1.22

Assume to the contrary that there does exist an $f$, that's continuous everywhere, and $f = \chi_{[0,1]}$ a.e... (From here on out, let's just call it $\chi$.)

We have, $\forall \epsilon > 0,\hspace{0.25cm} \exists \delta > 0$ s.t. $|x - y| < \delta \hspace{0.25cm} \Rightarrow |f(x) - f(y)| < \epsilon$.

Choose $x = 1$ and $\epsilon = \frac{1}{4}$.

We know $\exists \delta > 0$ such that $|f(y) - f(1)| < \frac{1}{4}$ $\forall y \in (1,1+\delta) \hspace{0.25cm} (\dagger$).

Since $f$ is continuous everywhere, we have $\lim_{x \to 1}f(x) = 1 = f(1)$. It follows from $\dagger$ that: $$-\frac{1}{4} < f(y) - 1 < \frac{1}{4}$$ ...thus $$f(y) \in (\frac{3}{4},\frac{5}{4})$$ ...on $(1, 1+\delta)$. This is a contradiction, since $m\big((1, 1+\delta)\big)$ = $\delta$ > 0, and $\chi\big((1, 1+\delta)\big) = \lbrace 0 \rbrace$.

1.21

Let $f: \mathcal{C} \to [0,1]$ be the Cantor-Lebesgue function. It's continuous, and surjective. Certainly, for any interval in $\mathbb{R}$, $\exists \mathcal{N}$, non-measurable. (Simply scale the non-measurable set constructed on p.24.)

Since $f$ is continuous, and $\mathcal{N} \subset [0,1]$, the pre-image, $f^{-1}(\mathcal{N}) \subset \mathcal{C}$.

Notice: $$0 \leq m^*(f^{-1}(\mathcal{N})) \leq m^*(\mathcal{C}) = 0$$ ...and thus, $$m^*(f^{-1}(\mathcal{N})) = 0$$ ...and since all sets of outer-measure zero are measurable (by Property 2, in Section 1.3) $f^{-1}(\mathcal{N})$ is a measurable set. Thus, the continuous map $f$ maps a measurable set to a non-measurable set.

1.18

The exercise didn't specify anything about $f = \infty$ on sets with positive measure. I'll assume $f$ is a.e. finite. From here, it, of course, suffices to consider only non-negative measurable functions. For any such $f$, we need to find a sequence of continuous functions $\lbrace f_n \rbrace_{n = 1}^{\infty}$ such that the set: $$E = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} |f(x) - f_n(x)| > 0 \Big\rbrace$$ has measure zero. First, define the $d-$dimensional cubes: $$Q_n = [-n,n]\times \ldots \times[-n,n]$$ And the functions: $$g_n(x) = \min{\lbrace n,f(x) \rbrace}\chi_{Q_n}$$ It follows from Lusin's Theorem that for any of our $g_n$'s, we can find a set $F_n$ such that $g_n \chi_{F_n}$ is continuous, and $m(Q_n \backslash F_n)\leq \frac{1}{2^{n}}$. Now, we can simply define $f_n = g_n \chi_{F_n}$.

Lastly, define: $E_n = Q_n \backslash F_n$. Notice that: $$\sum_{n = 1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} \leq 1$$ Now, of course, we simply need to show that $E$ is $\sup{\lbrace E_n \rbrace}$. This actually isn't all that hard. First, notice: $$\bigcup_{n \geq k} E_n = \Big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \bigg| \hspace{0.25cm} |f(x) - f_n(x)| > 0, \hspace{0.25cm} n \geq k \Big\rbrace$$ Naturally, we arrive that the result that $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k} E_n$, and thus, by the Borel-Cantelli Lemma, $m(E) = 0$, as desired.

1.17

The goal of this exercise is to show that if we choose $c_n$'s as the hint prescribes, the set: $$E = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \lim_{n \to \infty} \Big| \frac{f_n(x)}{c_n} \Big| > 0 \Big\rbrace$$ has measure zero. Define: $$S_c = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} |f_n(x)| > \frac{c}{n} \Big\rbrace$$ Notice that $m(S_1) \leq 1$. Also, notice $S_{c+1} \subset S_{c} \hspace{0.25cm} \forall c$, and $\bigcap_{c=1}^{\infty}S_c = \emptyset$. Thus, by Corollary 3.3, $\lim_{n \to \infty} m(S_n) = 0$.

