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\title{Functional analysis problem sheet II}
\author{Max Klinger and Kilian Klebes}
\begin{document}
\maketitle
\setcounter{thm}{14}
\thm{ Let $\mathcal{C}$ be the Cantor set, $\mu_{\mathcal{L}}$ the lebesgue Measure on $[0,1]$.
$$\mathcal{C}_0 = [0,1]; \; \mathcal{C}_n = \frac{\mathcal{C}_{n-1}}{3} \cup \left(\frac{2}{3} + \frac{\mathcal{C}_{n-1}}{3}\right), \mathcal{C} := \bigcap^\infty_{n=0} \mathcal{C}_n$$
where $ c+\mathcal{J} := [a+c, b+c],\; c \mathcal{J} := [ac, bc]\; \forall \mathcal{J} = [a,b]$ closed.
\begin{enumerate}[1)]
\item $\mathcal{C}$ is a compact Borel set and $\mu_{\mathcal{L}}(\mathcal{C})=0$.
\item $\mathcal{C} \neq \emptyset$ is clear, since $\frac{1}{3}, \frac{2}{3}, \frac{1}{4}$ und $\frac{3}{10}$ belong to $\mathcal{C}$.
\item The Cantor set is uncountable.
\item $\displaystyle x\in \mathcal{C}\Leftrightarrow x=\sum^\infty_{k=1}\frac{a_k}{3^k}; \; a_k \in \{0,2\} \Leftrightarrow x=\sum^\infty_{i=1}\frac{2}{3^{n_i}}$ for some $0< n_i < n_{i+1}$. This implies that the Cantor set has no interior points and that the characteristic is $$\displaystyle\chi_\mathcal{C}(x) = \prod^\infty_{n=0} \chi_\mathcal{J} (3^n x mod 1)$$, where $\chi_\mathcal{J}: [0,1]\rightarrow \RR$ is the characterstic function of $\mathcal{J}=[0,\frac{1}{3}]\cup [\frac{2}{3},1]$
\item Define the Devil's staircase like on the sheet. This converges uniformly to some $f:[0,1]\rightarrow [0,1]$ with certain probperties
\item compute the explicit actio fo f on the points in the Cantor set.
\item f is differentiable almost everywhere and $f^{'}$ is zero almost everywhere.
\end{enumerate}
}
\proof{
\begin{enumerate}[1)]
\item It is clear, that $\mathcal{C}$, as a subset of the bounded set [0,1], is itself bounded by the monotonicity of the Lebesgue measure. Since $c + \mathcal{J}$ and $c \mathcal{J}$ are again closed intervals for any closed interval $\mathcal{J}$, it follows by induction that $\mathcal{C}_n$ is closed for all n. Thus $\mathcal{S}_n$ as a complement of a closed set is open, thus giving the openness of $\mathcal{S}$ as a union of open sets and therefore we conclude that $\mathcal{C}$ is closed. By Heine-Borel and the fact that $\mathcal{C} \subset \RR$ we get that the Cantor set is therefore compact. \\\hfill \\
From the above argument it is also clear that $\mathcal{C}$ is also Borel-measurable as a countable union/intersection of open/closed and thus Borel-measurable sets.\\\\
We see immmediately from the definition of the Lebesgue measure (Recall:\\ $\mu_{\mathcal{L}}\left([a,b]\right) = b-a$), that $\mu_{\mathcal{L}}\left(c\mathcal{J}\right) = \mu_{\mathcal{L}}c\left(\mathcal{J}\right)$ and $\mu_{\mathcal{L}}\left(c+\mathcal{J}\right) = \mu_{\mathcal{L}}\left(\mathcal{J}\right)$ is clear by translational invariance of measures. Thus together with the disjointness of the involved sets and the resulting additivity of $\mu_\mathcal{L}$ we see that $$\mu_{\mathcal{L}}\left(\mathcal{C}_n\right)= \frac{1}{3}\mu_{\mathcal{L}}\left(\mathcal{C}_{n-1}\right) + \frac{1}{3}\mu_{\mathcal{L}}\left(\mathcal{C}_{n-1}\right) = \frac{2}{3}\mu_{\mathcal{L}}\left(\mathcal{C}_{n-1}\right)$$ $$\stackrel{Induction}{\Rightarrow} \mu_{\mathcal{L}}\left(\mathcal{C}_n\right)= \left(\frac{2}{3}\right)^n* \mu_{\mathcal{L}}\left(\mathcal{C}_0\right) = \left(\frac{2}{3}\right)^n$$ Thus:
