Ch2. Integration Theory (Ex.12 ~ Ex.24)

12. Show that there exist \(f \in L^1\) and a sequence \(\{f_n\} \subset L^1\) such that \(\|f_n - f\|_1 \to 0\) but \(\lim_{n \to \infty} f_n(x) \ne f(x)\).

proof

① Definition of \(I_n\) and \(f_n\)

For every \(n \in \mathbb{N}\), there exists \(k \in \mathbb{Z}^+\) such that \(2^k \le n < 2^{k+1}\).
Let \(\ell = n - 2^k\) and
\[I_n = \left[\frac{\ell}{2^{k+1}}, \frac{\ell+1}{2^{k+1}}\right], \quad f_n = \chi_{I_n}.\] (e.g. when \(n=9\), \(2^3 \le n < 2^4\), \(\ell = 1\), \(I_n = \left[\frac{1}{16}, \frac{2}{16}\right]\).)

② \(\lim_{n \to \infty} f_n(x) \ne 0\) for \(x \in [0,1]\)

Since \([0,1] = \bigcup I_n\), we have \(x \in I_n\) for some \(n\) with \(2^k \le n < 2^{k+1}\) (\(k = 0,1,2,\dots\)).

For a given integer \(M \in \mathbb{N}\) such that \(M < 2^k\) for some \(k \in \mathbb{N}\),
\(x \in I_n\) for some proper \(N \in \mathbb{N}\) with \(2^k \le N < 2^{k+1}\) and \(f_n(x) = 1 > 0\) when \(n = N\).

This holds for all \(M \in \mathbb{N}\),
so \(\limsup_{n \to \infty} f_n(x) \ne 0\).

③ \(\lim_{n \to \infty} \|f_n\|_{L^1} = 0\)

When \(2^k \le n < 2^{k+1}\), \[ \int |f_n(x)|\,dx = \frac{1}{2^{k+1}}. \] As \(n \to \infty\) implies \(k \to \infty\),
so \(\lim_{n \to \infty} \|f_n\|_1 = \lim_{k \to \infty} \frac{1}{2^{k+1}} = 0\).

12. Show that there exist \(f \in L^1\) and a sequence \(\{f_n\} \subset L^1\) such that \(\|f_n - f\|_1 \to 0\) but \(\lim_{n \to \infty} f_n(x) \ne f(x)\).

proof

① Definition of \(I_n\) and \(f_n\)

② \(\lim_{n \to \infty} f_n(x) \ne 0\) for \(x \in [0,1]\)

Since \([0,1] = \bigcup I_n\), we have \(x \in I_n\) for some \(n\) with \(2^k \le n < 2^{k+1}\) (\(k = 0,1,2,\dots\)).

This holds for all \(M \in \mathbb{N}\),
so \(\limsup_{n \to \infty} f_n(x) \ne 0\).

③ \(\lim_{n \to \infty} \|f_n\|_{L^1} = 0\)

14. Let \(V_d\) be a measure of a unit ball in \(\mathbb{R}^d\).

(a) Show that \(V_d = 2 \int_0^1 (1 - x^2)^{\frac{d-1}{2}} dx\).

proof We’ll just accept that \(m(A^t) = m(A)\), where \(A^t = \{(x, -y) : (x, y) \in A\}\) (by reflection symmetry).

① \(\mathbb{R}^d \times \{0\}\) has measure 0 (\(d \in \mathbb{N}\))

Let \(E_n = Q_n \times \{0\}\), where \(Q_n \subset \mathbb{R}^d\) is a cube centered at the origin with side length \(2n\): \[Q_n = [-n, n]^d.\] Since \(m(Q_n) = (2n)^d\) and \(m(\{0\}) = 0\),
\[m(E_n) = (2n)^d \cdot 0 = 0 \quad (\text{Prop 2.3.6}).\]
Then \(E_n \subset E_{n+1}\) for all \(n \in \mathbb{N}\) and \(\bigcup_{n=1}^\infty E_n = \mathbb{R}^d \times \{0\}\),
so \[ m(\mathbb{R}^d \times \{0\}) = \lim_{n \to \infty} m(E_n) = 0 \quad (\text{Cor 1.3.3}). \]

② \(E = \{(x, 0) \in \mathbb{R}^{d+1} : |x| > 1\}\), then \(m(B_1^{(d+1)}) = m(B_1^{(d)} \cup E)\)

\(B_1^{(d+1)}\) can be written as \(B \cup B^t \cup E\),
where \(B = \{(x, y) \in \mathbb{R}^{d+1} : |x| \le 1, 0 \le y \le (1 - x^2)^{1/2}\}\).