With that out of the way, notice that for the sets: $$E_n = \Big\lbrace x \in [0,1] \hspace{0.25cm} \bigg| \hspace{0.25cm} \Big| \frac{f_n(x)}{c_n} \Big| > \frac{1}{n} \Big\rbrace$$ ...we can choose $c_n$ large enough such that for every $n$, $m(E_n) \leq \frac{1}{2^n}$. Certainly, it follows that ($\dagger$): $$\sum_{n=1}^{\infty} m(E_n) \leq \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 < \infty$$ Now, observe that $E = \sup{ \lbrace E_n \rbrace }$, i.e. (recalling problem 1.16), $E = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k}E_n$. Thus, applying the Borel-Cantelli Lemma, we have that $$(\dagger) \Rightarrow m(E) = 0$$

1.16

This exercise is asking us to prove the Borel-Cantelli Lemma. In the measure theory settings, it states:

Suppose $\lbrace E_n \rbrace_{n=1}^{\infty}$ is a countable family of subsets of $\mathbb{R}^d$, and ($\dagger$) $$\sum_{k=1}^{\infty} m(E_k) < \infty$$ The set $E = \bigcap_{n=1}^{\infty} \bigcup_{k \geq n} E_k$ has measure zero.

Part a.) We first need to show $E$ is a measurable set. Recall that the measurable sets form a $\sigma$-algebra, $\mathcal{M}$, which is closed under countable unions and intersections.

Since each $E_k \in \mathcal{M}$, certainly $A_n = (\bigcup_{k \geq n} E_k) \in \mathcal{M} \hspace{0.25cm} \forall n$. It follows that since $E = \bigcap_{n=1}^{\infty} A_n$, that $E \in \mathcal{M}$.

Part b.) Assume to the contrary that $m(E) = \delta > 0$. Notice that if we define: $$S_N = \bigcap_{k=1}^N \bigcup_{n \geq k} E_n = \bigcup_{n \geq N} E_n$$ Then, certainly, $S_N \searrow E$, and $\forall N$, $$\delta \leq m(S_N) = m(\bigcup_{n \geq N} E_n) \leq \sum_{n=N}^{\infty} m(E_n)$$ Which certainly contradicts $\dagger$, completing the proof.

1.14

Part a.) First notice that for any set $E \subset \mathbb{R}$, any finite closed covering, $E \subset \bigcup_{k=1}^{N}I_k$, must be a closed set. For any $E \subset \mathbb{R}$, choose an arbitrary convergent sequence, $\lbrace x_n \rbrace \subset E$. By definition, $x_n \to x \in \overline{E}$. Notice also, that $\lbrace x_n \rbrace \subset E \Rightarrow \lbrace x_n \rbrace \subset \bigcup_{k=1}^{N}I_k$. Since $\bigcup_{k=1}^{N}I_k$ is closed, $x_n \to x \in \bigcup_{k=1}^{N}I_k$.

Thus, any finite closed covering of an arbitrary set $E$ must also contain $\overline{E}$. Therefore, $J_*{(E)} = J_*{(\overline{E})}$.

Part b.) Consider $\mathbb{Q} \cap [0,1]$. Certainly, $m(\mathbb{Q} \cap [0,1]) = 0$. However, $\overline{\mathbb{Q} \cap [0,1]} = [0,1]$. The result from part (a) gives $J_*{(\mathbb{Q} \cap [0,1])} = J_*{([0,1])} = 1$.

1.13

Part a.) Let $F \subset \mathbb{R}^{d}$ be an arbitrary closed set. First, define: $$O_n = \lbrace x \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} \mathbb{d}{(x,F)} < \frac{1}{n} \rbrace$$ Notice $\overline{F} = \bigcap_{n = 1}^{\infty} O_n$. Since $F$ is closed, $F = \overline{F}$. Thus $F \in F_\sigma$.

Part b.) There's a well-known corollary to Baire's Category Theorem that states:

$\dagger$ In a complete metric space, the intersection of any countable collection of dense $G_\delta$'s is again a dense $G_\delta$.

First notice that we can write $\mathbb{Q} = \bigcup_{n=1}^{\infty} \lbrace q_n \rbrace$, where $\lbrace q_n \rbrace_{n \in \mathbb{N}}$ is an enumeration of the rationals. Since each $\lbrace q_n \rbrace$ is closed, it's easy to see that $\mathbb{Q} \in F_\sigma$.

Next, observe that $\mathbb{Q} \in F_\sigma \Rightarrow \mathbb{Q}^c \in G_\delta$. Now it is clear from $\dagger$ that $\mathbb{Q}$ can not be $G_\delta$, since $\mathbb{Q} \cap \mathbb{Q}^c = \emptyset$, which is nowhere-dense. Thus, $\mathbb{Q}$ is $F_\sigma$ but not $G_\delta$, as desired.

