Ch2. Integration Theory (Ex.01 ~ Ex.11)

1. Given a collection of sets $F_1, F_2, \cdots, F_n$, construct another collection $E_1, E_2, \cdots, E_N$ $(N = 2^n - 1)$ so that $E_i \cap E_j = \varnothing$ $(i \neq j)$, and $F_k = \bigcup_{i \in I_k} E_i$.

proof

(1) construction of $\{E_i\}_{i=1}^N$

For $i = 1, 2, \dots, 2^n - 1$, write $i$ as an $n$-digit binary number :
$i = (\hat{i}_n, \dots, \hat{i}_1)_2$

(ex) $n = 3 \Rightarrow N = 2^3 - 1 = 7 \Rightarrow 5 = 101_{(2)}$
Next, for $k = 1, \dots, n$, let $G_{i,k} = F_k$ if $\hat{i}_k = 1$ and $G_{i,k} = F_k^c$ if $\hat{i}_k = 0$.
Then, let $E_i = \bigcap_{k=1}^n G_{i,k}$.
Note that there is at least one $F_k$ such that $E_i \subset F_k$ (if not, $E_i = \bigcap_k F_k^c$ so $\hat{i} = 0 \Rightarrow$ contradiction).

(2) $E_i$, $E_j$ are disjoint ($i \neq j$)

If $i \neq j$, so there is at least one digit in binary number such that $\hat{i}_k \neq \hat{j}_k$ ($k = 1, \dots, n$).
Let $\hat{i}_k = 0$ and $\hat{j}_k = 1$ w.o.l.g. then
$E_i \subset F_k^c$ and $E_j \subset F_k$.
Each $E_i$ and $E_j$ are included in disjoint sets $(F_k^c, F_k)$ respectively, so $E_i \cap E_j = \varnothing$.

(3) $F_k = \bigcup_{i: \hat{i}_k = 1} E_i$ ($k = 1, \dots, n$)

Let the $E_i$ included in $F_k$ be denoted by $E_j^k$ ($j = 1, \dots, 2^{n-1}$)

$\Leftarrow$
Since $E_j^k \subset F_k$ for all $j$, $\bigcup_j E_j^k \subset F_k$ (trivial)
$\Rightarrow$
$x \in F_k$
Check whether $x \in E_i$ ($1 \le i \le 2^n - 1$), then $x \in E_I = \bigcap_{r=1}^n G_{I,r}$ for some $I$.
Since $G_{I,k} = F_k$, $E_I \subset F_k$ and $E_I = E_j^k$ for some $j$.
Thus, $x \in \bigcup_j E_j^k$.

(4) $\bigcup_{k=1}^n F_k = \bigcup_{i=1}^N E_i$

$\Leftarrow$
$x \in \bigcup_i E_i$, then there exists $I \in N$ such that $x \in E_I$ $(1 \le I \le N)$.
Then, there is $K \in N$ such that $1 \le K \le n$ and $E_I \subset F_K$.
So $x \in F_K$ and thus $x \in \bigcup_{k=1}^n F_k$.
$\Rightarrow$
$y \in \bigcup_{k=1}^n F_k$, so there exists $K \in N$ such that $y \in F_K$.
Due to (3), $F_K$ can be written as $\bigcup_{i \in I_K} E_i$.
Then, $y \in E_i$ for some $i$ $(1 \le i \le N)$.
This leads to $y \in \bigcup_{i=1}^N E_i$.

(5) conclusion

The collection $\{E_i\}_{i=1}^N$ satisfies
\[ E_i \cap E_j = \varnothing \ (i \ne j), \quad F_k = \bigcup_{i \in I_k} E_i, \quad \text{and} \quad \bigcup_{k=1}^n F_k = \bigcup_{i=1}^N E_i. \]

2. If $f$ is integrable on $\mathbb{R}^d$ and $\delta > 0$, then show that $\|f(\delta x) - f(x)\|_1 \to 0$ as $\delta \to 1$. (Let $F(x) = f(\delta x)$)

proof

(1) continuous function $g$ of compact support

For given $\varepsilon > 0$, there is a continuous function $g$ of compact support $K$ such that $\|f - g\|_1 < \varepsilon$.
Using this function, $f^{(\delta)} - f$ can be written as $(f^{(\delta)} - g^{(\delta)}) + (g^{(\delta)} - g) + (g - f)$.
Then, \[ \|f^{(\delta)} - f\|_1 \le \|f^{(\delta)} - g^{(\delta)}\|_1 + \|g^{(\delta)} - g\|_1 + \|g - f\|_1 \]

(2) $\|f^{(\delta)} - g^{(\delta)}\|_1$

\[ \|f(\delta x) - g(\delta x)\|_1 = \delta^d \|f(x) - g(x)\|_1 \]

$\delta^d$ is a continuous real function, so there is $\beta_0 > 0$ such that $|\delta - 1| < \beta_0$ implies $\delta^d < 1$.
For such $\delta$, $\|f^{(\delta)} - g^{(\delta)}\|_1 < \|f - g\|_1$.

(3) $\|g^{(\delta)} - g\|_1$

1) Continuous $g^{(\delta)}$

Let $\{x_n\}$ be a sequence in $\mathbb{R}^d$ such that $\lim_{n \to \infty} x_n = x$.
$g$ is continuous on $K$, so $\lim_{n \to \infty} g(x_n) = g(x)$ by definition.
Then, $\lim_{n \to \infty} g^{(\delta)}(x_n) = g^{(\delta)}(x)$ as well (PMA 4.2).
Note that $g$ is supported on some $B_R$, where $K \subset B_R$.