Let \(E = \{(x, 0) \in \mathbb{R}^{d+1} : |x| > 1\}\),
then \(m(E) = 0\) (\(E \subset \mathbb{R}^d \times \{0\}\)) and \(E \cap B_1^{(d)} = \varnothing\).
Then,
\[m(B_1^{(d)} \cup E) = m(B_1^{(d)}) + m(E) = m(B_1^{(d)}).\]

③ \(A = \{(x, y) \in \mathbb{R}^{d+1} : 0 \le y \le f(x)\}\), then \(m(B_1^{(d+1)}) = 2m(A)\)

Let
\[ f(x) = \begin{cases} (1 - x^2)^{1/2}, & |x| \le 1, \\ 0, & \text{o.w.} \end{cases} \] and
\[A = \{(x, y) \in \mathbb{R}^{d+1} : 0 \le y \le f(x)\} = B \cup E.\]

Then \(A^t = B^t \cup E\).
Hence,
\[ m(B_1^{(d+1)}) = m(B \cup B^t \cup E) = m((B \cup E) \cup (B^t \cup E)) = m(A \cup A^t) = m(A) + m(A^t) - m(A \cap A^t). \] Since
\[A \cap A^t = \{(x, 0) : x \in \mathbb{R}^d\},\]
which has measure 0 (by ①),
\[m(B_1^{(d+1)}) = 2m(A).\]

Thus,
\[ m(B_1^{(d+1)}) = V_{d+1} = 2 \int_{-1}^1 f(x)\,dx = 2 \int_0^1 (1 - x^2)^{d/2} dx. \] (by Cor. 2.3.8)

(b) Show that \(v_d = 2v_{d-1} \int_0^1 (1 - x^2)^{\frac{d-1}{2}} dx\)

\(B^{(d)}\) can be written as \(B^{(d-1)}_{1-x^2} \times B^{(1)}_x\).

proof

\[ B^{(d)} = \{ (z, x) \in \mathbb{R}^{d-1} \times \mathbb{R} : |x| \le 1 \text{ and } |z| \le (1 - x^2)^{1/2} \}. \]

Then,

\[ m(B^{(d)}) = \int_{B^{(d)}} 1 \, dz \, dx = \int \int_{\{z : (z, x) \in B^{(d)}\}} 1 \, dz \, dx \quad (\text{by Fubini's theorem}) \]

\[ = \int_{-1}^1 \int_{B^{(d-1)}_{\sqrt{1 - x^2}}} 1 \, dz \, dx = \int_{-1}^1 v_{d-1} (1 - x^2)^{\frac{d-1}{2}} dx = 2 v_{d-1} \int_0^1 (1 - x^2)^{\frac{d-1}{2}} dx \]

(c) Show that \(v_d = \pi^{d/2} / \Gamma(\frac{d}{2} + 1)\)

proof

① \(v_1\)

In \(\mathbb{R}^1\), \(B^{(1)}_1\) is an open interval of length 2.
So, \(v_1 = 2\).

