a bounded set. Prove that if there exist $c>0$ such that for all $x\in A$ we have that $f(x)\geq c$ then $A$ is Jordan measurable set.
Recall that a set $A$ is Jordan measurable iff the indicator function of $A$ is integrable over a rectangle containing $A$ iff the boundary of $A$ has zero content.
My first idea was to prove that $h(x)=c\chi_{A}(x)$ is integrable over $R$ (and therefore $\chi_A(x)$), i.e., consider a rectangle $R$ such that $A\subseteq R$ and proving that the differente of upper and lower sums are small. As $f$ is integrable over $A$, the zero extension of $f$ to all rectangle, called $f_A$ (i.e., $f_A$ is defined as $f$ if $x\in A$ and $0$ if $x\in R\setminus A$), is integrable. Then, there exist a partition $P$ of $R$ such that $U(f_A,P)-L(f_A,P)$ is less than a small number. But now, if we take a rectangle $R_i$ generated by $P$ we can see the next relation between the infimum and supremum of $f_A$ and $h$ over $R_i$: $$\inf(h(x),R_i )\leq \inf(f_A,R_i) \ \text{and} \ \sup(h(x),R_i)\leq \sup(f_A,R_i)$$This is not a good inequality because with this I can't bound (or not directly) $U(c\chi_A,P)-L(c\chi_A,P)$. So, I don´t know how to continue…
My next idea was to use the Lebesgue criterion of integrability to prove directly that $h$ is integrable. But, again, clearly, the set of discontinuities of $c\chi_A$ is $\operatorname{bd}(A)$ (boundary of $A$). If I knew that the boundary of $A$ has zero content or zero Lebesgue measure (they are equivalents as $A$ is bounded) I have finished, but this is equivalent to Jordan measurability of $A$. So, again, I'm stucked.
Any hint or idea? I really appreciate. Thanks!
Best Answer
You are given that $f$ is Riemann integrable over the bounded set $A$. By definition, for any rectangle $Q$ containing $A$ it must hold that $f_A = f\chi_A$ is Riemann integrable over $Q$ and
$$\int_Af = \int_Qf_A,$$
where it can be shown that the value of the integral is independent of the choice for $Q$.
If $x \in \text{bd}(A)$, then for every open neighborhood $U$ of $x$ there exist points $y_1 \in U \cap \text{ext}(A)$ where $f_A(y_1)=0$ and $y_2 \in U\cap \text{int}(A)$ where $f_A(y_2) = f(y_2)\geqslant c > 0$. Thus, $f_A$ is discontinuous on $\text{bd}(A)$.
Now you simply invoke the theorem that a bounded function $g$ is Riemann integrable if and only if the set of points $D_g$ where $g$ is discontinuous is a set of Lebesgue measure zero. Hence, $\text{bd}(A)\subseteq D_{f_A}$ has Lebesgue measure zero. Since $\text{bd}(A)$ is bounded and closed (no point on the boundary is an interior point of the boundary), it follows that $\text{bd}(A)$ is compact and has zero content.
Alternatively using Darboux sums, since $f_A$ is Riemann integrable on a rectangle $Q \supseteq A$, given $\epsilon >0$ there is a partition $P$ such that
$$c\epsilon > U(f_A,P) - L(f_A,P) =\sum_{R \in P}\omega_R(f_A) \,v(R)\geqslant \sum_{R\cap \text{bd}(A) \neq \emptyset}\omega_R(f_A) \, v(R),$$
where $v(R)$ is the volume of rectangle $R$ and $\omega_R(f)$ is the oscillation given by
$$\omega_R(f_A) = \sup\{|f_A(u) -f_A(v)| \, : u,v \in R\}$$
For any rectangle $R$ that intersects the boundary of $A$ we have $\sup_{x \in R}f_A(X) \geqslant c$ and $\inf_{x \in R} f_A(x) = 0$. This implies that $\omega_{R}(f_A)\geqslant c$, and
$$c\epsilon > \sum_{R\cap \text{bd}(A) \neq \emptyset}c \, v(R) \geqslant c\sum_{R\cap \text{bd}(A) \neq \emptyset} \, v(R)$$
Hence, there is a finite collection of rectangles that cover $\text{bd}(A)$ with total volume less than $\epsilon$ and, therefore, $\text{bd}(A)$ has zero content.