2) $g^{(\delta)}$ is supported on $\overline{B_{DR}}$

$g^{(\delta)}(x) = g(\delta x)$ is continuous on a compact set $\frac{1}{\delta} K \subset \overline{B_R}$.
Since $\frac{1}{\delta} \to 1$ as $\delta \to 1$, there is $D > 1$ such that $|\delta| \le D$ for all $\delta$ near 1.
Then, $g^{(\delta)}$ is supported on $D \cdot \overline{B_R} = \overline{B_{DR}}$.

3) $\|g_n - g\|_1 \to 0$

$g_n$ is a multiplication of measurable functions ($g, \chi_{B_{R_n}}$), so it is measurable.
$|g_n| \le |g|$ and $g$ is continuous on $K$, so $\exists M > 0$ such that $|g| \le M$ and thus $|g_n| \le M$.
$g_n$ is supported on $\overline{B_R}$ with finite measure.
\[ \lim_{n \to \infty} g_n = g \]
Thus, $\lim_{n \to \infty} \|g_n - g\|_1 = 0$ (BCT).

4) $\|g^{(\delta)} - g\|_1 \to 0$

$\lim_{\delta \to 1} g^{(\delta)} = g$ holds for all $\{x_n\}$ such that $\lim_{n \to \infty} \delta_n = 1$.
So, $\lim_{\delta \to 1} \|g^{(\delta)} - g\|_1 = 0$.
For given $\varepsilon > 0$, there is $\beta_\varepsilon > 0$ s.t. $|\delta - 1| < \beta_\varepsilon$ implies $\|g^{(\delta)} - g\|_1 < \varepsilon$.

(4) $\lim_{\delta \to 1} \|f^{(\delta)} - f\|_1 = 0$, conclusion

Let $\beta = \min(\beta_0, \beta_\varepsilon)$.
Then, \[ \|f^{(\delta)} - f\|_1 \le \|f^{(\delta)} - g^{(\delta)}\|_1 + \|g^{(\delta)} - g\|_1 + \|f - g\|_1 < 3\varepsilon \quad \text{for } |\delta - 1| < \beta, \] and this holds for all $\varepsilon > 0$.
Thus, \[ \lim_{\delta \to 1} \|f^{(\delta)} - f\|_1 = 0. \]

3. $f$ is integrable on $[-\pi, \pi]$ and extended to $\mathbb{R}$ by making it periodic of $2\pi$. Then, show that $\int_I f(x)\,dx = \int_{-\pi}^{\pi} f(x)\,dx$, where $I$ is any interval of length $2\pi$.

proof

(1)

\[\begin{align*} \int_0^{2\pi} f(x)\,dx - \int_{-\pi}^{\pi} f(x)\,dx &= \left\{ \int_0^{\pi} f(x)\,dx + \int_{\pi}^{2\pi} f(x)\,dx \right\} - \left\{ \int_{-\pi}^0 f(x)\,dx + \int_0^{\pi} f(x)\,dx \right\} \\ &= \int_{\pi}^{2\pi} f(x)\,dx - \int_{-\pi}^0 f(x)\,dx \\ &= \int_0^{\pi} f(x+\pi)\,dy - \int_0^{\pi} f(x)\,dx \quad (x = T + y) \\ &= \int_0^{\pi} f(y)\,dy - \int_0^{\pi} f(x)\,dx = 0 \end{align*}\]

(2)

Let $T = 2\pi$. Then \[ \int_0^{2\pi} f(x)\,dx = \int_a^{a+2\pi} f(x)\,dx \] for all $a \in \mathbb{R}$, and in particular \[ \int_0^{2\pi} f(x)\,dx = \int_{-\pi}^{\pi} f(x)\,dx \quad (a = -\pi). \] Thus, \[ \int_I f(x)\,dx = \int_{-\pi}^{\pi} f(x)\,dx \] for all intervals $I$ with length $2\pi$.

4. $f$ is integrable on $[0,b]$ and $g(x) = \int_0^x f(t)\,dt$ $(x \in (0,b])$. Show that $g$ is integrable on $[0,b]$ and $\int_0^b g(x)\,dx = \int_0^b f(t)(b-t)\,dt$.

proof

(1) $h(t) = b - t$ is measurable $(t > 0)$

If $a \le 0$, $\{t \mid h(t) < a\} = \varnothing$.
If $a > 0$, $\{t \mid h(t) < a\} = \{t \in \mathbb{R} : t > b - a\}$, which is open and consequently measurable.
Since $\{t \mid h(t) < a\}$ is a measurable set for all $a \in \mathbb{R}$, $h$ is a measurable function.

(2) $g$ is integrable on $[0,b]$

\[\begin{align*} \int_0^b |g(x)|\,dx &= \int_0^b \left| \int_0^x f(t)\,dt \right| dx \le \int_0^b \int_0^x |f(t)|\,dt\,dx. \end{align*}\]

$h(x,t) = |f(t)| \, \chi_{t < x}$ is measurable, and $|f(t)|$ is measurable (since $f$ is measurable),
so the sequence of the last integral can be changed (Thm 2.3.2(iii)):

\[\begin{align*} \int_0^b \int_0^x |f(t)|\,dt\,dx &= \int_0^b \int_t^b |f(t)|\,dx\,dt = \int_0^b |f(t)|(b - t)\,dt < \infty \end{align*}\]

(since $f$ is integrable).
Thus, $\int_0^b |g(x)|\,dx < \infty$, so $g$ is integrable on $[0,b]$.

