Ch1. Measure Theory (Ex.19 ~ Ex.38)

19. \(A + B\) is defined as follows: \(A + B = \{a + b : a \in A \text{ and } b \in B \}\) (\(\subset \mathbb{R}^d\))

(a) If \(A\) or \(B\) is open, show that \(A + B\) is open.

proof

① \(B_r(a)\), \(B_r(x)\)

Let \(A\) be an open set (w.o.l.g.)
For \(x = a + b \in A + B\) such that \(a \in A\) and \(b \in B\), there is \(r > 0\) such that \(B_r(a) \subset A\)
Then \(B_r(a + b) = \{ y + b : y \in B_r(a) \}\)

② \(B_r(x) \subset A + B\)

For \(z \in B_r(x)\), \(z = y + b\) where \(y \in B_r(a)\)
Since \(B_r(a) \subset A\), \(y \in A\) as well
Then, \(z = y + b \in A + B\)
This (\(z \in A + B\)) holds for all \(z \in B_r(x)\), so \(B_r(x) \subset A + B\)
So, \(x\) is an interior point of \(A + B\)
This holds for all \(x \in A + B\), so \(A + B\) is open

(b) \(A\) and \(B\) are closed, then show that \(A + B\) is \(F_\sigma\)

proof

① \(K\) is compact and \(B\) is closed, then \(K + B\) is closed.

  1. \(z_n = k_n + b_n \in K + B\), \(k_n \to k \in K\)
    Consider a sequence \(\{ z_n \} \subset K + B\), where \(\lim_{n \to \infty} z_n = z \in \mathbb{R}^d\), \(z_n = k_n + b_n\) and \(k_n \in K\), \(b_n \in B\)
    \(\{ k_n \}\) is a sequence in a compact set \(K\), then there is a subsequence \(\{ k_{n_k} \}\) such that \(\lim_{k \to \infty} k_{n_k} = k \in K\) (PMA 3.6(a))

  2. \(z \in K + B\), so closed
    It is obvious that \(z_{n_k} \to z\) (\(\because z_n \to z\))
    Then \(\lim_{k \to \infty} b_{n_k} = \lim_{k \to \infty} (z_{n_k} - k_{n_k}) = z - k\)
    \(B\) is closed, so \(z - k \in B\)
    This means \(z = k + (z - k) \in K + B\), so a limit of a sequence in \(K + B\) is in \(K + B\)
    This holds for all \(\{ z_n \} \subset K + B\), so \(K + B\) is closed

② \(\bigcup_n K_n + B = \bigcup_n (K_n + B)\)

(\(\Rightarrow\)) \(x \in \bigcup_n K_n + B\), so \(x = k + b\) where \(k \in \bigcup_n K_n\) and \(b \in B\)
Then there exists an integer \(N\) such that \(k \in K_N\)
Naturally, \(k \in K_N + B\)
Thus, \(x \in \bigcup_n (K_n + B)\)

(\(\Leftarrow\)) \(x \in \bigcup_n (K_n + B)\), so \(x = k + b\) where \(k \in K_n\) and \(b \in B\)
Naturally, \(k \in \bigcup_n K_n\)
Thus, \(k + b \in \bigcup_n K_n + B\)

③ \(A + B\) is measurable, conclusion.

A closed set \(A\) can be expressed as \(\bigcup_n K_n\), where \(K_n\) is compact
Since \(A + B = \bigcup_n (K_n + B)\) and \(K_n + B\) is closed, \(A + B\) is a countable union of closed sets, which is measurable (\(F_\sigma\))

cf) \(F\) is closed iff every convergent sequence in \(F\) converges to a point in \(F\)

(c) Show that \(A + B\) might not be closed even though \(A\) and \(B\) are closed.

proof

① \(A = -\mathbb{N}\), \(B = \{ n + \frac{1}{n+1} : n \in \mathbb{N} \}\) are closed.

\(\{ (n+1) + \frac{1}{n+2} \} - \{ n + \frac{1}{n+1} \} = 1 - \frac{1}{n+1} + \frac{1}{n+2} - 1 = \frac{1}{n+2} - \frac{1}{n+1}\)
So, for all \(\epsilon\) with \(0 < \epsilon < \frac{1}{n+1} - \frac{1}{n+2}\), \(B_\epsilon(n + \frac{1}{n+1})\) does not contain any point of \(B\)
So, \(n + \frac{1}{n+1}\) is not a limit point of \(B\)
This means \(B\) has no limit point, so \(B\) is closed

It is trivial that \(A\) is closed

② \(0 \notin A + B\), so not closed.