② \(v_d\)

Let \(x^2 = t\) so that \(2x dx = dt\) \((0 \le t \le 1)\). Then,

\[\begin{align*} \int_0^1 (1 - x^2)^{\frac{d-1}{2}} dx &= \int_0^1 (1 - t)^{\frac{d-1}{2}} \cdot \frac{1}{2} t^{-1/2} dt \\ &= \frac{1}{2} \int_0^1 t^{\frac{1}{2} - 1} (1 - t)^{\frac{d+1}{2} - 1} dt \\ &= \frac{1}{2} \mathrm{Beta}\left( \frac{1}{2}, \frac{d+1}{2} \right) \\ &= \frac{1}{2} \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2} + 1)}. \end{align*}\]

Then,

\[ v_d = 2 v_{d-1} \cdot \frac{1}{2} \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2} + 1)} = v_{d-1} \pi^{1/2} \frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2} + 1)}. \]

So,

\[ v_d = v_1 (\pi^{1/2})^{d-1} \frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2} + 1)} = 2 (\pi^{1/2})^{d-1} \frac{1}{2} \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2} + 1)} = \frac{(\pi)^{d/2}}{\Gamma(\frac{d}{2} + 1)}. \]

(PMA 8.18, 8.20)

15. Consider the function \(f(x) = \begin{cases}x^{1/2}, & 0 < x < 1 \\0, & \text{o.w.}\end{cases}\) For a fixed enumeration \(\{r_n\}\) of the rationals \(\mathbb{Q}\), let \(F(x) = \sum_{n=1}^\infty 2^{-n} f(x - r_n)\). Then, show that

(1) \(\int_{\mathbb{R}} f(x) dx = 2\)

proof

①
\(f\) is continuous on \([a, b]\) \((0 < a < b < 1)\), so \(f \in \mathbb{R}\) on \([a,b]\) (PMA 6.8).
Then

\[ \int_{[a,b]} f(x) dx = \int_a^b x^{1/2} dx = \left[\frac{1}{\frac{3}{2}}x^{3/2}\right]_a^b = 2(b^{1/2} - a^{1/2}) = \int_{[a,b]} f(x) dx \]

(thm 2.1.5).
Let the number \(b'\) greater than \(1/2\).

②
Let \(f_n(x) = \begin{cases} f(x), & n \le x \le b \\ 0, & \text{o.w.} \end{cases}\)
and \(f(x) = 0\) for \(x \in \mathbb{R}\).

For \(x \in [\frac{1}{n+1}, \frac{1}{n})\), \(f_{n,m}(x) > 0\) and \(f_{n,m}(x) = 0\).

This leads to \(f_{n,m}(x) \ge f_n(x)\) \((x \in \mathbb{R})\).
\(f_n\) is measurable, and \(\lim_{n \to \infty} f_n(x) = f(x) \chi_{(0,b)}(x)\) for every \(x \in \mathbb{R}\).

\[ \lim_{n \to \infty} \int f_n(x) dx = \int f(x)\chi_{(0,b)}(x) dx \]

(thm 2.1.9, monotone convergence theorem).

\[ \int_{a}^{b} f_n(x) dx = 2(b - \sqrt{\frac{1}{n}}) \]

\[ \int f(x) \chi_{(0,b)}(x) dx = 2b. \]

Similar approach to the right limit of the interval \([a,b]\) leads to \(\int_0^1 f(x) dx = 2.\)

(2) \(|F(x)| < \infty\) a.e. \(x\)

proof

By translation invariance,
\[\int f(x - r_n) dx = \int f(x) dx = 2.\]

\(f\) is non-negative and measurable, so

\[\begin{align*} \int F(x) dx &= \int \sum_{n=1}^\infty 2^{-n} f(x - r_n) dx \\ &= \sum_{n=1}^\infty 2^{-n} \int f(x - r_n) dx \quad (\text{MCT}) \\ &= \sum_{n=1}^\infty 2^{-n} \times 2 = 2. \end{align*}\]

Since \(\int F < \infty\), \(|F(x)| < \infty\) a.e. \(x \in \mathbb{R}\).

(3) Show that \(F\) is unbounded on every open interval.

proof

① \(r_n \in I\)

Let \(I = (a, b)\) be some open interval of \(\mathbb{R}\),
then \(I\) contains some rational \(r_n\): \(a < r_n < b\)
(\(\because \mathbb{Q}\) is dense in \(\mathbb{R}\)).

② \(f(x - r_n)\) is unbounded on \(I\).