(3) conclusion

\[\begin{align*} \int_0^b g(x)\,dx &= \int_0^b \int_0^x f(t)\,dt\,dx \quad \text{(note that $0 \le t \le x \le b$)} \\ &= \int_0^b \int_t^b f(t)\,dx\,dt \quad (\text{since } f, h \text{ are measurable, Thm 2.3.2}) \\ &= \int_0^b f(t)(b - t)\,dt. \end{align*}\]

5. $F$ is a closed set in $\mathbb{R}$, where $m(F) < \infty$. Let $\delta(x) = d(x, F) = \inf\{|x - z| : z \in F\}$, and $I(x) = \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy$.

(a) Show that $\delta$ is continuous

proof

① $|\delta(x) - \delta(y)| \le |x - y|$ for $x, y \in \mathbb{R}$

For given $\eta > 0$, there is $z \in F$ such that $|x - z| < \delta(x) + \eta$ (by definition of $\delta$).
Then, \[ \delta(x) \le |x - z| \le |x - y| + |y - z| < |x - y| + \delta(y) + \eta. \] So, $\delta(x) - \delta(y) < |x - y| + \eta$.
Switching $x$ and $y$ yields $\delta(y) - \delta(x) < |x - y| + \eta$.
Then, $|\delta(x) - \delta(y)| < |x - y| + \eta$ holds for all $\eta > 0$,
so $|\delta(x) - \delta(y)| \le |x - y| \quad (x, y \in \mathbb{R})$.

② $\delta$ is continuous

Then, for every $\varepsilon > 0$, $|\delta(x) - \delta(y)| < \varepsilon$ holds if $|x - y| < \varepsilon$.
Thus, $\delta$ is (uniformly) continuous on $\mathbb{R}$.

(b) Show that $I(x) = \infty$ for $x \in F^c$

proof

① $\delta(x) = 2\varepsilon > 0$

$d(x, F) = 0$ implies $x \in \overline{F} = F$.
So, $d(x, F) > 0$ if $x \notin F$.
Let $\delta(x) = 2\varepsilon > 0$.

② $\delta(y) > \varepsilon$

Consider the interval $|y - x| < \varepsilon$: $x - \varepsilon < y < x + \varepsilon$.
This leads to $|\delta(y) - \delta(x)| < \varepsilon$ due to (a).
Then, $\delta(y) > \delta(x) - \varepsilon = \varepsilon$ holds.

③ $I(x) = \infty$, conclusion

Then, \[ I(x) = \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy \ge \int_{x - \varepsilon}^{x + \varepsilon} \frac{\delta(y)}{|x - y|^2} \, dy \ge \varepsilon \int_{x - \varepsilon}^{x + \varepsilon} \frac{1}{|x - y|^2} \, dy. \] (Let $y - x = t \Rightarrow dy = dt$)

\[\begin{align*} I(x) &\ge \varepsilon \int_{-\varepsilon}^{\varepsilon} \frac{dt}{t^2} = \varepsilon \left[ -\frac{1}{t} \right]_{-\varepsilon}^{\varepsilon} = \varepsilon \left( \frac{1}{\varepsilon} + \frac{1}{\varepsilon} \right) = \infty. \end{align*}\]

Thus, $I(x) = \infty$ for $x \notin F$.

(c) Show that $I(x) < \infty$ for a.e. $x \in F$: $\int_F I(x)\,dx < \infty$.

proof

① $\delta(y)$ is a measurable function

We’ve shown in Ch1 Ex13 that $\{y \in \mathbb{R} : d(y,F) < a\} = \{y \in \mathbb{R} : \delta(y) < a\}$ is an open set in $\mathbb{R}$ for $a > 0$.
$\{y \mid \delta(y) < a\} = \varnothing$ when $a \le 0$, so $\{y \mid \delta(y) < a\}$ is a measurable set for all $a \in \mathbb{R}$.
So, $\delta$ is a measurable function on $\mathbb{R}$.

② $h(y) = \frac{1}{|x - y|^2}$ is a measurable function on $\mathbb{R}$

$\{y \mid h(y) < a\} = \varnothing$ when $a \le 0$.
When $a > 0$, $\{y \mid h(y) < a\} = \{y \in \mathbb{R} : \frac{1}{|x - y|^2} < a\} = \{y \in \mathbb{R} : |x - y| > \frac{1}{\sqrt{a}}\} = (x - \infty, x - \frac{1}{\sqrt{a}}) \cup (x + \frac{1}{\sqrt{a}}, \infty)$,
a union of open sets, which is open.
Since $\{y \mid h(y) < a\}$ is measurable for all $a \in \mathbb{R}$, $h(y) = \frac{1}{|x - y|^2}$ is a measurable function.

③ If $f(x)$ is measurable, then $f(x + b)$ is measurable ($b \in \mathbb{R}$)

$f_b(x) = f(x + b)$.
$\{x \mid f_b(x) < a\} = \{x \in \mathbb{R} : f(x + b) < a\} = \{y \in \mathbb{R} : f(y) < a\} - b = b + \{f(x) < a\}$.
Since $\{f(x) < a\}$ is measurable, $f_b(x)$ is measurable (translation invariance).
This holds for all $a > 0$, so $f_b$ is measurable.

④ $\int_F I(x)\,dx = \iint_{F \times \mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx$

Then, \[ \int_F I(x)\,dx = \int_F \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx. \] $\delta(y)$ is measurable (by ①), $\frac{1}{|x - y|^2}$ is also measurable (by ②, ③),
so their multiplication $\frac{\delta(y)}{|x - y|^2}$ is measurable and non-negative.
Hence, by Tonelli’s Theorem,
\[ \int_F \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx = \int_{\mathbb{R}} \int_F \frac{\delta(y)}{|x - y|^2} \, dx\,dy. \]

⑤ $\int_F \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx= \int_F \delta(y) \int_{\mathbb{R}} \frac{1}{|x - y|^2} \, dx\,dy$

When $y \in F$, $\delta(y) = 0$. So, \[ \int_F \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx = \int_F \delta(y) \int_{\mathbb{R}} \frac{1}{|x - y|^2} \, dx\,dy. \]

6.