Then, \(\{ -n + n \}_{n \in \mathbb{N}} \subset A + B\) and \(0\) is the limit point of \(K\)
However, \(0 \notin A + B\) (\(\because\) there is no such \(m, n \in \mathbb{N}\) that \(-m + n + \frac{1}{n+1} = 0\))
Since the limit of \(\{ -n + n \}_{n \in \mathbb{N}} \subset A + B\) is not in \(A + B\), \(A + B\) is not closed

20. Show that there exist closed sets \(A, B\) such that \(m(A) = m(B) = 0\) and \(m(A + B) > 0\)

(a) In \(\mathbb{R}\), let \(A = C\) (Cantor set) and \(B = \frac{1}{2}C\). Show that \([0,1] \subset A + B\)

① binary expansion

Every element in \(A\) can be represented as a binary expansion with 0 and 2. Then, every element in \(B\) is a binary expansion with 0 and 1.

② \([0,1] \subset A + B\)

For \(x \in [0,1]\), the ternary expansion of \(x\) is \(0.x_1x_2x_3\dots\), where \(x_n = 0,1,\text{or }2\). Let \(a_n = 2\) when \(x_n = 2\), and \(a_n = 0\) o.w. Also, let \(b_n = 1\) when \(x_n = 1\), and \(b_n = 0\) o.w. Then, \(x = a + b\), where \(a = 0.a_1a_2a_3\dots\) and \(b = 0.b_1b_2b_3\dots\). Since \(a \in A\) and \(b \in B\), \(x \in A + B\). Thus, \([0,1] \subset A + B\) and \(m(A + B) \geq 1\). Note that \(m(A) = m(B) = 0\).

(b) \(A = I \times \{0\}\) and \(B = \{0\} \times I\) (\(I = [0,1]\)), then \(A + B = I \times I\)

proof

\(x \in A + B \Leftrightarrow x = a + b\) with \(a = (a_1, 0)\) and \(b = (0, b_1)\) (\(a_1, b_1 \in I\)) \(\Leftrightarrow x = (a_1, b_1) \Leftrightarrow x \in I \times I\). Note that \(m(A) = m(I)m(\{0\}) = 0\), \(m(B) = 0\), and \(m(I \times I) = (m(I))^2 = 1\) (Prop 2.3.6).

21. Show that there is a continuous function that maps a Lebesgue measurable set to a non-measurable set

proof

Q.32(b) says a set \(G\) in \(\mathbb{R}\) has \(m(G) > 0\), then it has a non-measurable subset.
Let \(f\) be a Cantor-Lebesgue function which is continuous.
If \(C\) is a Cantor set, \(f(C)\) is compact (\(\because f\) is continuous).
\(f(C)\) is measurable, so there is a subset \(E \subset f(C)\) that is non-measurable.
Since \(f^{-1}(E) \subset C\) and \(m(C) = 0\), \(f\) maps \(f^{-1}(E)\) (measure 0) to non-measurable \(E\).

22. Show that there is no continuous function \(f\) on \(\mathbb{R}^1\) such that \(f(x) = \chi_{[0,1]}(x)\) for a.e. \(x \in \mathbb{R}^1\).

proof

① \(U = \{x \in \mathbb{R} : f(x) \neq \chi_{[0,1]}(x)\}\)

The value of \(\chi_{[0,1]}(x)\) is only 0 and 1.
Then, the set \(\{x \in \mathbb{R} : f(x) \neq 0 \text{ and } f(x) \neq 1\} = \{x : f(x) \neq \chi_{[0,1]}(x)\}\). Let \(U\) has measure zero.
Assume that \(f\) is continuous.

② \(U\) is open

\(U\) can be expressed as follows: \(U = f^{-1}((-\infty, 0) \cup (0,1) \cup (1, \infty))\).
\(f\) is continuous and \((-\infty,0) \cup (0,1) \cup (1, \infty)\) is open, so \(U\) is open. (PMA 4.8)

③ \(m(U) > 0\), contradiction (\(\because U \neq \emptyset\))

Each point of \(U\) is an interior point \(p\), so \(U\) contains an open interval \(I\) centered at \(p\) with positive radius \(r\).
Since \(I \subset U\) and \(|I| = 2r\), \(m(U) \geq m(I) = 2r > 0\), which is a contradiction.

④ \(\nexists\) continuous \(f\) s.t. \(f(x) = \chi_{[0,1]}(x)\) a.e.

23. \(f(x, y)\) is a separately continuous function on \(\mathbb{R}^2\); for fixed \(x\), \(f\) is continuous on \(y \in \mathbb{R}^1\), and vice versa. Show that \(f\) is measurable.

proof

① def of \(f_n\), \(\{f_n > a\}\)

Let \(f_n(x, y) = f(\frac{k}{2^n}, y)\), where \(\frac{k}{2^n} \leq x < \frac{k+1}{2^n}\), \(n \in \mathbb{N}\) and \(k \in \mathbb{Z}\).
For \(a \in \mathbb{R}\), \(\{f_n \geq a\} = \bigcup_{k=0}^{\infty} \{(x, y) : \frac{k}{2^n} \leq x < \frac{k+1}{2^n}, f(\frac{k}{2^n}, y) \geq a\}\).
\(\{x \in \mathbb{R} : \frac{k}{2^n} \leq x < \frac{k+1}{2^n}\} \times \{y \in \mathbb{R} : f(\frac{k}{2^n}, y) \geq a\}\).

