Prove that $W=\{(x,y)\in\mathbb{R}^{n+1}:x\in P,−|f(x)|\leq y\leq |f(x)|\}$ is Jordan measurable

Question

Given $f:\mathbb{R}^n\to\mathbb{R}$ integrable over $P\subset\mathbb{R}^n$
Does the set $W=\{(x,y)\in\mathbb{R}^{n+1}:x\in P,-|f(x)|\leq y\leq|f(x)$|} is Jordan measurable?
I've shown that W is bounded and i thought maybe i can show that $\partial W$ is measure zero because it's union of two set that they are graph of a function but i am not sure it it's true

Answer 1

It is not entirely clear what $P$ is, so I am presuming there is some compact rectangle $B \subset \mathbb{R}^n$ such that $P \subset B$ and $f^*=f \cdot 1_P$ is integrable.

Assuming this to be the case, we can write $W = \{ (x,y) | \ |y| \le |f(x)| \}$.

To show that $W$ is Jordan measurable it is sufficient to show that $\partial W$ has measure zero.

Note that if $O \subset W \subset C$ with $C$ closed and $O$ open then $\partial W \subset C \cap O^c$, so if we can show that we can choose $O,C$ such that $C \setminus O = C \cap O^c$ has arbitrarily small measure then we are finished.

If $f$ is integrable then so is a continuous function composed with $f$ hence $|f|$ and $-|f|$ are integrable.

Let $\epsilon>0$ and choose a partition $\pi$ such that $L(|f|,\pi) \le \int |f| \le U(|f|,\pi)$ and $U(|f|,\pi) - L(|f|,\pi) < \epsilon$. Let ${\cal R}$ be the collection of closed rectangles defined by the partition $\pi$. (Note that the collection is essentially disjoint, that is, if $R_1,R_2 \in {\cal R}$ then $R_1 \cap R_2$ has measure zero.)

Let $O = \bigcup_{R \in {\cal R}} R^\circ \times (-l_R,l_R)$ and $C = \bigcup_{R \in {\cal R}} R \times [-u_R,u_R]$, where $l_R = \inf_{x \in R} |f(x)|$, $u_R = \sup_{x \in R} |f(x)|$. Note that $) $ is open, $C$ is closed, $O \subset W \subset C$ and $C \setminus O \subset \bigcup_{R \in {\cal R}} R \times ( [-u_R, -l_R] \cup [l_R,u_R])$.

Hence $m (\partial W) \le m(C \setminus O ) \le 2(U(|f|,\pi) - L(|f|,\pi)) < 2 \epsilon$.