(a) Show that there is a positive continuous function $f$ on $\mathbb{R}$ such that $f$ is integrable on $\mathbb{R}$ but $\limsup_{x \to \infty} f(x) = \infty$.

proof

$f(x)$ is defined as follows:

\[ f(x) = \begin{cases} e^x, & (-\infty < x < 0) \\ 1, & (0 \le x < 1) \\ \text{straight line from } (n - \frac{1}{n^2}, 1) \text{ to } (n, n), & (n - \frac{1}{n^2} \le x < n), \, n \ge 2 \\ \text{straight line from } (n, n) \text{ to } (n + \frac{1}{n^2}, 1), & (n \le x < n + \frac{1}{n^2}), \, n \ge 2 \\ \frac{1}{x^2}, & \text{(otherwise)} \end{cases} \]

It is obvious that $f$ is continuous on $\mathbb{R}$ (by construction).
$\int_{-\infty}^0 f(x)\,dx = \int_{-\infty}^0 e^x\,dx = 1$, and $\int_0^1 f(x)\,dx = 1$,
so it is enough to consider the case $x \ge 1$ for integrability.

The area between $f(x)$ and the $x$-axis when $x \ge 1$ is overlaid by two regions:
one of them is made by $y = \frac{1}{x^2}$, and the other is a triangle centered at $n \in \mathbb{N}$
with height $n$ and base $\frac{2}{n^2}$.
The area of each triangle is $\frac{1}{n^3}$, so

\[ \int_1^\infty f(x)\,dx \le \int_1^\infty \frac{1}{x^2}\,dx + \sum_{n=2}^\infty \frac{1}{n^3} < \infty. \]

To sum up, $\int_{-\infty}^\infty |f(x)|\,dx$ is finite, so $f$ is integrable.

However, $f(n) = n$ for $n \in \mathbb{N}$ while $f(x) = \frac{1}{x^2}$ for $x \in [n + \frac{1}{n^2}, n + \frac{1}{(n+1)^2})$ $(x \ge 1, n \ge 2)$.
So, $\limsup_{x \to \infty} f(x) = \infty$, while $\liminf_{x \to \infty} f(x) = 0$.

(b) $f$ is uniformly continuous on $\mathbb{R}$ and integrable, then show that $\lim_{|x| \to \infty} f(x) = 0$.

proof

① Assume $\limsup_{x \to \infty} f(x) \ne 0$

This proof will cover the case $\limsup_{x \to \infty} f(x) = 0$ first,
and then the other case $\liminf_{x \to \infty} f(x) = 0$ can be proved similarly.
To start with, let us assume that $\limsup_{x \to \infty} f(x) \ne 0$.

② $|f(x_k)| \ge 2\varepsilon$ for $k \in \mathbb{N}$

Since $\limsup_{x \to \infty} f(x) \ne 0$, for every non-negative sequence $\{y_n\}$ with $y_n \to \infty$,
there is $\varepsilon > 0$ such that for every $k \in \mathbb{N}$ there is $n_k \ge k$ with $|f(y_{n_k})| > 2\varepsilon$ (PMA 3.1).

Among those $n_k$’s with $y_{n_k} > 1$, choose $\{n_k\}$ with $y_{n_k} > y_{n_{k+1}} + 1$.
(If such $n_k$ does not exist, $y_{n_k} \le y_{n_{k+1}}$ for all $k$ so $\{y_{n_k}\}$ becomes a bounded sequence,
which is a contradiction to $y_n \to \infty$.)
Let $x_k = y_{n_k}$, then $|f(x_k)| \ge 2\varepsilon$ for $k \in \mathbb{N}$.

③ $|f(x)| > \varepsilon$

$f$ is uniformly continuous, so there is $\delta_\varepsilon > 0$ such that $|t - s| < \delta_\varepsilon$ implies $|f(t) - f(s)| < \varepsilon$.
Let $\delta = \min(\frac{1}{2}, \delta_\varepsilon)$. Then $|x - x_k| < \delta$ implies
$|f(x_k)| - |f(x)| < \varepsilon$, so $|f(x)| > |f(x_k)| - \varepsilon \ge 2\varepsilon - \varepsilon = \varepsilon$.

④ $\int_0^\infty |f(x)|\,dx = \infty$, conclusion.

Then, \[\int_0^\infty |f(x)|\,dx \ge \sum_{k=1}^\infty \int_{x_k - \delta}^{x_k + \delta} |f(x)|\,dx > \sum_{k=1}^\infty \varepsilon \cdot \int_{x_k - \delta}^{x_k + \delta} dx = \sum_{k=1}^\infty \varepsilon \cdot 2\delta = \infty.\]

This means $f$ is not integrable, so it is a contradiction.
Thus, $\lim_{x \to \infty} f(x) = 0$ for such $f$.