For given \(M > 0\),

\[ f(x - r_n) > M \Leftrightarrow (x - r_n)^{1/2} > M \Leftrightarrow M < \frac{1}{x - r_n} \Leftrightarrow M^2 < \frac{1}{x - r_n} \Leftrightarrow 0 < x - r_n < \frac{1}{M^2} \]

\(\Rightarrow r_n < x < r_n + \frac{1}{M^2}\).

Let \(b_M = \min(b, r_n + \frac{1}{M^2})\),
then \(f(x - r_n) > M\) holds when \(x \in (r_n, b_M) \subset (a, b)\).

This means for every \(M > 0\), there is \(x \in (r_n, b_M)\) such that \(f(x - r_n) > M\).
So, \(f(x - r_n)\) is unbounded on \(I\).

③ \(F\) is unbounded on \(I\).

Thus, for every \(M' > 0\), there is \(x \in I\) such that
\(F(x) \ge 2^{-n} f(x - r_n) > 2^{-n} M \ge M'\).

This means \(F\) is unbounded on \(I\).
Since \(I\) is an arbitrary interval on \(\mathbb{R}\), \(F\) is unbounded on every interval.

(4) \(\tilde{F} = F\) for a.e. \(x \in \mathbb{R}\), then show that \(\tilde{F}\) is unbounded in any interval.

proof

\(\tilde{F}(x) = F(x)\) for a.e. \(x \in I\).
There is \(x \in (r_n, b_M) \subset I\) such that \(F(x) > M'\) for every \(M'\).
Then, for a.e. \(x \in (r_n, b_M)\), \(\tilde{F}(x) > M'\) holds.
Thus, \(\tilde{F}\) is unbounded.

Let \(\tilde{I} = \bigcup I_x\) such that \(\tilde{F} = F\) on \(I_x\) and \(I_x \cap I_{x'} = \emptyset\) \((x \ne x')\).
\(F\) is unbounded on \(I_x \subset I\), so \(\tilde{F}\) is unbounded on \(I\).

16. \(f\) is integrable on \(\mathbb{R}^d\), and \(\delta = (\delta_1, \dots, \delta_d)\) is a d-tuple of non-zero real numbers. Let \(f^\delta(x) = f(\delta x) = f(\delta_1 x_1, \dots, \delta_d x_d)\). Then show that \(f^\delta\) is integrable and \[\int_{\mathbb{R}^d} f^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int_{\mathbb{R}^d} f(x) dx.\](\(\delta x = (\delta_1 x_1, \dots, \delta_d x_d)\))

proof

① \(\chi_{E^\delta}(x) = \chi_E(\delta x)\) (measurable \(E\))

\(\delta x \in E \Leftrightarrow x \in \delta^{-1}E\), so \(\chi_E(\delta x) = \chi_{\delta^{-1}E}(x)\).
Then,

\[ \int \chi_E(\delta x) dx = \int \chi_{\delta^{-1}E}(x) dx = m(\delta^{-1}E) = |\delta_1|^{-1} \cdots |\delta_d|^{-1} m(E) \]

(ex 1-9)

\[ = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int \chi_E(x) dx. \]

② For simple function \(\varphi\),

\[ \int \varphi^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int \varphi(x) dx. \]

Let \(\varphi(x) = \sum_{k=1}^m a_k \chi_{E_k}(x)\), then \(\varphi^\delta(x) = \sum_{k=1}^m a_k \chi_{E_k}(\delta x)\).

So,

\[\begin{align*} \int \varphi^\delta(x) dx &= \sum_{k=1}^m a_k \int \chi_{E_k}(\delta x) dx = \sum_{k=1}^m a_k m(\delta^{-1} E_k) \\ &= |\delta_1|^{-1} \cdots |\delta_d|^{-1} \sum_{k=1}^m a_k m(E_k) = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int \varphi(x) dx. \end{align*}\]

③ For measurable & non-negative \(f\),

\[ \int f^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \]

