0
$\begingroup$

According to Spivak (before thrm 3.5) ,

A subset of $\ \mathbb{R^n}$ has ($n$ - dimensional) content zero if for every $\epsilon > 0$ there is a $finite$ cover $ \ \{U_1 , U_2 , \cdots,U_m \}$ by closed rectangles such that $\sum_{i=1}^m v(U_i) < \epsilon$ .

Also according to Spivak (after thrm 3.9 ),

A bounded set C whose boundary has measure 0 is called Jordan Measurable . The integral $\displaystyle \int_C 1$ is called the ($n$ - dimensional ) content of C

Are these two definitions equivalent ie ; is the following true :

A set is of content zero iff content of the set is zero ?

$\endgroup$
0
$\begingroup$

Here is a proof of the forward implication.

Since $C$ is of content zero, it is bounded and is contained inside a bounded rectangle $Q$. The content of $C$ is the Riemann integral of $f \equiv 1$ over $C$, defined as

$$\int_Cf= \int_Q g,$$

where $g(x) = 1$ for $x \in C$ and $g(x) = 0$ for $x \in Q \setminus C$.

Note that $g$ is Riemann integrable, since it is continuous everywhere except possibly at points of $C$ and the content of $C$ is zero.

Now take any partition $P$ of $Q$. Any subrectangle $R$ of $P$ has non-zero content (by definition of a partition). Hence $R$ is not a subset of $C$ and must contain at least one point where $g(x) = 0$. This implies that $\inf_R g(x) \leqslant 0 \leqslant \sup_R g(x)$. Forming upper and lower Riemann sums we have for any partition $P$,

$$L(P,g) \leqslant 0 \leqslant U(P,g).$$

Since $g$ is integrable we have

$$\int_Q g = \sup_P L(P,g) \leqslant 0 \leqslant \inf_P U(P,g) = \int_Q g,$$

And this implies that $\int_Q g = 0$, i.e., the content of C is zero.

For the reverse implication, if the content of $C$ is zero then $\int_Q g =0$ for any enclosing rectangle $Q$. Hence, for any $\epsilon >0$ there exists a partition $P$ such that $0 \leqslant U(P,g) < \epsilon$. Considering only the subrectangles that intersect $C$, we can show that these subrectangles form a finite covering of $C$ with total volume less than $\epsilon$. See if you can fill in the details of this last step in the proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.