$$\mu_{\mathcal{L}}\left(\mathcal{C}\right) =\mu_{\mathcal{L}}\left(\LN \bigcap_{n=1}^N\mathcal{C}_n\right) =\footnote{continuity from below of measures and the fact that $\mathcal{C}_n\subset\mathcal{C}_{n-1}$ (clear I hope)} \Ln \left(\mu_{\mathcal{L}}\left(\mathcal{C}_n\right)\right) = \Ln \left(\frac{2}{3}\right)^n = 0 $$ Thus the Cantor set is a Lebesgue null set.
\item After the first step $\frac{1}{3}$ and $\frac{2}{3}$ are left in the set $\mathcal{C}_1$ and they are points on the boundary of the set. Since the intervals removed in each step are open and completely contained in the interior of the set we know that both these numbers are left in the remaining set $\mathcal{C}$ and consequently the Cantor set is nonempty! \\
The numbers $\frac{1}{4}$ und $\frac{3}{10}$ are different in that they are not boundary points of any of the sets. $\frac{1}{4}$ is in the ``bottom `` third in the first step, in the ''top`` third in the second step and alternating like this the whole process (Similiary for $\frac{3}{10}$), which is the reason why they are not in the boundary.
\item You can show the noncountability by a diagonal argument. Let $s^{(n)}_m$ be a countable listing of all the sequences in $\mathcal{C}$, where n is the index of the sequence and m the index of its components. Now construct $s^0_{l} = \overline{s^m_m}$ where the bar means 2 if the component was 0 and 0 if the component was 2. This sequence is obviously not equal to any of the sequences in the listing. This is a contradiction and thus the Cantor set is uncountable.
\item Let $x \in \mathcal{C}$. Begin with the first step. Here x needs to lie in either the ''bottom`` or ''top`` interval. Since the corresponding borders $\frac{1}{3}$ and $\frac{2}{3}$ have a ternary expansion of 0.1 and 0.2 we know that $x\in [0,0.1]$ and $x \in [0.2,1]$. In the next step you are left with numbers laying in [0,0.01], [0.02, 0.1], [0.20, 0.21] and [0.22, 1]. If you repeat this process it is clear that there is a one to one correspondence between the x in the cantor set and x with a base-3 expansion consisting only of 0 and 2's.\\
The equivalence of the two sums seems obvious, just leave out the terms where $a_k = 0 $ and the terms that are left give you the $n_i$ needed. \\
Now assume the Cantor set had an interior point x. Since we established that it has the abovementioned representations, e.g. $x = \sum^\infty_{k=1} \frac{a_k}{3^k}$ it is obvious that for any $\epsilon$ given we can choose $N \in \NN$ with the property that $\frac{1}{3^N} < \epsilon$. Then choose $y = \sum^\infty_{k=1} \frac{b_k}{3^k}$, where $b_k = a_k;\; \forall k \neq N, \; b_{N+1} = 1 $ and then $\abs{x-y} < \epsilon$ and since $\epsilon$ is arbitrarily small x can't be interior. \\
Define $\chi_C(x) = \prod^\infty_{n=0} \chi_j (3^n x \mod 1); \chi_j $ the characteristic function of $[0,\frac{1}{3}]\cup[\frac{2}{3},1]$. This function is always zero if any $3^n x$ has a ternary representation starting with a one, which is equivalent to saying, that none of the coefficients in the base-3 expansion are allowed to be one. Thus this must be the characteristtic function of the cantor set.