② \(X_k\) is measurable

\(X_k\) can be expressed as \(\{x : \frac{k}{2^n} \leq x \leq \frac{k+1}{2^n}\} \cap \{\frac{k+1}{2^n}\}^c\).
\(\{x : \frac{k}{2^n} \leq x \leq \frac{k+1}{2^n}\}\) is closed, and \(\{\frac{k+1}{2^n}\}^c\) is a complement of closed set.
Then, \(X_k\) is an intersection of measurable sets, which is measurable.

③ \(Y_k\) is measurable

For fixed \(\frac{k}{2^n}\), \(f\) is continuous of \(y \in \mathbb{R}\).
Since \(\{f_n : f_n \geq a\}\) is open, its inverse \(Y_k\) is open, which is measurable.

④ \(f_n\) is measurable

By prop 2.3.6, \((X_k \times Y_k)\) is measurable.
Also, \(\{f_n \geq a\}\) is a countable union of measurable sets \((X_k \times Y_k)\), so \(f_n \geq a\) is measurable.
This holds for all \(a \in \mathbb{R}\), so \(f_n\) is measurable.

⑤ \(\forall \epsilon, \exists \delta\) s.t. \(|x - x_0| < \delta \Rightarrow |f(x, y) - f(x_0, y)| < \epsilon\)

For fixed \(y \in \mathbb{R}\), \(f(x, y)\) is continuous at \(x \in \mathbb{R}^1\).
Then, for given \(\epsilon > 0\), there is \(\delta_\epsilon > 0\) such that \(|x - x_0| < \delta_\epsilon\) implies \(|f(x_0, y) - f(x, y)| < \epsilon\).

⑥ \(f_n \to f\) p.w.

There is an integer \(N_\epsilon\) such that \(n \geq N_\epsilon\) implies \(\frac{1}{2^n} < \delta_\epsilon\).
When \(\frac{k}{2^n} \leq x < \frac{k+1}{2^n}\) with such \(n \geq N_\epsilon\), \(|\frac{k}{2^n} - x| < \delta_\epsilon\).
Putting \(\frac{k}{2^n}\) into \(x_0\), \(n \geq N_\epsilon\) implies \(|f_n(x, y) - f(x, y)| < \epsilon\).
This holds for all \(\epsilon > 0\), so \(f_n \to f\) p.w.

⑦ conclusion

\(f\), the limit of measurable functions, is measurable.

24. Does there exist an enumeration \(\{r_n\}\) of the rationals such that the complement of \(\bigcap_{n=1}^{\infty} (r_n - \frac{1}{n}, r_n + \frac{1}{n})\) in \(\mathbb{R}\) is non-empty?

proof

Let \(\alpha\) be some irrational and \(\{q_n\}\) be a sequence of all rationals. It is obvious that for \(n \in \mathbb{N}\) there exists a rational \(q\) such that \(|q - \alpha| \geq \frac{1}{n}\). Using this property, a sequence of rationals \(\{p_n\}\) can be made.

Let \(n_1 = \min\{n \in \mathbb{N} : |q_n - \alpha| \geq \frac{1}{1}\}\), \((p_1 = q_{n_1})\), and \(N_1 = \mathbb{N} - \{n_1\}\). Provided that \(n_m, p_m\) and \(N_m\) are defined \((m \geq 1, m \in \mathbb{N})\), let \(n_{m+1} = \min\{n \in N_m : |q_n - \alpha| \geq \frac{1}{m+1}\}\), \(p_{m+1} = q_{n_{m+1}}\), and \(N_{m+1} = N_m - \{n_{m+1}\}\). After repeating this inductively, a sequence of rationals \(\{p_n\}\) is made, with the property that \(|p_n - \alpha| \geq \frac{1}{n}\) for all \(n \in \mathbb{N}\).

Thus,

\[ \alpha \in \bigcap_{n=1}^{\infty} (p_n - \frac{1}{n}, p_n + \frac{1}{n})^c = \left( \bigcup_{n=1}^{\infty} (p_n - \frac{1}{n}, p_n + \frac{1}{n}) \right)^c. \]

25. Show that \(E\) is measurable if for every \(\varepsilon > 0\) there exists a closed set \(F\) such that \(F \subset E\) and \(m^*(E - F) < \varepsilon\).

proof

Let \(\varepsilon > 0\) be fixed. \(F\) is closed, so \(F^c \overset{def}{=} O\) is open. Then, \(O - E^c = O \cap E = E \cap F^c = E - F\). So \(m_k(O - E^c) = m_k(E - F) < \varepsilon\). This holds for all \(\varepsilon > 0\), so \(E^c\) is measurable. Thus, its complement \(E\) is measurable.