Part c.) $$S = (\mathbb{Q} \cap (-\infty, 0]) \cup (\mathbb{Q}^c \cap (0, \infty))$$

1.11

Notice that we can construct $\mathcal{A}$ in a similar fashion to how we construct $\mathcal{C}$. Proceed as follows. Define: $$\mathcal{A}_1 = [0, 0.4) \cup [0.5, 1].$$ $$\mathcal{A}_2 = \bigcup_{k=0}^{10^1 - 1} 10^{-1}([0 + k, 0.4 + k) \cup [0.5 + k, 1 + k])$$ $$\vdots$$ $$\mathcal{A}_n = \bigcup_{k=0}^{10^{n-1}-1} 10^{-n + 1}([0 + k, 0.4 + k) \cup [0.5 + k, 1+k])$$ .

Now, just define $\mathcal{A} = \bigcap_{n=1}^{\infty} \mathcal{A_n}$. By corollary 3.3, since $\bigcap_{n=1}^{N} \mathcal{A_n} \searrow \mathcal{A}$ as $N \to \infty$, and $m(\mathcal{A_1}) = 0.9$, then: $$m(A) = \lim_{N \to \infty} m(\bigcap_{n=1}^{N} \mathcal{A_n}) = \lim_{N \to \infty} (0.9)^N = 0$$

1.10

Part a.) Choose an arbitrary $x \in [0,1]$. Clearly, the sequence $\lbrace f_n(x) \rbrace$ is monotonic-decreasing, and bounded below by $0$. Thus, $\lim_{n \to \infty} f_n(x)$ exists $\forall x \in [0,1]$. Simply define $f$ to be the point-wise limit of $f_n$.

Part b.) Recall from the construction of $\hat{\mathcal{C}}$ that: $$[0,1] \cap \hat{\mathcal{C}}^c = \bigcup_{k=1}^{\infty} \bigcup_{n = 1}^{3^{k-1}-1} O_n^k$$ Where the length of the open intervals $O_n^k$ depend on how we construct our specific $\hat{\mathcal{C}}$. Let's choose $x \in \hat{\mathcal{C}}$ arbitrarily. Define: $$S = \lbrace x \in [0,1] | x = \frac{b_{n,k} - a_{n,k}}{2}, O_n^k = (a_{n,k}, b_{n,k}) \rbrace$$ Notice, much like what we observed in (1.9), $\forall r > 0$, $B_r(x) \cap S \neq \emptyset$. Construct a sequence from $S$ in this fashion. (I.e. $\forall n$ choose any $x_n \in B_{\frac{1}{n}}(x) \cap S$.)

Certainly, we observe $x_n \to x$ as $n \to \infty$. However, by construction of $f$, notice $f(x_n) = 0$ for every $n$. Thus, $\lim_{n \to \infty} f(x_n) = 0 \neq f(x) = 1.$ Therefore $f$ must be discontinuous $\forall x \in \hat{\mathcal{C}}$.

Finally, the author wants you to observe that since $f_n$ is both Riemann integrable $\forall n$, and decreasing, that the sequence $\lbrace \int f_n \rbrace$ is both decreasing and bounded below (by, say, $0$). Therefore $\lim_{n \to \infty} \int f_n$ exists. However, we have $\int \lim_{n \to \infty} f_n = \int f$, but $f$ clearly isn't Reimann integrable, since it's discontinuous on $\hat{\mathcal{C}}$, a set with positive measure. Thus, the point-wise limit of a sequence of Riemann-integrable functions isn't necessarily Riemann integrable in general.

1.9

Consider a Cantor-like set $\hat{\mathcal{C}}$ such that $m(\hat{\mathcal{C}}) > 0$. Similar to how the hint prescribes, let $O$ be the union of all of the open intervals removed during the odd stages of the construction of $\hat{\mathcal{C}}$. Next, let $O'$ be the union of all of the open intervals removed during the even stages of the construction of $\hat{\mathcal{C}}$.

Notice $(\dagger)$: $$[0,1] \backslash \hat{\mathcal{C}} = O \cup O'$$ Now, given the construction of $\hat{\mathcal{C}}$, $\forall x \in \hat{\mathcal{C}}$, it's easy to see that $\forall r > 0, B_r{(x)} \cap O \neq \emptyset.$

Thus, since every element of $\hat{\mathcal{C}}$ is a limit point of $O$, (given $\dagger$), $\hat{\mathcal{C}} = \overline{O} \backslash O$. Thus, $m(\overline{O} \backslash O) > 0$, as desired.

1.6

$\dagger$ Note that it was simpler for me to first consider problem 1.7.