\item Let $f_n$ be defined like on the sheet and let f denote its limit function. The continuity of $f_n$ is implied in the definition, for the linear parts join the constant parts. Also the function is monotone increasing or non decreasing (I think the sheet is wrong here), since the function only has constant parts of increasing heights and linear parts joining those. Now use that $f_{k+1} = f_k $ on the $\mathcal{I}^{(n)}_j $ and that hence $\norm{f_k - f_{k+1}} < 2^{-k}$(Difference of the constant levels). But this means that $\sum^\infty_{k=1}f_k - f_{k+1}$ is uniformly convergent on [0,1] and therefore $\{f_k\}$ converge uniformly on [0,1] as well. The continouity of the limit is clear as a uniform limit of continous functions. $ f(0) =0, f(1)=1$ and monotone increasing are clearly inhabited from the approximating functions. For the last statement use that $\mathcal{S} = \bigcup^\infty_{n=0} \mathcal{S}_n$ and then it is obvious that f(x) is constand on every interval removed.
\item Take the expansion of $x = \sum^\infty_{i=0} \frac{2}{3^{n_i}}\; \forall x \in \mathcal{C}$. Then $f(x) = \sum^\infty_{i=0} \frac{1}{2^{n_i}}$ thus the devil's staircase maps the Cantor set to the space of all $\{0,1\}$-sequences which can be uniquely identified with the binary expansions of all elements in [0,1] and thus with the continuum. Thus the Cantor set has the same cardinality as the continuum.
\item This is now trivial: f(x) is constant on all $\mathcal{I}^{(n)}_j$ and thus on $\mathcal{S}$, but this means that f is differentiable on $\mathcal{S}$ and $f^{'}$ is 0 in all those points. Now use a) and the fact that $\mu_{\mathcal{L}}(\mathcal{C} =0$ to see that the set of points that constitute exceptions to the claim actually have measure zero and thus the claim holds almost everywhere.
\item Define $\mu_\mathcal{C}([a,b])=\lim_{\epsilon\rightarrow 0} f(b-\abs{\epsilon}) - f(a+\abs{\epsilon}$ and extend it to the whole of [0,1]. Those limits exist because of the monotonicity of f. (I believe this is what was meant in the exercise) $\mu_{\mathcal{C}}(\{p\}) = 0$ for any one point set since f is continous by 5). If you now write $\mu^{'}_\mathcal{C}([a,b])=\int^b_a f^{'} (x) dx$. By Calculus we know that $\mu^{'}_\mathcal{C}$ and $\mu_\mathcal{C}$ coincide on the intervals and thus on all of [0,1]. The Radon Nikodym derivative is therefore just $f^{'}$ like intended. \\
Welldefinedness:
\begin{itemize}
\item $f(\emptyset) \stackrel{5)}{=} 0$
\item $\displaystyle f\left([0,1]\right) = f(1) - f(0) \stackrel{5)}{=} 1 -0 = 1$
\item $\displaystyle f\left(\bigcup^\infty_{i=1} A_i\right) =\footnote{Linearity of the integral} \sum^\infty_{i=1} f(A_i)$ $\forall A_i \cap A_j \cap \emptyset\; \forall j\neq i$
\end{itemize}
Still $\mu_{\mathcal{C}}([0,1]\mathcal{C}) = \mu_{\mathcal{C}}(\mathcal{S})=0$ by construction of f. Nevertheless we showed in 1) that the Cantor set has Lebesgue measure 0 but these two statements just mean $\mu_{\mathcal{C}}$ is singular continous with respect to $\mu_{\mathcal{L}}$.
\end{enumerate}
}
\end{document}