26. Suppose \(A \subset E \subset B\), where \(A, B\) are measurable sets of finite measure. If \(m(A) = m(B)\), show that \(E\) is measurable (and \(m(E) = m(A)\)).

proof

① \(m(B - A) = 0\)

\(B\) is measurable and \(A^c\) is measurable, so \(B - A = B \cap A^c\) is measurable. Since \(m(B) = m(A) + m(B - A)\), \(m(B - A) = 0\).

② \(m(E - A) = 0\), conclusion

Since \(E - A \subset B - A\), \(m_k(E - A) \leq m_k(B - A) = 0\), so \(E - A\) is measurable with measure 0. Thus, \(E = A \cup (E - A)\) is a union of measurable sets, which is measurable, and \(m(E) = m(A) + m(E - A) = m(A)\) (\(\because A \cap (E - A) = \varnothing\)).

27. Suppose \(E_1\) and \(E_2\) are compact sets in \(\mathbb{R}^d\) with \(E_1 \subset E_2\), and let \(a = m(E_1)\) and \(b = m(E_2)\). Show that for any \(c\) with \(a < c < b\), there is a compact set \(E\) with \(E_1 \subset E \subset E_2\) and \(m(E) = c\).

proof

① For a compact set \(K\) with \(m(K) = \mu\) and \(\xi \in \mathbb{R}\) with \(0 < \xi < \mu\), there is a compact set \(K'\) such that \(K' \subset K\) and \(m(K') = \xi\).

  1. \(f(x) = m(K \cap B_{x}(0))\)
    Let \(f(x) = m(K \cap B_{x}(0)) \quad (x \geq 0)\).
    It is obvious that \(f(0) = 0 \quad (\because \text{measure of a point})\).
    \(K\) is bounded, so there is a positive number \(M > 0\) such that \(K \subset B_{M}(0)\).
    Let \(M_0\) be the minimum of such \(M\), then \(f(x) = \mu\) for all \(x \geq M_0\).

  2. \(f\) is continuous
    \(m(B_{x}(0)) = \nu_d x^d\) is continuous on \(\mathbb{R}^d\), so for given \(\varepsilon > 0\) there is \(\delta_{\varepsilon} > 0\) such that \(|x - x'| < \delta_{\varepsilon}\) implies \(|m(B_{x}(0)) - m(B_{x'}(0))| < \varepsilon\).
    Consider the case \(x \geq x'\). Then, \[ \begin{aligned} |f(x) - f(x')| &= m(K \cap B_{x}(0)) - m(K \cap B_{x'}(0)) \\ &= m\big( (K \cap B_{x}(0)) \setminus (K \cap B_{x'}(0)) \big) \\ &= m\big( K \cap (B_{x}(0) \setminus B_{x'}(0)) \big) \\ &\leq m\big( B_{x}(0) \setminus B_{x'}(0) \big) \\ &< \varepsilon. \end{aligned} \] The opposite case \(x' > x\) also leads to \(|f(x) - f(x')| < \varepsilon\).
    Thus, \(|x - x'| < \delta_{\varepsilon}\) implies \(|f(x) - f(x')| < \varepsilon\).
    This holds for all \(\varepsilon > 0\), so \(f\) is continuous.

  3. Intermediate Value Thm, conclusion
    \(f(0) = 0\), \(f(M_0) = \mu\), and \(f\) is continuous.
    Thus, there is \(x_{\xi}\) such that \(x_{\xi} \in (0, M_0)\) and \(f(x_{\xi}) = \xi\).
    Note that \(K \cap B_{x_{\xi}}(0)\) is compact, since \(K\) and \(B_{x_{\xi}}(0)\) are compact.
    Thus, \(K' := K \cap B_{x_{\xi}}(0)\).

② There is a compact \(K_0 \subset (E_2 \cap O^{c})\), and thus \(E = E_1 \cup K\) : conclusion

\(E_1\) is measurable, so there is an open set \(O\) such that \(E_1 \subset O\) and \(m(O - E_1) < \varepsilon\) for given \(\varepsilon > 0\).
When \(0 < c \leq b - a\), \(0 \leq m(E_1 \cap O^{c}) < b - a\).
Considering \(E_1 \cap O^{c}\) is compact (\(\because O^{c}\) is closed and bounded),
there is a compact set \(K_0\) such that \(K_0 \subset (E_2 \cap O^{c})\) and \(m(K_0) = c - a \quad (0 < c - a < b - a)\).
Let \(E = E_1 \cup K_0\), which is compact and disjoint.
Thus, \(m(E) = a + (c - a) = c\).

28. Let \(E\) be a subset of \(\mathbb{R}\) with \(m_{*}(E) > 0\). Show that for every \(\alpha\) with \(0 < \alpha < 1\), there is an open interval \(I_{\alpha}\) such that \(m_{*}(E \cap I_{\alpha}) \geq \alpha \cdot m_{*}(I_{\alpha})\).

proof

① Choose \(O\) s.t. \(E \subset O\)

Using the property \(m_{*}(E) = \inf\{m(O): E \subset O\}\), some open set \(O\) can be chosen so that \(m_{*}(O) < m_{*}(E) + \varepsilon\) for some \(\varepsilon > 0\).

