Let $A$ and $B$ be rectangles in $R^n$ and $R$,respectively.Let $S$ be a set contained in $A\times B$.For each $y\in B$,let
$$S_y=\{x|x\in A\ \text{and}\ (x,y)\in S\}.$$
We call $S_y$ a cross-section of $S$.Show that if $S$ is rectifiable,and if $S_y$ is rectifiable for each $y\in B$,then
$$v(S)=\int_{y\in B}v(S_y)$$
According to this book,I should give a explanation of some words .
Let $S$ be a bounded set in $R^n$.If the constant function 1 is Riemann integrable over $S$,we say that $S$ is rectifiable and we define the ($n$-dimensional)volume of $S$ by the equation$$v(S)=\int_{S}1.$$
I have sloved this problem with @Matematleta help,but I find another question in zorich's Analysis(11.4.3.(2)) which can be seen,to some extent, as a inverse proposition of the above problem.
If $S_y$ is rectifiable for each $y\in B$,and if $v(S_y)$ is integrable over $B$.Can we assert that in this case the set $S$ is rectifiable?
I think the answer is affirmative,but I have no idea how to prove it.Any help will be thanked.
Best Answer
The answer is no with the following counterexample.
Take $A = B =[0,1]$ and let $S \subset [0,1]^2$ be a dense set which intersects every vertical and horizontal line in exactly one point. The contruction of such a set is discussed here
For each $y \in [0,1]$, the set $S_y = \{x\in [0,1]: (x,y) \in S\} = \{x_y\}$ contains a single point, and, hence, is rectifiable with $v(S_y) = \int_{[0,1]} \chi_{S_y} = 0$. Since $v(S_y)$ is identically $0$ for every $y \in [0,1]$ it is Riemann integrable over $[0,1]$.
The indicator function $\chi_S$ has the property that every neighborhood $V$ in $[0,1]^2$ contains a point $x \in V\cap S$ where $\chi_S(x) = 1$ and a point $y \in V\cap S^c$ where $\chi_S(y) = 0$ .
Consequently, $\chi_S$ is not Riemann integrable since the lower and upper integrals are unequal:
$$0=\underline{\int}_{[0,1]} \chi_S \neq \overline{\int}_{[0,1]} \chi_S = 1,$$
and, therefore, $S$ is not rectifiable.