1.34

I'll prove such a function exists by actually constructing one.

Let $F:[0,1] \to [0,1]$ such that:

$F(x) = \Bigg\lbrace \begin{array}{cc} f(x) & x \in \mathcal{C}_\epsilon & \\ g(x) & x \in [0,1]\cap \mathcal{C}_\epsilon^c \\ \end{array}$

Our mapping $f(x):\mathcal{C}_\epsilon \to \mathcal{C}_\gamma$ is constructed by simply reassinging $x \in \mathcal{C}_\epsilon$'s binary expansion to $\mathcal{C}_\gamma$'s. To do this, simply observe that for any $x$ in $\mathcal{C}_\epsilon$,

$x = \sum_{n=1}^{\infty} \alpha_n \epsilon^n$ where $\hspace{0.25cm} \alpha_n \in \lbrace 0, 2 \rbrace$, and is unique to $x$.

Thus,

$f(x) = \sum_{n=1}^{\infty} \alpha_n \gamma^n$

Next, we need to construct $g(x)$. These functions are going to be continuous (linear) functions that map open disjoint intervals to corresponding open disjoint intervals.

First, lets define the set $\beta_{n,k}$ as the "binary representation of the integer $k$ with a string of length $n$". For example: $$\beta_{5,3} = (0,0,1,0,1)$$ Next, augment $\beta$ with a $1$ on the rightmost-side of the string. Call this set $\phi_{n,k}$. I.e. $$\phi_{5,3} = (0,0,1,0,1,1)$$ Now, let $\phi_{n,k}(i)$ be the $i^{th}$ element of the string, from the left.
(I.e. $\phi_{5,3}(3) = 1$.)

Moving forward, we'll use $\phi_{n,k}$ to define the sets $\mathcal{O}_{n,k}^\epsilon$ (and similarly for $\mathcal{O}_{n,k}^\gamma$) such that: $$ \bigcup_{n = 0}^\infty \bigcup_{k = 0}^{2^n - 1} \mathcal{O}_{n,k}^\epsilon = \hspace{0.25cm} [0,1]\cap \mathcal{C}_\epsilon^c $$ Without further ado, $$ \mathcal{O}_{n,k}^\epsilon = \Bigg( \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big)-\epsilon \Big(\frac{(1-\epsilon)}{2}\Big)^n, \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big) \Bigg)$$ While the expressions are regrettably unpleasant, notice that for each $n,k, \epsilon$, it successfully reproduces the correct $(k+1)^{th}$ open interval for the $(n+1)^{th}$ generation of an $\epsilon$-cantor set. For example, we know that the second interval in the third generation of the standard ($\epsilon = \frac{1}{3}$) Cantor set is $\big(\frac{7}{27}, \frac{8}{27}\big)$. We can easily check...

First, lets calculate: $$\sum_{i=1}^{4} \Big(\frac{(\frac{2}{3})^{i-1}(\frac{4}{3})}{2^i}\phi_{3,2}(i)\Big) = \frac{2}{9} + \frac{2}{27} = \frac{8}{27}$$ With that out of the way, it's easy to check now that: $$ \mathcal{O}_{2,1}^{\frac{1}{3}} = \Big(\frac{7}{27}, \frac{8}{27} \Big) $$ So... it works! This was constructed so that the collection of all $\mathcal{O}_{n,k}^\epsilon$'s are pairwise-disjoint. From here, it's actually easy to define a linear function from $\mathcal{O}_{n,k}^\epsilon$ $\to$ $\mathcal{O}_{n,k}^\gamma$. The slope from one set to the other is (obviously) $\big(\frac{\gamma}{\epsilon}\big)^{n+1}$. The $y$-intercept $b_{n,k}$ is fairly straightforward to calculate... (just pick the middle points of the corresponding intervals, and evaluate.)

Thus, we can finally fully define $g$ as follows: $$g(x) = \sum_{n=0}^{\infty} \sum_{k = 0}^{2^n - 1} \Big(\big(\frac{\gamma}{\epsilon}\big)^{n+1} x + b_{n,k}\Big)\chi_{\mathcal{O}_{n,k}^\epsilon} $$ Finally, we have our function $F(x) = f(x) + g(x)$, which satisfies all three properties outlined in the exercise.