7. Let $\Gamma = \{(x, f(x)) : x \in \mathbb{R}^d\}$ be a graph of measurable function $f$ on $\mathbb{R}^d$. Show that $\Gamma$ is a measurable set and $m(\Gamma) = 0$.

proof

① Definition of $Q_n$, $G_n$, and $F_{k,n}$, with $f: Q_n \to \mathbb{R}$

Let $Q_n$ be a closed cube in $\mathbb{R}^d$ centered at the origin with side length $n \in \mathbb{N}$,
\[ G_n = \{(x, f(x)) : x \in Q_n\} \subset \Gamma, \] and \[ F_{k,n} = G_n \cap (Q_n \times [k, k+1)) = \{(x, f(x)) : x \in Q_n, \, k \le f(x) < k+1\} \subset G_n. \]

This proof starts with $F_{n,0} = G_n \cap (Q_n \times [0,1))$ to show that $m(F_{n,0}) = 0$.
Also, $f$ is restricted to a measurable function on $Q_n$ ($f: Q_n \to \mathbb{R}$).

② $m(G_n \cap (Q_n \times I_{i,\ell})) = \frac{1}{\ell} m(f^{-1}(I_{i,\ell}))$

For fixed $\ell \in \mathbb{N}$, let $I_{i,\ell} = [\frac{i}{\ell}, \frac{i+1}{\ell})$ $(i = 0, 1, \dots, \ell - 1)$.
This is a partition of $[0,1)$.

\[ f^{-1}(I_{i,\ell}) = \{x \in Q_n : f(x) \in I_{i,\ell}\}, \] and \[ G_n \cap (Q_n \times I_{i,\ell}) = \{(x, f(x)) : x \in Q_n, \, f(x) \in I_{i,\ell}\} = f^{-1}(I_{i,\ell}) \times I_{i,\ell}. \]

$f^{-1}(I_{i,\ell})$ is measurable, so $m(f^{-1}(I_{i,\ell}) \times I_{i,\ell}) = \frac{1}{\ell} m(f^{-1}(I_{i,\ell}))$ (Prop 2.3.6).
Thus, \[ m(G_n \cap (Q_n \times I_{i,\ell})) = \frac{1}{\ell} m(f^{-1}(I_{i,\ell})). \]

③ $m(F_{n,0}) = \frac{1}{\ell} m\left(\bigcup_{i=0}^{\ell-1} f^{-1}(I_{i,\ell})\right)$

\[ \bigcup_{i=0}^{\ell-1} (G_n \cap (Q_n \times I_{i,\ell})) = G_n \cap \bigcup_{i=0}^{\ell-1} (Q_n \times I_{i,\ell}) = G_n \cap (Q_n \times [0,1)) = F_{n,0}, \] and each $G_n \cap (Q_n \times I_{i,\ell})$ is disjoint w.r.t. $i$.
So, \[\begin{align*} m(F_{n,0}) &= \sum_{i=0}^{\ell-1} m(G_n \cap (Q_n \times I_{i,\ell})) \\ &= \frac{1}{\ell} \sum_{i=0}^{\ell-1} m(f^{-1}(I_{i,\ell})). \end{align*}\]

Each $f^{-1}(I_{i,\ell})$ is disjoint w.r.t. $i$ and measurable,
so \[ \frac{1}{\ell} \sum_{i=0}^{\ell-1} m(f^{-1}(I_{i,\ell})) = \frac{1}{\ell} m\left(\bigcup_{i=0}^{\ell-1} f^{-1}(I_{i,\ell})\right). \]

④ $A = \bigcup_i A_i$ ($A_i \subset \mathbb{R}$), then $f^{-1}(A) = \bigcup_i f^{-1}(A_i)$

$x \in f^{-1}(A) \Leftrightarrow f(x) \in A \Leftrightarrow \exists i \in \mathbb{N}$ s.t. $f(x) \in A_i \Leftrightarrow x \in f^{-1}(A_i) \Leftrightarrow x \in \bigcup_i f^{-1}(A_i)$.

⑤ $m(F_{n,k}) = 0$ for all $k \in \mathbb{Z}$

Using the result of ④,

\[ \frac{1}{\ell} m\left(\bigcup_{i=0}^{\ell-1} f^{-1}(I_{i,\ell})\right) = \frac{1}{\ell} m(f^{-1}([0,1))) \le \frac{1}{\ell} m(Q_n) \quad (\because f^{-1}([0,1)) = \{x \in Q_n : f(x) \in [0,1)\} \subset Q_n) = \frac{1}{\ell} n^d. \]

To sum up, $m(F_{n,0}) \le \frac{1}{\ell} n^d$ and this holds for all $\ell \in \mathbb{N}$.
So $m(F_{n,0}) = 0$.
Using the same logic, $F_{n,k}$ has measure 0 for all $k \in \mathbb{Z}$.

⑥ $\bigcup_{k \in \mathbb{Z}} F_{n,k} = G_n$

\[ \bigcup_{k \in \mathbb{Z}} F_{n,k} = \bigcup_{k \in \mathbb{Z}} G_n \cap (Q_n \times [k, k+1)) = G_n \cap (Q_n \times \mathbb{R}) = G_n. \]

⑦ $\Gamma = \bigcup_{n=1}^{\infty} G_n$

$x \in \bigcup_{n=1}^{\infty} G_n$:
$x \in G_n$ for some $n \in \mathbb{N}$.
Since $G_n \subset \Gamma$ for all $n \in \mathbb{N}$, $x \in \Gamma$.

Conversely, $x = (x, f(x)) \in \Gamma$, there is $N \in \mathbb{N}$ such that $-N < |x| < N$.
Hence $x \in G_N$, thus $x \in \bigcup_{n=1}^{\infty} G_n$.

⑧ $m(G_n) = 0 \Rightarrow m(\Gamma) = 0$

$F_{n,k}$ is disjoint w.r.t. $k$, so $m(G_n) = \sum_{k \in \mathbb{Z}} m(F_{n,k}) = 0$.

