Exercise 7.3.9

Question

[Content Zero] A set $A \subseteq[a, b]$ has content zero if for every $\epsilon>0$ there exists a finite collection of open intervals $\left\{O_{1}, O_{2}, \ldots, O_{N}ight\}$ that contain $A$ in their union and whose lengths sum to $\epsilon$ or less. Using $\left|O_{n}ight|$ to refer to the length of each interval, we have
$$
A \subseteq \bigcup_{n=1}^{N} O_{n} \quad 	ext { and } \quad \sum_{n=1}^{N}\left|O_{n}ight| \leq \epsilon .
$$
\begin{enumerate}[label=(\alph*)] 
\item Let $f$ be bounded on $[a, b]$. Show that if the set of discontinuous points of $f$ has content zero, then $f$ is integrable.
\item Show that any finite set has content zero.
\item Content zero sets do not have to be finite. They do not have to be countable. Show that the Cantor set $C$ defined in Section $3.1$ has content zero.
\item Prove that
$$
h(x)= \begin{cases}1 & 	ext { if } x \in C \ 0 & 	ext { if } x 
otin C .\end{cases}
$$
is integrable, and find the value of the integral.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Let $\epsilon > 0$, let $M$ be the difference between the maximum and minimum values of $f$. Let $A$ be the set of discontinuous points of $f$, and choose $\left\{O_{1}, \ldots, O_{N}\right\}$ so that $\sum |O_n| < \epsilon / (2M)$. Assume without loss of generality that none of $O_i$ overlap; if there was an overlapping pair then of intervals then they can be merged into a single interval.

Start a partition $P$ with the endpoints of each $O_i$, and note that the sum of $(M_k - m_k)\Delta x_k$ over the subintervals from each $O_i$ is less than $\epsilon/2$. Moreover over the remaining subintervals (of which there are finitely many) $f$ is continuous and thus integrable, and so it is possible to refine $P$ so that over the remaining subintervals the sum of $(M_k - m_k)\Delta x_k$ is less than $\epsilon/2$.

Putting these together gets us that $f$ is integrable.

\item Let this finite set $A$ have $N$ elements, then take $V_\delta(x)$ where $x \in A$ and $\delta = \epsilon / N$.

\item Refer to Section 3.1 for the definition of $C_n$. Let $n$ be large enough so that the total length of the intervals in $C_n$ is less than $\epsilon/2$. Let $C_{n,i}$ refer to the $i$'th interval in $C_n$, and define $O_i$ so that $C_{n,i} \subset O_i$ and $|O_i| - |C_{n,i}| = \delta$, where $\delta < \epsilon/2^{n+1}$ is small enough so that $\sum^{2^n}_{i=1} |O_i| - \sum^{2^n}_{i=1} |C_{n,i}| < \epsilon/2$.

\item $C$ is closed and its complement is open; therefore the set of discontinuities of $h$ is $C$. Parts (a) and (c) imply $h$ is integrable.

Exercises 3.5.8 and 3.5.9 show that the complement of $C$ is dense, so for any partition $P$, $L(h, P) = 0$. Thus since $h$ is integrable, the value of the integral must be 0.
 \end{enumerate}

Exercise 7.3.9

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Exercise 7.3.9

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