1.33

Recall two things. First, that $m^*(E) = 0 \hspace{0.25cm} \Rightarrow E \in \mathcal{M}$. Second, recall from problem 32, part a, that any measurable subset to a non-measurable set must have measure zero. Thus, since $\mathcal{N}^c \cup \mathcal{N} = [0,1]$, it suffices to show that $m(\mathcal{N}^c) = 1$.

Assume to the contrary that for some $\epsilon \in (0,1)$, $\hspace{0.25cm} \mathcal{N}^c \subset U \subset [0,1]$, measurable, such that $m(U) = 1 - \epsilon$. It directly follows that $([0,1] \cap U^c) \subset \mathcal{N}$. This is a contradiction, because $m \big(([0,1] \cap U^c) \big) = \epsilon > 0$.

Therefore, $m^*(\mathcal{N}) + m^*(\mathcal{N}^c) > m(\mathcal{N} \cup \mathcal{N}^c)$.

1.28

Given $E \subset \mathbb{R}$, $m^*(E) > 0$, assume to the contrary that for any open interval $I$, $m^*(E \cap I) < \alpha m(I)$.

By the definition of the outer measure, we know that $\forall \epsilon > 0$ $\exists \lbrace Q_k \rbrace_{k=1}^\infty$ such that $E \subset \bigcup_{k=1}^\infty Q_k$, and: $$\sum_{k=1}^\infty |Q_k| \leq m^*(E) + \epsilon$$ Next, define $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ to be open intervals s.t. $Q_k \subset \mathcal{O}_k$ for all $k$, and $m(\mathcal{O}_k)) = m(Q_k) + \frac{\epsilon}{2^k}$. (Note, this can always be done by adding $\frac{\epsilon}{2^{k+1}}$ to each side of the interval $Q_k$.)

We constructed $\lbrace \mathcal{O}_k \rbrace_{k=1}^\infty$ with the following property in mind ($\dagger$): $$\sum_{k=1}^\infty m(\mathcal{O}_k) \leq \big(\sum_{k=1}^\infty m(Q_k)\big) + \epsilon \leq m^*(E) + 2\epsilon $$ Now, just observe that $E \subset \bigcup_{k=1}^\infty \mathcal{O}_k$, so, by ($\dagger$): $$m^*(E) \leq \sum_{k=1}^\infty m^*(E \cap \mathcal{O}_k) \leq \alpha \sum_{k=1}^\infty m(\mathcal{O}_k) \leq \alpha(m^*(E) + 2\epsilon) $$ So finally, we finally arrive at the result: $$ m^*(E) \leq \alpha m^*(E) $$ Which, of course, gives that $m^*(E) = 0$... a contradiction.

1.27

Since $E_1$ and $E_2$ are compact sets in $\mathbb{R}^d$, certainly $E_2 \backslash E_1$ is a bounded and measurable. For any $t > 0$, The sets $(E_2 \backslash E_1) \cap \overline{B_{t}(0)}$ are also bounded and measurable. For simplicity's sake, from here on out, let's say:

$$S_t = (E_2 \backslash E_1) \cap \overline{B_{t}(0)}$$ Define $f(t) = m(S_t)$. If I can prove $f$ is a continuous function, we're done --(by the intermediate value theorem).

Side note: Notice the function $g(t) = \alpha t^d$ is continuous, but not uniformly continuous. ($\dagger$)

Let $0 \leq \tau < t$. Notice $|f(t) - f(\tau)| = |m(S_t) - m(S_\tau)|$.

Certainly, $S_\tau \subset S_t$, thus:.

$$|m(S_t) - m(S_\tau)| = m(S_t) - m(S_\tau) = m(S_t \backslash S_\tau)$$ Now, notice that $$(S_t \backslash S_\tau) \subset \overline{B_t(0)} \backslash \overline{B_\tau(0)}$$ so: $$m(S_t \backslash S_\tau) \leq m(\overline{B_t(0)} \backslash \overline{B_\tau(0)}) \leq \alpha(d)\big(t^d - \tau^d)$$ where $\alpha(d) = \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}$, the volume of the $d$-dimensional unit ball. (Recall we proved the relation between $B_1$ and $B_t$ in problem 6 of this chapter.) Thus, by $\dagger$, $$|t - \tau| < \delta \Rightarrow \alpha(d)\big(t^d - \tau^d) \leq \epsilon(t)$$ Therefore, $f$ is a continuous function, and by the IVT, given any $c$ with $a < c < b$, we can find a $t^* > 0$ such that $m(E) = m\big(E_1 \cup S_{t^*}\big) = c$, where $E_1 \cup S_{t^*}$ is compact (by construction).