Thus, \[ m(\Gamma) = \lim_{n \to \infty} m(G_n) = \lim_{n \to \infty} 0 = 0 \quad (\because G_n \subset G_{n+1}, \, \forall n). \]

8. $f$ is integrable on $\mathbb{R}$, then show that $F(x) = \int_{-\infty}^x f(t)\,dt$ is uniformly continuous.

proof

For given $\varepsilon > 0$, there is $\delta > 0$ such that $m(E) < \delta$ implies $\int_E |f(t)|\,dt < \varepsilon$
(Prop 2.1.12(ii), absolute continuity).

Choose $x, y \in \mathbb{R}$ with $x < y$ and $x - y < \delta$. Then, \[ |F(y) - F(x)| = \left|\int_x^y f(t)\,dt\right| \le \int_x^y |f(t)|\,dt < \varepsilon \quad (\because |x - y| < \delta). \] This holds for all $\varepsilon > 0$, so $F(x)$ is uniformly continuous on $\mathbb{R}$.

9. (Chebyshev inequality) Let $f \ge 0$ and integrable. If $\alpha > 0$ and $E_\alpha = \{x : f(x) > \alpha\}$, show that $m(E_\alpha) \le \frac{1}{\alpha} \int f$.

proof

① $\alpha \cdot \chi_{E_\alpha}(x) \le f(x)$

If $x \in E_\alpha$, then $f(x) > \alpha$.
Since $\chi_{E_\alpha}(x) = 1$, $\alpha \chi_{E_\alpha}(x) < f(x)$.
If $x \notin E_\alpha$, then $\chi_{E_\alpha}(x) = 0$.
Since $f$ is non-negative, $f(x) \ge 0 = \alpha \cdot \chi_{E_\alpha}(x)$.

② Conclusion

Then, \[ \alpha \int \chi_{E_\alpha} = \alpha \cdot m(E_\alpha) \le \int f. \] Thus, \[ m(E_\alpha) \le \frac{1}{\alpha} \int f. \]

10. $f \ge 0$, $E_k = \{x : f(x) \ge 2^k\}$, and $F_k = \{x : 2^k < f(x) \le 2^{k+1}\}$ on $\mathbb{R}^d$.

(1) If $f < \infty$ a.e., then show that $\bigcup_{k=-\infty}^{\infty} F_k = \{x : 0 < f(x) < \infty\}$.

proof

① $\Rightarrow$ (trivial)

$\exists k \in \mathbb{Z}$ s.t. $x \in F_k$, then $2^k < f(x) \le 2^{k+1}$.
So $x \in \{x : 0 < f(x) < \infty\}$.

② $\Leftarrow$

There uniquely exists $y \in \mathbb{R}$ s.t. $f(y) = 2^x$, since $y \mapsto 2^y$ is 1–1 correspondence.
Then, there is $k \in \mathbb{Z}$ such that $k < x \le k+1$.
Thus, $y \in F_k$ and $y \in \bigcup_k F_k$.

(2) Show that $f$ is integrable iff $\sum_{k=-\infty}^{\infty} 2^k m(F_k) < \infty$.

proof

① $g \le f \le h$ for some $g, h$.

Let $g(x) = \sum_{k=-\infty}^{\infty} 2^k \chi_{F_k}(x)$.
When $x \in F_k$ ($k \in \mathbb{Z}$), $g(x) = 2^k$, while $2^k < f(x) \le 2^{k+1}$.
So $g(x) < f(x)$ and this holds for all $k \in \mathbb{Z}$.

Let $h(x) = \sum_{k=-\infty}^{\infty} 2^{k+1} \chi_{F_k}(x)$.
When $x \in F_k$, $h(x) = 2^{k+1}$ and $2^k < f(x) \le 2^{k+1}$.
So $f(x) \le h(x)$ and this holds for all $k \in \mathbb{Z}$.

② conclusion

When $\int f(x)\,dx < \infty$, \[ \int g(x)\,dx = \sum 2^k m(F_k) < \int f(x)\,dx < \infty. \] In reverse, \[ \sum 2^k m(F_k) \le \int g(x)\,dx = \frac{1}{2} \int h(x)\,dx < \infty \] means $\int f(x)\,dx < \infty$.
So $f$ is integrable.

(3) Show that $f$ is integrable iff $\sum_{k=-\infty}^{\infty} 2^k m(E_k) < \infty$.

proof

① $\phi$ with $f \le \phi \le 2f$

Let $\phi(x) = \sum_{k=-\infty}^{\infty} 2^k \chi_{E_k}(x)$.
When $2^k < f(x) \le 2^{k+1}$,
\[ \phi(x) = \sum_{i=-\infty}^k 2^i = 2^{k+1} - 1 < 2^{k+1}. \] So $f(x) \le \phi(x)$.
Since $2^k < 2f(x)$, $\phi(x) \le 2f(x)$.

② conclusion

If $\int f(x)\,dx < \infty$, then $\int \phi(x)\,dx < \infty$ ($\because \phi \le 2f$).
So $\sum 2^k m(E_k) = \sum 2^k \int \chi_{E_k}(x)\,dx < \infty$.
Conversely, $\sum 2^k m(E_k) < \infty$ means $\phi(x) < \infty$ and $f \le \phi$,
so $\int f(x)\,dx < \infty$.

(4) Let $f(x) = \begin{cases} |x|^{-\alpha}, & |x| \le 1 \\0, & \text{o.w.} \end{cases}$ Then $f$ is integrable on $\mathbb{R}^d$ if $\alpha < d$.

proof The case $\alpha \le 0$ is trivial, so the following will cover $\alpha > 0$.