Since the Lebesgue measure is translation-invariant, it suffices to only consider balls centered at the origin. Theorem 1.4 gives that $B_1{(0)} = \bigcup_{k=1}^{\infty} Q_k$, where $\left\lbrace Q_n \right\rbrace_{n \in \mathbb{N}}$ is a countable collection of pairwise almost-disjoint cubes. Thus, $$v_d = m(B_1{(0)}) = \sum_{k=1}^{\infty} m(Q_k)$$ By $\dagger$, any dilation of the unit ball in $\mathbb{R}^d$ is equivalent to letting $\delta = (r, r, \ldots, r)$. It follows from the main argument in 1.7 that:

$m(B_r{(0)}) = m(\delta B_1{(0)}) = r^d m(B_1{(0)}) = r^d v_d$, as desired.

1.7

It suffices to show that $\forall \epsilon > 0, \exists \mathcal{O}$, open, such that $m^*(\mathcal{O}-\delta E) < \epsilon$. Before we move ahead, first define: $$\Delta = (\delta_1 \cdot \ldots \cdot \delta_d)$$ Recall that if $Q_k$ is a closed cube, $$Q_k = [a_1, b_1]\times[a_2,b_2]\times \ldots \times[a_d,b_d]$$ ...where $a_k < b_k \in \mathbb{R}$, and: $$m(Q_k) = \prod_{k = 1}^{d}(b_k - a_k)$$
$\dagger$ Observe that for any closed cube $Q$, $$\delta Q = [\delta_1 a_1, \delta_1 b_1]\times[\delta_2 a_2,\delta_2 b_2]\times \ldots \times[\delta_d a_d,\delta_d b_d]$$ $$\Rightarrow m(\delta Q) = \prod_{k = 1}^{d}\delta_k(b_k - a_k) = \Delta m(Q)$$
Next, observe that if $\mathcal{O}$ is open, $\delta \mathcal{O}$ is open. (This is somewhat trivial... just notice that $x \in \mathcal{O} \Rightarrow \exists r > 0$ s.t. $B_r{(x)} \subset \mathcal{O}$.

Now, just let $\tau = \min\left\lbrace \delta_1, \ldots, \delta_d \right\rbrace$, and observe that $x \in \mathcal{O} \Rightarrow \delta x \in \delta \mathcal{O}$ and $B_{\tau r}{(\delta x)} \subset \delta \mathcal{O}$.)

Next, since $E$ is measurable, $\forall \epsilon > 0, \exists \mathcal{O}$, open, such that $E \subset \mathcal{O}$, and $m^*(\mathcal{O}-E) < \frac{\epsilon}{\Delta}$. Certainly, $E \subset \mathcal{O} \Rightarrow \delta E \subset \delta \mathcal{O}.$

Lastly, notice: $$m^*(\delta \mathcal{O} - \delta E) = \inf \Big\lbrace \sum_{k=1}^{\infty} m(\delta Q_k) \hspace{0.25cm} \Big| \hspace{0.25cm} (\mathcal{O} - E) \subset \bigcup_{k=1}^{\infty} Q_k \Big\rbrace$$ ...which, by ($\dagger$): $$= \inf \Big\lbrace \Delta \sum_{k=1}^{\infty} m(Q_k) \hspace{0.25cm} \Big|\hspace{0.25cm} (\mathcal{O} - E) \subset \bigcup_{k=1}^{\infty} Q_k \Big\rbrace = \ldots$$
$$\ldots = \Delta m^*(\mathcal{O}-E) < \Delta \frac{\epsilon}{\Delta} = \epsilon$$ ...as desired.

1.5

Part a.) Since $E$ is compact, it's clear that $O_n \searrow E$, and $\forall n, m(O_n) < \infty$. Thus, it follows from corollary 3.3 that $m(E) = \lim_{n \to \infty} m(O_n)$.

Part bi.) Let $E = \bigcup_{k=1}^{\infty} [k - \frac{1}{2^{k+1}}, k + \frac{1}{2^{k+1}}].$ Notice that $E^c$ is clearly an open set, so $E$ must be closed. Certainly, $$m(E) = \sum_{k = 1}^{\infty} \frac{1}{2^k} = 1.$$

However, $\forall n > 5$, (I say 5, but it's probably 2 or 3... who cares, haha.) $$m(O_n) > \sum_{k = 1}^{\infty} \frac{2}{n} = \infty$$ .

Thus, $\lim_{n \to \infty} m(O_n) = \infty \neq m(E)$.

Part bii.) We see in a later exercise (1.9) that there exist bounded open sets $S$ such that $m(\overline{S} \backslash S) > 0.$ Since for any bounded open set $S$, $O_n \searrow \overline{S},$ certainly $\lim_{n \to \infty} m(O_n) \neq m(S).$

Test!

This is a test: $$\cos{(x + y)} = 1$$ This is also a test: $$\mu(E_n) = 0$$ Nice! We seem to be up and running!