1.23

First, $\forall x \in \mathbb{R}$ and $\forall n \in \mathbb{N}$ define $x_{r,n} = \frac{r}{2^n}$ where $x \in \big(\frac{r}{2^n}, \frac{r+1}{2^n} \big]$.

Define $f_n(x,y) = f(x_{r,n},y)$. Are these $f_n$'s measurable? Observe that: $$ \lbrace (x,y) \in \mathbb{R}^2 \big| f_n(x,y) > \alpha \rbrace $$ $$ = \bigcup_{r = -\infty}^{\infty} \bigg(\frac{r}{2^n}, \frac{r+1}{2^n} \bigg] \times \lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace $$ Certainly, the intervals $\big(\frac{r}{2^n}, \frac{r+1}{2^n} \big]$ are measurable, and further, the sets $\lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace$ are measurable, given that $f(x,y)$ is continuous in the $y$-direction for any fixed $x$. Thus, their product is measurable. Lastly, since the measurable sets form a $\sigma$-algebra, we have that $f_n$ must be a measurable function $\forall n \hspace{0.25cm} \in \mathbb{N}$.

Next, if we can confirm that $f_n \to f$ as $n \to \infty$ on $\mathbb{R}^2$, we're done (given property 4 of measurable functions on p. 29.)

Fix an arbitrary $(x,y) \in \mathbb{R}^2$. Since $f$ is continuous in both directions, we have that since $x_{r,n} \to x$ as $n \to \infty$, that $f_n(x,y) = f(x_{r,n},y) \to f(x,y)$ as $n \to \infty$. Thus, for any $(x,y)$ in $\mathbb{R}^2$, $f_n$ converges to $f$ point-wise.

Thus, $f$ is measurable on $\mathbb{R}^2$.

1.26

It's fairly easy to see that since $B = A \cup (B \backslash A)$, that since both $A$ and $B$ are measurable, $m(B) = m(A) + m(B \backslash A)$. Thus, $m(B \backslash A) = 0$. Similarly, we observe $m^{*}(E \backslash A) = 0$, and thus $E \backslash A$ is measurable, with measure zero.

Certainly, $E = A \cup (E \backslash A)$. Thus, since $A$ is measurable, and the union of any two measurable sets is measurable, we have $E$ must be measurable.

1.24

First, break the integers into two countably infinite subsets: $$S = \lbrace n \in \mathbb{N} \hspace{0.25cm} | \hspace{0.25cm} n = 2^{m+1} \hspace{0.25cm} \forall m \in \mathbb{N} \rbrace$$ and $$\mathbb{N} \backslash S$$ Next, let $\lbrace a_n \rbrace_{n \in S}$ be an enumeration of $\mathbb{Q} \cap [0,1]^c$, and $\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}$ be an enumeration of $\mathbb{Q} \cap [0,1]$.

Certainly, the set $\lbrace r_n \rbrace_{n \in \mathbb{N}} = \big(\lbrace a_n \rbrace_{n \in S}\big) \bigcup \big(\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}\big)$ enumerates $\mathbb{Q}$.

Notice the set: $$A = \bigcup_{n=1}^{\infty} \Big( r_n - \frac{1}{n}, r_n + \frac{1}{n} \big) = T \cup V$$ where $$T = \bigcup_{n=1}^{\infty} \Big( a_n - \frac{1}{n}, a_n + \frac{1}{n} \big)$$ and $$V = \bigcup_{n=1}^{\infty} \Big( b_n - \frac{1}{n}, b_n + \frac{1}{n} \big)$$ Observe that $m(A) \leq m(T) + m(V)$, however, by construction, both $m(T) \leq 3$ and $m(V) \leq 2$. (Those numbers, 2 and 3 are both just "safe" over-estimates.) Thus, since $m(A)$ is finite, we have that $A^c$ must be non-empty.