There is a sequence of non-negative simple functions \(\varphi_k\) such that \(\varphi_k \uparrow f\) (thm 1.4.1).
Then it is obvious that \(\varphi_k^\delta \uparrow f^\delta(x)\).
\(\varphi_k^\delta\) is a sequence of non-negative measurable functions, so

\[ \lim_{k \to \infty} \int \varphi_k^\delta(x) dx = \int f^\delta(x) dx \quad (\text{monotone convergence thm 2.1.9}). \]

Thus,

\[ \int f^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \lim_{k \to \infty} \int \varphi_k(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \quad (\text{: MCT}) \]

④ For real-valued and measurable \(f\),

\[ \int f^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \]

\(f^\delta(x) = f^{\delta+}(x) - f^{\delta-}(x)\), where \(f^{\delta+} = \max(f^\delta, 0)\) and \(f^{\delta-} = \max(-f^\delta, 0)\).
\(f^{\delta+}\) and \(f^{\delta-}\) are non-negative and measurable. Then,

\[\begin{align*} \int f^\delta(x) dx &= \int f^{\delta+}(x) dx - \int f^{\delta-}(x) dx \\ &= |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f^+(x) dx - |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f^-(x) dx \\ &= |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \end{align*}\]

⑤ If \(f = u + iv \in L^1\), then

\[ \int f^\delta(x) dx = |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \]

\[\begin{align*} \int f^\delta(x) dx &= \int u^\delta(x) dx + i \int v^\delta(x) dx \\ &= |\delta_1|^{-1} \cdots |\delta_d|^{-1} \left( \int u(x) dx + i \int v(x) dx \right) \\ &= |\delta_1|^{-1} \cdots |\delta_d|^{-1} \int f(x) dx. \end{align*}\]

17. \(f\) is defined on \(\mathbb{R}^2\) as follows: for \(n \in \mathbb{Z}^+\), \[f(x, y) =\begin{cases}a_n, & n \le y < n+1 \text{ and } n \le x < n+1 \\-a_n, & n \le x < n+1 \text{ and } n+1 \le y < n+2 \\0, & \text{otherwise.}\end{cases} \] Here, \(a_n = \sum_{k=0}^n b_k\) where \(\{b_k\}\) is a sequence of positive real numbers such that \(\sum_{k=0}^\infty b_k = s < \infty\). Let \(a_0 = 0\) for simplicity.

(a)-1 Show that \(f^y\) is integrable

proof

① \(y < 0\): \(f(x, y) = 0\) for all \(x \in \mathbb{R}\), so \(f^y(x) = 0\) which is integrable.

② \(n \le y < n+1\) \((n \in \mathbb{Z}^+)\)

\[ f(x, y) = \begin{cases} a_n, & n \le x < n+1 \\ -a_{n-1}, & n-1 \le x < n \\ 0, & \text{o.w.} \end{cases} \]

Then, \(f^y(x) = a_n \chi_{[n, n+1)}(x) - a_{n-1} \chi_{[n-1, n)}(x)\), and
\(\int f^y(x) dx = a_n - a_{n-1} < \infty.\)

(a)-2 Show that \(f_x\) is integrable & \(\int f_x(y) dy = 0\) \(\forall x \in \mathbb{R}\)

proof

① \(x < 0\): \(f(x, y) = 0\) for all \(y \in \mathbb{R}\), so \(f_x(y) = 0\) and \(\int f_x(y) dy = 0 < \infty\).

② \(n \le x < n+1\) \((n \in \mathbb{Z}^+)\)

\[ f(x, y) = \begin{cases} a_n, & n \le y < n+1 \\ -a_n, & n+1 \le y < n+2 \\ 0, & \text{o.w.} \end{cases} \]

Then, \(f_x(y) = a_n \chi_{[n, n+1)}(y) - a_n \chi_{[n+1, n+2)}(y)\),
so \(\int f_x(y) dy = a_n - a_n = 0.\)

(a)-3 Show that \(\int_{\mathbb{R}} \int_{\mathbb{R}} f(x, y) dy dx = 0\)

proof

\[ \int_{\mathbb{R}} \int_{\mathbb{R}} f(x, y) dy dx = \int_{\mathbb{R}} 0 dx = 0 \quad (\text{by (a)-2}) \]

(b) Show that \(\int f^y(x) dx\) is integrable on \((0, \infty)\) and \(\iint f(x, y) dx dy = s\)

proof

We’ve shown in (a)-1 that \(\int f^y(x) dx = a_n - a_{n-1}\) for \(y \in [n, n+1)\).
Since \(a_n\) is finite for all \(n \in \mathbb{Z}^+\), \(\int f^y(x) dx < \infty\) for \(y \in \mathbb{R}\) (as well as \((0, \mathbb{R}^+)\)).