① $E_k$ ($k = -1, k \ge 0$)

$f(x) \ge 1$ for all $x$ with $|x| \le 1$,
so $E_{-1} = \{x : |x| \le 1\}$ for $k = -1$.
Then $m(E_{-1}) = v_d$.

When $k \ge 0$, $|x|^{-\alpha} > 2^k$ is same as $|x| < 2^{-k/\alpha}$.
So $E_k = \{x : |x| < 2^{-k/\alpha}\}$ and
$m(E_k) = v_d \cdot 2^{-kd/\alpha}$.

② Conclusion

\[ \sum_{k=-\infty}^{\infty} 2^k m(E_k) = \sum_{k=-1}^{\infty} 2^k v_d + \sum_{k=0}^{\infty} 2^k v_d 2^{-kd/\alpha}. \]

$v_d \sum_{k=-1}^{\infty} 2^k = v_d$ is convergent.
The series $\sum_{k=0}^{\infty} 2^{k(1 - d/\alpha)}$ converges when $1 - \frac{d}{\alpha} < 0$, i.e. $\alpha < d$.
So $\alpha < d$ implies $\sum 2^k m(E_k)$ converges, so $f$ is integrable.
Reversely, if $\int f < \infty$, then $\sum 2^k m(E_k)$ converges and thus $\alpha < d$.

(5) Let $f(x) = \begin{cases}|x|^{-b}, & |x| \ge 1 \ (b > 0) \\0, & \text{o.w.}\end{cases}$ Then $f$ is integrable iff $b > d$.

proof

① $E_k$ ($k \ge 0, k \le -1$)

When $k \ge 0$, there is no $x \in \mathbb{R}^d$ such that $f(x) > 2^k$.
So $E_k = \varnothing$ and $m(E_k) = 0$.

When $k \le -1$, $|x|^{-b} > 2^k$ means $|x| < 2^{-k/b}$.
So $E_k = \{x : |x| < 2^{-k/b}\}$ and
$m(E_k) = v_d \cdot 2^{-kd/b}$.

② Conclusion

\[ \sum_{k=-\infty}^{\infty} 2^k m(E_k) = \sum_{k=-\infty}^{0} 2^k v_d 2^{-kd/b} = v_d \sum_{k=-\infty}^{0} 2^{k(1 - d/b)}. \]

This series converges if $(1 - \frac{d}{b}) > 0$, i.e. $b > d$.
So $b > d$ implies $\int f < \infty$.
Reversely, $\int f < \infty$ implies $b > d$.

11.

(1) $f$ is integrable on $\mathbb{R}^d$, real-valued, and $\int_E f \ge 0$ for every measurable set $E$. Then, show that $f(x) \ge 0$ a.e. $x \in \mathbb{R}^d$.

proof We’ll assume that $m(\{f < 0\}) > 0$ and yield a contradiction.

① $\{f < 0\} = \bigcup_{n=1}^{\infty} \{f \le -\frac{1}{n}\}$

If $x \in \{f \le -1\}$, then $x \in \{f \le -1\}$.
When $-1 < f(x) < 0$, there is $N \in \mathbb{N}$ such that $-\frac{1}{N} < f(x) \le -\frac{1}{N-1}$ $(N \ge 2)$.
Then, $x \in \{f \le -\frac{1}{N}\}$, so $x \in \bigcup_{n=1}^{\infty} \{f \le -\frac{1}{n}\}$.
If $x \in \bigcup_{n=1}^{\infty} \{f \le -\frac{1}{n}\}$, then there is $N \in \mathbb{N}$ such that $f(x) \le -\frac{1}{N} < 0$,
so $x \in \{f < 0\}$.

② $m(E_N) > 0$, conclusion

Using sub-additivity, \[ m(\{f < 0\}) \le \sum_{n=1}^{\infty} m(\{f \le -\tfrac{1}{n}\}) = \sum_{n=1}^{\infty} m(E_n). \] The measure of $\{f < 0\}$ is assumed to be positive,
so there is $N \in \mathbb{N}$ such that $m(E_N) > 0$.
(If not, $m(\{f < 0\}) \le \sum_{n=1}^{\infty} m(E_n) = 0$.)
Note that $E_N$ is a measurable set.

Then, \[ \int_{E_N} f(x)\,dx = \int f(x) \chi_{E_N}(x)\,dx \le -\tfrac{1}{N} \int \chi_{E_N}(x)\,dx = -\tfrac{1}{N} m(E_N) < 0, \] a contradiction.

(2) $\int_E f(x)\,dx = 0$ for every measurable set $E$, then show that $f = 0$ a.e. $x$.

proof

$\int_E f = 0$ means $\int_E f \ge 0$ and $\int_E f \le 0$ simultaneously.
$\int_E (-f) \ge 0$ means $(-f) \ge 0$ a.e. $x$, which is $f \le 0$ a.e. $x$.
Since $f \ge 0$ a.e. $x$, $f = 0$ a.e. $x$.

Ch2. Integration Theory (Ex.01 ~ Ex.11)

1. Given a collection of sets \(F_1, F_2, \cdots, F_n\), construct another collection \(E_1, E_2, \cdots, E_N\) \((N = 2^n - 1)\) so that \(E_i \cap E_j = \varnothing\) \((i \neq j)\), and \(F_k = \bigcup_{i \in I_k} E_i\).