② Assume \(\Rightarrow m_{*}(E) < \alpha \cdot m_{*}(O)\)

Assume that none of the intervals satisfies the inequality: \(m_{*}(E \cap I) < \alpha \cdot m_{*}(I)\) for all open intervals in \(\mathbb{R}\).
Since \(O = \bigcup_{n} I_{n}\) where \(I_{n} \cap I_{m} = \varnothing\) \((m \neq n)\),

  • \(E = E \cap O = E \cap \left( \bigcup_{n} I_{n} \right) = \bigcup_{n} (E \cap I_{n})\)
  • \(m_{*}(E) \leq \sum_{n} m_{*}(E \cap I_{n}) < \alpha \sum_{n} m_{*}(I_{n})\)

Each \(I_{n}\) is disjoint, so \(\sum_{n} m_{*}(I_{n}) = m(O)\).
Then, \(m_{*}(E) < \alpha \cdot m_{*}(O)\).

③ Let \(\varepsilon = (1 - \alpha) \cdot m_{*}(O)\), contradiction

\[ \begin{aligned} m_{*}(E) &> m_{*}(O) - \varepsilon \\ &= m_{*}(O) - (1 - \alpha) \cdot m_{*}(O) \\ &= \alpha \cdot m_{*}(O) \end{aligned} \] \(\Rightarrow m_{*}(E) > \alpha \cdot m_{*}(O)\), which is contradiction.

④ \(\therefore \exists I_{\alpha}\) such that \(m_{*}(E \cap I_{\alpha}) \geq \alpha \cdot m_{*}(I_{\alpha})\)

29. \(E\) is a measurable set of \(\mathbb{R}\) with \(m(E) > 0\). The difference set of \(E\), denoted by \(E - E\), is defined as \(E - E = \{x - y: x, y \in E\}\). Show that \(E - E\) contains an open interval centered at the origin.

proof

① Express \(E\) using Q. 28.

Let \(E_{0}\) be a measurable subset with \(m(E_{0}) > 0\).
Then, there is an open interval \(I\) such that \(m(E_{0} \cap I) \geq \alpha \cdot m(I)\) for every \(\alpha \in (0,1)\) by Q. 28.
Let \(E = E_{0} \cap I\), which is still measurable.
The following statements are based on the assumption that \(E - E\) has no open interval centered at zero.

② Only 0 exists around it

It is obvious that \(0 \in (E - E)\) (since \(x \in E\), \(x - x = 0 \in (E - E)\)).
The elements of \(E - E\) is symmetric; if \(z \in E - E\), then \(-z \in E - E\) as well.
So there is no open interval containing 0 even not centered at 0.
So only 0 exists around it; there exists \(R > 0\) such that \(B_{R}(0) \cap (E - E) = \varnothing\) for all \(r < R\).

③

\(E \cap (E + \delta) = \varnothing \quad \forall 0 < \delta_1 < R, x \in E \ \& \ x \in E + \delta_1,\)
\(x - \delta \in E, \ \& \ x \in E \Rightarrow \delta_1 \in E - E \Rightarrow\) contradiction.

④

\(E\) and \(E + \delta\) are disjoint for some \(\delta\).
So, \(E \cap (E + \delta) = \varnothing\) for all \(\delta \in \mathbb{R}\) with \(0 < \delta < R\).

⑤ \(E \cup (E + \delta) \subset I \cup (I + \delta) \Rightarrow\) contradiction

Choose \(\delta < \min(|\delta_1, (2\alpha - 1)|I|)\) where \(\frac{1}{2} < \alpha < 1\).
\(E \subset I\) (since \(E = E_{0} \cap I\)) and \(E + \delta \subset I + \delta\),
so \((E \cup (E + \delta)) \subset (I \cup (I + \delta))\).
\(I \cup (I + \delta)\) is an open interval with length \(|I| + \delta\),
and \(E \cap (E + \delta) = \varnothing\), so \(|I| + \delta \geq 2 \cdot m(E)\).
Since \(m(E) \geq \alpha \cdot |I|\) in Q28, then \(\delta \geq (2\alpha - 1)|I|\).
This is a contradiction because \(\delta < (2\alpha - 1)|I|\).

⑥ \(\therefore E - E\) has an open interval centered at origin.