Then,

\[\begin{align*} \iint f(x, y) dx dy &= \sum_{n=0}^\infty \int_n^{n+1} \int_{\mathbb{R}} f^y(x) dx \, dy = \sum_{n=0}^\infty \int_n^{n+1} (a_n - a_{n-1}) dy \\ &= \sum_{n=0}^\infty (a_n - a_{n-1}) = \sum_{k=0}^\infty b_k = s. \end{align*}\]

(c) Show that \(\iint_{\mathbb{R}} |f(x, y)| \, dx \, dy = \infty\)

proof

① \(g(x, y) \overset{\text{def}}{=} |f(x, y)|\) is measurable

Let \(g(x, y) = |f(x, y)|\).
For \(n \le y < n+2\),
\[ g(x, y) = \begin{cases} a_n, & n \le y < n+2 \\ 0, & \text{o.w.} \end{cases} \quad (n \ge 0) \]

For \(a' \in [a_n, a_{n+1})\),
\(\{ g > a' \} = \bigcup_{k=n}^\infty \{ g = a_k \} = \bigcup_{k=n}^\infty ([k, k+1) \times [k, k+2))\).

\(\{ g = a_k \}\) is a product of two measurable sets \([k, k+1)\) and \([k, k+2)\),
so \(\{ g = a_k \}\) is measurable, as well as \(\{ g > a' \}\).

If \(a < 0\), \(\{ g > a \} = \mathbb{R}^2\), which is measurable.
Thus, \(g\) is a measurable function on \(\mathbb{R}^2\).

② \(\int_{\mathbb{R}^2} g = \infty\)

By Tonelli’s theorem,

\[\begin{align*} \int_{\mathbb{R}^2} g &= \int_{\mathbb{R}^2} |f(x, y)| \, dy \, dx = \sum_{n=0}^\infty \int_n^{n+1} \int_{\mathbb{R}} g(x, y) \, dy \, dx \\ &= \sum_{n=0}^\infty \int_n^{n+1} 2a_n \, dx = \sum_{n=0}^\infty 2a_n \ge \sum_{n=0}^\infty 2a_0 = \infty. \end{align*}\]

18. \(f\) is a measurable finite-valued function on \([0,1]\), and \(g(x,y) = |f(x) - f(y)|\) is integrable on \([0,1] \times [0,1]\). Show that \(f(x)\) is integrable on \([0,1]\).

proof

\(g\) is integrable on \([0,1] \times [0,1]\), so \(g^y\) is integrable for a.e. \(y \in [0,1]\) (Fubini’s Thm).
Let \(y_0\) be such that \(\int_0^1 g(x, y_0) dx < \infty\).

Since \(|f(x)| - |f(y_0)| \le |f(x) - f(y_0)| = g(x, y_0) = g^{y_0}(x)\),

\[|f(x)| \le |f(y_0)| + g^{y_0}(x).\]

Then, \[\begin{align*} \int_0^1 |f(x)| dx &\le \int_0^1 |f(y_0)| dx + \int_0^1 g^{y_0}(x) dx \\ &= |f(y_0)| + \int_0^1 g^{y_0}(x) dx. \end{align*}\]

\(f\) is finite for all \(x \in [0,1]\), so \(|f(y_0)| < \infty.\)
\(g^{y_0}\) is integrable, so \(\int_0^1 g^{y_0}(x) dx < \infty.\)

Thus, \(\int_0^1 |f(x)| dx < \infty\) and \(f\) is integrable on \([0,1].\)

19. \(f\) is integrable on \(\mathbb{R}^d\), and let \(E_\alpha = \{ x \in \mathbb{R}^d : |f(x)| > \alpha \}\) for \(\alpha > 0\). Show that \[\int_{\mathbb{R}^d} |f(x)| dx = \int_0^\infty m(E_\alpha) d\alpha\]

proof

① \(\chi_{E_\alpha}\) is measurable

When \(b < 0\), \(\{ \chi_{E_\alpha} > b \} = \mathbb{R}^d\).
When \(b \ge 1\), \(\{ \chi_{E_\alpha} > b \} = \emptyset\).