(1) construction of \(\{E_i\}_{i=1}^N\)

(2) \(E_i\), \(E_j\) are disjoint (\(i \neq j\))

(3) \(F_k = \bigcup_{i: \hat{i}_k = 1} E_i\) (\(k = 1, \dots, n\))

(4) \(\bigcup_{k=1}^n F_k = \bigcup_{i=1}^N E_i\)

(5) conclusion

2. If \(f\) is integrable on \(\mathbb{R}^d\) and \(\delta > 0\), then show that \(\|f(\delta x) - f(x)\|_1 \to 0\) as \(\delta \to 1\). (Let \(F(x) = f(\delta x)\))

(1) continuous function \(g\) of compact support

(2) \(\|f^{(\delta)} - g^{(\delta)}\|_1\)

(3) \(\|g^{(\delta)} - g\|_1\)

1) Continuous \(g^{(\delta)}\)

2) \(g^{(\delta)}\) is supported on \(\overline{B_{DR}}\)

3) \(\|g_n - g\|_1 \to 0\)

4) \(\|g^{(\delta)} - g\|_1 \to 0\)

(4) \(\lim_{\delta \to 1} \|f^{(\delta)} - f\|_1 = 0\), conclusion

3. \(f\) is integrable on \([-\pi, \pi]\) and extended to \(\mathbb{R}\) by making it periodic of \(2\pi\). Then, show that \(\int_I f(x)\,dx = \int_{-\pi}^{\pi} f(x)\,dx\), where \(I\) is any interval of length \(2\pi\).

(1)

(2)

4. \(f\) is integrable on \([0,b]\) and \(g(x) = \int_0^x f(t)\,dt\) \((x \in (0,b])\). Show that \(g\) is integrable on \([0,b]\) and \(\int_0^b g(x)\,dx = \int_0^b f(t)(b-t)\,dt\).

(1) \(h(t) = b - t\) is measurable \((t > 0)\)

(2) \(g\) is integrable on \([0,b]\)

(3) conclusion

5. \(F\) is a closed set in \(\mathbb{R}\), where \(m(F) < \infty\). Let \(\delta(x) = d(x, F) = \inf\{|x - z| : z \in F\}\), and \(I(x) = \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\).

(a) Show that \(\delta\) is continuous

① \(|\delta(x) - \delta(y)| \le |x - y|\) for \(x, y \in \mathbb{R}\)

② \(\delta\) is continuous

(b) Show that \(I(x) = \infty\) for \(x \in F^c\)

① \(\delta(x) = 2\varepsilon > 0\)

② \(\delta(y) > \varepsilon\)

③ \(I(x) = \infty\), conclusion

(c) Show that \(I(x) < \infty\) for a.e. \(x \in F\): \(\int_F I(x)\,dx < \infty\).

① \(\delta(y)\) is a measurable function

② \(h(y) = \frac{1}{|x - y|^2}\) is a measurable function on \(\mathbb{R}\)

③ If \(f(x)\) is measurable, then \(f(x + b)\) is measurable (\(b \in \mathbb{R}\))

④ \(\int_F I(x)\,dx = \iint_{F \times \mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx\)

⑤ \(\int_F \int_{\mathbb{R}} \frac{\delta(y)}{|x - y|^2} \, dy\,dx= \int_F \delta(y) \int_{\mathbb{R}} \frac{1}{|x - y|^2} \, dx\,dy\)

6.

(a) Show that there is a positive continuous function \(f\) on \(\mathbb{R}\) such that \(f\) is integrable on \(\mathbb{R}\) but \(\limsup_{x \to \infty} f(x) = \infty\).

(b) \(f\) is uniformly continuous on \(\mathbb{R}\) and integrable, then show that \(\lim_{|x| \to \infty} f(x) = 0\).

7. Let \(\Gamma = \{(x, f(x)) : x \in \mathbb{R}^d\}\) be a graph of measurable function \(f\) on \(\mathbb{R}^d\). Show that \(\Gamma\) is a measurable set and \(m(\Gamma) = 0\).

8. \(f\) is integrable on \(\mathbb{R}\), then show that \(F(x) = \int_{-\infty}^x f(t)\,dt\) is uniformly continuous.

9. (Chebyshev inequality) Let \(f \ge 0\) and integrable. If \(\alpha > 0\) and \(E_\alpha = \{x : f(x) > \alpha\}\), show that \(m(E_\alpha) \le \frac{1}{\alpha} \int f\).

10. \(f \ge 0\), \(E_k = \{x : f(x) \ge 2^k\}\), and \(F_k = \{x : 2^k < f(x) \le 2^{k+1}\}\) on \(\mathbb{R}^d\).

(1) If \(f < \infty\) a.e., then show that \(\bigcup_{k=-\infty}^{\infty} F_k = \{x : 0 < f(x) < \infty\}\).

(2) Show that \(f\) is integrable iff \(\sum_{k=-\infty}^{\infty} 2^k m(F_k) < \infty\).

(3) Show that \(f\) is integrable iff \(\sum_{k=-\infty}^{\infty} 2^k m(E_k) < \infty\).

(4) Let \(f(x) = \begin{cases} |x|^{-\alpha}, & |x| \le 1 \\0, & \text{o.w.} \end{cases}\) Then \(f\) is integrable on \(\mathbb{R}^d\) if \(\alpha < d\).

(5) Let \(f(x) = \begin{cases}|x|^{-b}, & |x| \ge 1 \ (b > 0) \\0, & \text{o.w.}\end{cases}\) Then \(f\) is integrable iff \(b > d\).

11.

(1) \(f\) is integrable on \(\mathbb{R}^d\), real-valued, and \(\int_E f \ge 0\) for every measurable set \(E\). Then, show that \(f(x) \ge 0\) a.e. \(x \in \mathbb{R}^d\).

(2) \(\int_E f(x)\,dx = 0\) for every measurable set \(E\), then show that \(f = 0\) a.e. \(x\).