30. \(E\) and \(F\) are measurable sets in \(\mathbb{R}\), and \(0 < m(E), m(F) < \infty\). Show that \(E + F\) contains an open interval.

proof

① \(O = \{x \in \mathbb{R} : g(x) > 0 \}\), then \(I \subset O\)

Let \(g(x) = \chi_{E} * \chi_{F}(x) = \int_{\mathbb{R}} \chi_{E}(x + t) \chi_{F}(t) dt\), which is continuous by ch2 ex 24.
Then, the set \(O = \{x \in \mathbb{R}: g(x) > 0 \}\) is open.

\[ \begin{aligned} \int_{\mathbb{R}} g(x) dx &= \int_{\mathbb{R}} \int_{\mathbb{R}} \chi_{E}(x + t) \chi_{F}(t) dt dx \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \chi_{E}(x + t) dx \cdot \chi_{F}(t) dt \quad (\text{by Fubini's Thm}) \\ &= m(E) \int_{\mathbb{R}} \chi_{F}(t) dt \quad (\text{by translation invariance}) \\ &= m(E) m(F) > 0 \end{aligned} \]

\((m(g) > 0 \Rightarrow O \neq \varnothing)\)
This means \(m(O) > 0\). So, there is an open interval \(I \subset O\).

② \(O \subset E + F\)

Consider \(x \notin E + F\). Assume there is \(y \in (-E + x) \cap F\).
Then \(y = f\) for some \(f \in F\) and \(y = -e + x\) for some \(e \in E\).
Then \(x = e + f\) and \(x \in E + F\), contradiction.
This leads to \((-E + x) \cap F = \varnothing\).

So, for \(\forall x \in E\), \(x - t \notin E\).
Also, for all \(t\) with \(x - t \in E\), \(x - (x - t) = t \notin F\).
Then, \(\chi_{E}(x - t) \chi_{F}(t) = 0\) for \(x \notin E + F\) and \(g(x) = \int_{\mathbb{R}} \chi_{E}(x - t) \chi_{F}(t) dt = 0\), which means \(x \notin O\).
Thus, in converse, \(x \in O\) implies \(x \in E + F\), so \(O \subset E + F\).

③ \(\exists I \subset E + F\), conclusion

In summary, \(I \subset O \subset E + F\). Thus, \(E + F\) has an open interval \(I\).

31. Let \(V'\) denote a set that consists of only one element in each equivalence class of \(\mathbb{R} / \mathbb{Q}\) in \(\mathbb{R}\). Show that \(V'\) is non-measurable using Q29.

proof

① \(V'_n = V' + g_n\), then \(V'_n \cap V'_m = \varnothing\) for \(n \neq m\)

Assume that \(V'\) is measurable.
Then \(V' + g\) has same measure as \(V'\) for any \(g \in \mathbb{Q}\).
Let \(V'_n = V' + g_n\), where \(\{g_n\}\) is an enumeration of \(\mathbb{Q}\).
If \(y \in V'_n \cap V'_m\), then \(y = x_1 + g_n = x_2 + g_m\) for some \(x_1, x_2 \in V'\).
This leads to \(x_1 - x_2 = g_m - g_n \in \mathbb{Q}\), contradiction to def of \(V'\).
So \(V'_n \cap V'_m = \varnothing\).

② \(\bigcup V'_n = \mathbb{R}\), so \(m(V') > 0\)

\(V'_n \subset \mathbb{R}\) for all \(n \in \mathbb{N}\), so it is obvious that \(\bigcup V'_n \subset \mathbb{R}\).
Let \(x \in \mathbb{R}\). Then there is a rational \(g_n\) such that \(x - g_n \in V'\).
This means \(x \in V' + g_n = V'_n\), so \(x \in \bigcup V'_n\).
So, \(\bigcup V'_n = \mathbb{R}\).
If \(m(V') = m(V'_n) = 0\), \(m(\bigcup V'_n) = \sum m(V'_n)\) (since \(V'_n\) are disjoint) \(= 0\).
Then, \(m(\mathbb{R}) = 0\), which is a contradiction.
So \(m(V') > 0\).

③ \(V'\) has an open interval, contradiction.

Then, Q29 says \(V'\) contains an open interval centered at the origin, say \(B(0, r)\) where \(r\) is the radius of the interval.
Due to the denseness of \(\mathbb{Q}\), there exists a rational \(g\) with \(0 < g < r\).
This is a contradiction, because \(0\) and \(g\) are in \(V'\) and \(0 \nsim g\).

④ \(\therefore V'\) is non-measurable.

32. Let \(V\) denote the non-measurable set in \([0,1]\) (Vitali set).

(a) Show that if \(E\) is a measurable subset of \(V\), then \(m(E) = 0\).

proof

Let \(E_g = E + g\), where \(g\) is a rational with \(-1 \leq g \leq 1\).
Since \(E \subset [0,1]\), \(E_g \subset [-1,2]\) and this holds for all \(g \in \mathbb{Q} \cap [-1,1]\).
Then \(\bigcup_{g \in \mathbb{Q} \cap [-1,1]} E_g \subset [-1,2]\).
As previously mentioned, \(E_g \cap E_{g_m} = \varnothing\) for \(g_n \neq g_m\).
This leads to \(\sum m(E_g) < 3\).
If \(m(E) = m(E_g) > 0\), \(\sum m(E_g)\) becomes infinite, which is contradiction.
Thus, \(m(E) = 0\).