When \(0 \le b < 1\),
\[\begin{align*} \{ \chi_{E_\alpha} > b \} &= \{ \chi_{E_\alpha} = 1 \} \\ &= \{ x \in E_\alpha \} = \{ |f| > \alpha \} \\ &= \{ f > \alpha \} \cup \{ f < -\alpha \}, \end{align*}\] which is a union of measurable sets.

Since \(\{ \chi_{E_\alpha} > b \}\) is measurable for all \(b \in \mathbb{R}\), \(\chi_{E_\alpha}\) is measurable.

② \(\int_0^\infty m(E_\alpha) d\alpha\)

Then, \[\begin{align*} \int_0^\infty m(E_\alpha) d\alpha &= \int_0^\infty \int_{\mathbb{R}^d} \chi_{E_\alpha}(x) dx \, d\alpha \\ &= \int_{\mathbb{R}^d} \int_0^\infty \chi_{E_\alpha}(x) d\alpha \, dx \quad (\text{Tonelli’s thm}) \end{align*}\]

③ Conclusion

For given \(x \in \mathbb{R}^d\),
\[ \chi_{E_\alpha}(x) = \begin{cases} 1, & 0 < \alpha < |f(x)|, \\ 0, & \alpha \ge |f(x)|. \end{cases} \]

(as a function of \(\alpha\)). Then, \[ \int_0^\infty \chi_{E_\alpha}(x) d\alpha = \int_0^{|f(x)|} 1 \, d\alpha = |f(x)|. \]

Thus, \[ \int_0^\infty m(E_\alpha) d\alpha = \int_{\mathbb{R}^d} |f(x)| dx. \]

21. \(f\) and \(g\) are measurable functions on \(\mathbb{R}^d\).

(a) Show that \(f(x - y) g(y)\) is measurable on \(\mathbb{R}^{2d}\).

proof

\(f(x - y)\) is measurable on \(\mathbb{R}^{2d}\) (Thm 2.3.9),
and \(g(y) = g(\pi_2(x, y))\) is measurable on \(\mathbb{R}^{2d}\) (Cor 2.3.7).
Then, \(f(x - y) g(y)\) is measurable (Prop 1.2.7).

(b) If \(f\) and \(g\) are integrable on \(\mathbb{R}^d\), show that \(f(x - y) g(y)\) is integrable on \(\mathbb{R}^{2d}\).

proof

\(f(x - y) g(y)\) is measurable on \(\mathbb{R}^{2d}\), as well as \(|f(x - y) g(y)|\).
Then,

\[\begin{align*} \iint_{\mathbb{R}^{2d}} |f(x - y) g(y)| \, dy \, dx &= \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x - y)| |g(y)| \, dy \, dx \\ &= \int_{\mathbb{R}^d} |g(y)| \left( \int_{\mathbb{R}^d} |f(x - y)| \, dx \right) dy \\ &= \int_{\mathbb{R}^d} |f(x)| dx \int_{\mathbb{R}^d} |g(y)| dy < \infty \quad (\because f, g \in L^1). \end{align*}\]

Thus, \(f(x - y) g(y)\) is integrable.

(c) The convolution of \(f, g \in L^1\) is defined as follows:\[(f * g)(x) = \int_{\mathbb{R}^d} f(x - y) g(y) \, dy.\] Show that \(f(x - y) g(y)\) is integrable on \(\mathbb{R}^d\) a.e. \(x\).

proof

\[\begin{align*} \int |(f * g)(x)| dx &\le \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x - y) g(y)| \, dy \, dx \\ &= \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x - y) g(y)| \, dx \, dy < \infty \quad (\text{by (b)}). \end{align*}\]

So, \((f * g)(x) = \int f(x - y) g(y) dy < \infty\) a.e. \(x\),
and thus \(f(x - y) g(y)\) is integrable on \(\mathbb{R}^d\) a.e. \(x\).

