8.4 $^*$The Ruler Function

8.38   Example (Ruler function.) We now present an example of an integrable function that is not monotonic on any interval of positive length. Define $R:[0,1]\to\mbox{{\bf R}}$ by

$$R(x)=\cases{ 1 &if $x=0 \mbox{ or }x=1$\cr {1\over {2^n}} &if... ...x{${\mbox{{\bf Z}}}^{+}$}$ and $q$ is odd\cr 0 &otherwise.\cr}$$

This formula defines $R(x)$ uniquely: If $${ {q\over {2^n}}={p\over {2^m}} }$$where $p$ and $q$ are odd, then $m=n$. (If $m>n$, we get $2^{m-n}q=p$, which says that an even number is odd.) The set $S_0^1R$ under the graph of $R$ is shown in the figure.

This set resembles the markings giving fractions of an inch on a ruler, which motivates the name ruler function for $R$. It is easy to see that $R$ is not monotonic on any interval of length $>0$. For each $p\in\mbox{{\bf R}}$ let $\delta_p\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ be defined by

$$\delta_p(t)=\cases{ 1 &if $p=t$\cr 0 &otherwise.\cr}$$

We have seen that $\delta_p$ is integrable on any interval $[a,b]$ and $${\int_a^b \delta_p=0}$$. Now define a sequence of functions $F_j$ by

$$\begin{array}{l} F_0=\delta_0+\delta_1 \\ \mbox{} \\ F_1=F_... ...\sum_{j=1}^{2^{n-1}}}\delta_{{2j-1 \over {2^{n}}}}. \end{array}$$

Each function $F_j$ is integrable with integral $0$ and

$$\vert R(x)-F_j(x)\vert\leq {1\over {2^{j+1}}} \mbox{ for } 0\leq x\leq 1.$$

I will now show that $R$ is integrable.

Let $(\{P_n\},\{S_n\})$ be a partition-sample sequence for $[0,1]$. I'll show that $\{\sum(R,P_n,S_n)\}\to 0$.

Let $\epsilon$ be a generic element in $\mbox{${\mbox{{\bf R}}}^{+}$}$. Observe that if $M \in \mbox{${\mbox{{\bf Z}}}^{+}$}$ then

$$\left( {1 \over 2^M} < \epsilon\right) \hspace{1ex}\Longleftr... ...ace{1ex}\left(M > {\ln(\epsilon) \over \ln({1\over 2})}\right).$$

Hence by the Archimedian property, we can choose $M \in \mbox{${\mbox{{\bf Z}}}^{+}$}$ so that $${ {1\over {2^M}}<\epsilon}$$. Then

$$\sum(R,P_n,S_n) = \sum(R-F_M+F_M,P_n,S_n)$$ (8.39)

$$= \sum(R-F_M,P_n,S_n)+\sum(F_M,P_n,S_n).$$ (8.40)

Now since $${ 0\leq R(x)-F_M(x)\leq {1\over {2^{M+1}}}<{1\over 2}\epsilon}$$ for all $x\in [0,1]$, we have

$$\sum (R-F_M,P_n,S_n)\leq {1\over {2^{M+1}}}<{1\over 2}\epsilon \mbox{ for all } n\in\mbox{${\mbox{{\bf Z}}}^{+}$}.$$

Since $F_M$ is integrable and $\int F_M=0$, we have $\{\sum (F_M,P_n,S_n)\}\to 0$ so there is an $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that $${\vert\sum (F_M,P_n,S_n)\vert<{\epsilon\over 2}}$$ for all $n\in\mbox{{\bf Z}}_{\geq N}$. By equation (8.40) we have

$$\begin{eqnarray*} 0 &\leq & \sum (R,P_n,S_n)=\sum (R-F_M,P_n,S_n)+\sum(F_M,P_n,S... ...epsilon = \epsilon \mbox{ for all } n\in\mbox{{\bf Z}}_{\geq N}. \end{eqnarray*}$$

Hence $\{\sum(R,P_n,S_n)\}\to 0$, and hence $R$ is integrable and $${\int_0^1 R=0}$$.

8.41   Exercise. A Let $R$ be the ruler function. We just gave a complicated proof that $R$ is integrable and $${\int_0^1 R=0}$$. Explain why if you assume $R$ is integrable, then it is easy to show that $${\int_0^1 R=0}$$.

Also show that if you assume that the non-integrable function $D$ in equation (8.37) is integrable then it is easy to show that $${\int_0^1 D=0}$$.