(b) \(G\) is a subset of \(\mathbb{R}\) with \(m(G) > 0\). Show that there is a subset of \(G\) that is non-measurable.

proof

① \(m(G) \leq \sum_{n=1}^\infty m(G \cap V'_n)\)

We’ll use the same notation as #31: \(V'\), \(V'_n\).
Also, \(\bigcup V'_n = \mathbb{R}\).
\(G\) can be written as \(G \cap \mathbb{R} = G \cap (\bigcup V'_n) = \bigcup (G \cap V'_n)\).
By the sub-additivity, \(m(G) \leq \sum m(G \cap V'_n)\).

Assume that \(G \cap V'_n\) is measurable for all \(n \in \mathbb{N}\).

② \(G \cap V'_n\) has no open interval centered at \(0\).

Suppose \(G \cap V'_n\) has \(B(0,r)\).
By the denseness of \(\mathbb{Q}\), there exists \(g \in \mathbb{Q}\) with \(0 < g < r\).
This means \(0\) and \(g\) are in \(G \cap V'_n\) simultaneously, but \(G \cap V'_n\) can have only one rational at most (\(G \cap V'_n \subset V'_n = V' + g_n\)).
This is impossible, so \(G \cap V'_n\) can’t have an open interval centered at \(0\).

③ \(m(G \cap V'_n) = 0\) and

By #29, \(m(G \cap V'_n) = 0\) and \(m(G) = 0\) (\(m(G) \leq \sum m(G \cap V'_n)\)), which is a contradiction.
Thus, there is \(N \in \mathbb{N}\) such that \(G \cap V'_N\) is non-measurable.

cf) If every subset of \(G\) is measurable, then \(m(G) = 0\) (converse)

33. Show that \(V^c = I - V\) satisfies \(m_k(V^c) = 1 \ (I = [0,1])\) and thus \(m_k(V \cup V^c) < m_k(V) + m_k(V^c)\)

proof

① \(V^c\) is non-measurable

Assume that \(V^c\) is measurable, then \(V\) is measurable, contradiction.

② \(m_k(V^c) = 1\)

Since \(V^c \subset I\), \(m_k(V^c) \leq 1\). Assume that \(m_k(V^c) < 1\).
Choose a measurable set \(U\) such that \(V^c \cup U \subset I\) and \(m_k(U) = \varepsilon\) (\(0 < \varepsilon < 1\)). \(U^c\) is measurable, and \[ m(U^c) = 1 - m(U) = \varepsilon > 0. \] Since \(U^c \subset V\), \(m(U^c) = 0\) by #32(a), which is a contradiction. So \(m_k(V^c) = 1\).

③ \(m_k(V) + m_k(V^c) > 1\), conclusion.

\(m_k(V) = 0\) implies \(V\) is measurable, so \(m_k(V) > 0\).
Thus, \(m_k(V) + m_k(V^c) > 1 = m_k(V \cup V^c) = m_k(I)\).

37. \(\Gamma\) is a curve (or graph) of \(f(x)\); \(\Gamma = \{ (x, f(x)) : x \in \mathbb{R} \}\). Show that \(m(\Gamma) = 0\) if \(f\) is continuous.

proof

① \(m(\Gamma_g) = 0\) for continuous \(g: [a, b] \to \mathbb{R}\)

Let \(g: [a, b] \to \mathbb{R}\) be a continuous function and \(\Gamma_g = \{ (x, g(x)) : x \in [a, b] \}\) be its graph.
Due to PMA 4.19, \(g\) is uniformly continuous (\(\because [a, b]\) is compact).
Then, for every \(\varepsilon > 0\) there is \(\delta_\varepsilon > 0\) such that \(|g(x) - g(y)| < \varepsilon\) for all \(x, y \in [a, b]\) with \(|x - y| < \delta_\varepsilon\) (PMA 4.18).
The following will prove \(m(\Gamma_g) = 0\).

1) \(\delta_\varepsilon < b - a\), partition \([a, b]\)

Consider the case \(\delta_\varepsilon < b - a\).
Choose a smallest integer \(n \in \mathbb{N}\) such that \(b - a < n \delta_\varepsilon\).
That is, \((n - 1)\delta_\varepsilon \leq b - a < n \delta_\varepsilon\).
Then, make a partition of \([a, b]\) with same \(n\) intervals:
\(a = x_0 < x_1 < \cdots < x_n = b\) such that \(\frac{b - a}{n} = x_i - x_{i - 1} < \delta_\varepsilon\).

2) \(\Gamma_g \subset \bigcup_{i = 1}^{n} [x_{i - 1}, x_i] \times (g(x_i) - \varepsilon, g(x_i) + \varepsilon)\)

For every \(x \in [a, b]\), there is \(i\) such that \(x \in [x_{i - 1}, x_i]\) (\(i = 1, 2, \dots, n\)).
\(|x - x_i| < \delta_\varepsilon\), so this satisfies \(|g(x) - g(x_i)| < \varepsilon\) and \(g(x) \in (g(x_i) - \varepsilon, g(x_i) + \varepsilon)\).
Then, \((x, g(x)) \in [x_{i - 1}, x_i] \times (g(x_i) - \varepsilon, g(x_i) + \varepsilon)\).
This holds for all \(x \in [a, b]\), so \(\Gamma_g \subset \bigcup_{i = 1}^{n} [x_{i - 1}, x_i] \times (g(x_i) - \varepsilon, g(x_i) + \varepsilon)\).

3) \(m(\Gamma_g) < 2(b - a)\varepsilon\)

The above relation leads to: \[ m(\Gamma_g) \leq \sum_{i = 1}^{n} m([x_{i - 1}, x_i]) \cdot m((g(x_i) - \varepsilon, g(x_i) + \varepsilon)) < n\delta_\varepsilon \cdot 2\varepsilon \] Since \((n - 1)\delta_\varepsilon \leq b - a\), \[ m(\Gamma_g) < 2\varepsilon(b - a + \delta_\varepsilon) < 2\varepsilon(b - a + b - a) = 4(b - a)\varepsilon \]

4) \(\delta_\varepsilon \geq b - a\)

When \(\delta_\varepsilon \geq b - a\), we can randomly choose \(\delta'_\varepsilon\) with \(\delta'_\varepsilon < b - a\) because the uniform continuity still works under \(\delta'_\varepsilon\).
Let \(\delta'_\varepsilon = \delta_\varepsilon\) and follow the same procedure from 1), then the inequality \(m(\Gamma_g) < 4(b - a)\varepsilon\) still holds.

5) \(m(\Gamma_g) = 0\).

Since \(m(\Gamma_g) < 4(b - a)\varepsilon\) holds for all \(\varepsilon > 0\),
\(m(\Gamma_g) = 0\).

② extend to \(f: \mathbb{R} \to \mathbb{R}\)

Let \(f_n: [-n, n] \to \mathbb{R}\) be a continuous function such that \(f_n(x) = f(x)\) for \(x \in [-n, n]\).

1) \(\Gamma_f = \bigcup_{n = 1}^{\infty} \Gamma_{f_n}\)

\(z \in \bigcup_{n = 1}^{\infty} \Gamma_{f_n}\), then there is an integer \(N \in \mathbb{N}\) such that
\(z \in \Gamma_{f_N} = \{ (x, f_N(x)) : x \in [-N, N] \}\).
For the \(z = (x, f_N(x))\), \(x \in [-N, N] \subset \mathbb{R}\) and \(f_N(x) = f(x)\).
Then, \(z \in \Gamma_f\) and thus \(\bigcup_{n = 1}^{\infty} \Gamma_{f_n} \subset \Gamma_f\).

\(y \in \Gamma_f\), then \(y = (x, f(x))\) for some \(x \in \mathbb{R}\).
There exists a positive integer \(N\) such that \(x \in [-N, N]\), then \(f(x) = f_N(x)\).
This leads to \(y \in \Gamma_{f_N}\) and thus \(\Gamma_f \subset \bigcup_{n = 1}^{\infty} \Gamma_{f_n}\).

Thus, \(\Gamma_f = \bigcup_{n = 1}^{\infty} \Gamma_{f_n}\).

2) conclusion

Then,
\[ m(\Gamma_f) \leq \sum_{n = 1}^{\infty} m(\Gamma_{f_n}) = \sum 0 = 0. \] Thus, \(m(\Gamma_f) = 0\).

38. Show that \((a+b)^r \geq a^r + b^r\) when \(r \geq 1\), and \((a+b)^r \leq a^r + b^r\) when \(0 \leq r \leq 1\) (\(a,b > 0\)).

proof

① \(r \geq 1\)

When \(t \geq 0\), \((a+t)^{r-1} \geq t^{r-1}\), so \[\int_a^b (a+t)^{r-1} dt \geq \int_a^b t^{r-1} dt.\] Then, \[ \frac{1}{r} \left( (a+b)^r - a^r \right) \geq \frac{1}{r} \left( b^r - a^r \right), \] that is \((a+b)^r - a^r \geq b^r\). Thus, \((a+b)^r \geq a^r + b^r\).

② \(0 \leq r \leq 1\)

When \(t \geq 0\), \((a+t)^{r-1} \leq t^{r-1}\), so \[\int_a^b (a+t)^{r-1} dt \leq \int_a^b t^{r-1} dt.\] Then, \[ \frac{1}{r} \left( (a+b)^r - a^r \right) \leq \frac{1}{r} \left( b^r - a^r \right), \] that is \((a+b)^r - a^r \leq b^r\). Thus, \((a+b)^r \leq a^r + b^r\).