(d)-1 Show that \(f * g\) is integrable on \(\mathbb{R}^d\) when \(f, g\) are integrable:

\[ \| f * g \|_1 \le \| f \|_1 \, \| g \|_1. \]

proof

\[\begin{align*} \int |(f * g)(x)| dx &\le \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x - y) g(y)| \, dy \, dx \\ &= \int_{\mathbb{R}^d} |f(x - y)| dx \, |g(y)| dy = \| f \|_1 \, \| g \|_1 < \infty. \end{align*}\]

(d)-2 Show that the equality in (d)-1 holds when \(f\) and \(g\) are non-negative.

proof

The inequality \[ \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x - y) g(y) \, dy \, dx \le \int_{\mathbb{R}^d} f(x - y) g(y) \, dy \, dx \] becomes an equality when \(f(x - y) g(y)\) is non-negative (Thm 2.1.11 (iv)).

24. Consider the convolution \((f * g)(x) = \int_{\mathbb{R}^d} f(x + y) g(y) \, dy \quad (x \in \mathbb{R}^d)\)

(a) Show that \(f * g\) is uniformly continuous when \(\| f \|_1 < \infty\) and \(|g| \le M\) (bounded).

proof

① \(f(x + y) g(y)\) is integrable

\[ \int_{\mathbb{R}^d} |f(x + y) g(y)| dy \le M \int_{\mathbb{R}^d} |f(x + y)| dy = M \int_{\mathbb{R}^d} |f(y)| dy = M \| f \|_1 < \infty. \]

② \(|(f * g)(x) - (f * g)(x')| \le M \| f_{x'} - f \|_1\)

For \(x, x' \in \mathbb{R}^d\),

\[\begin{align*} |(f * g)(x) - (f * g)(x')| &= \left| \int_{\mathbb{R}^d} f(x + y) g(y) dy - \int_{\mathbb{R}^d} f(x' + y) g(y) dy \right| \\ &= \left| \int_{\mathbb{R}^d} (f(x + y) - f(x' + y)) g(y) dy \right| \\ &\le \int_{\mathbb{R}^d} |f(x + y) - f(x' + y)| |g(y)| dy \\ &\le M \int_{\mathbb{R}^d} |f(x + y) - f(x' + y)| dy \quad (\because |g| \le M) \\ &= M \int_{\mathbb{R}^d} |f(y) - f(y - (x - x'))| dy \quad (\text{translation invariance}) \\ &= M \int_{\mathbb{R}^d} |f(y) - f(y - (x - x'))| dy \quad (\text{relative invariance, p.13}) \end{align*}\]

③ Conclusion

For given \(\varepsilon > 0\), there is \(\delta_0 > 0\) such that
\(|x - x'| < \delta_0\) implies \(\| f_{x - x'} - f \|_1 = \int_{\mathbb{R}^d} |f(y - (x - x')) - f(y)| dy < \varepsilon / M\)
(\(\because f \in L^1\), Prop 2.2.5).

Then, \(|x - x'| < \delta_0\) means \(|(f * g)(x) - (f * g)(x')| < M \varepsilon\).
This holds for all \(\varepsilon > 0\), so \(f * g\) is uniformly continuous.

(b) If \(g\) is integrable as well, show that \(\lim_{|x| \to \infty} (f * g)(x) = 0\)

proof

For given \(\varepsilon > 0\), there is a simple function \(f_\varepsilon\) and \(g_\varepsilon\) such that
\(\| f - f_\varepsilon \| < \frac{\varepsilon}{4 \| g \|_1}\) and \(\| g - g_\varepsilon \| < \frac{\varepsilon}{4 \| f_\varepsilon \|_1}\).

Then,

\[\begin{align*} |(f * g)(x)| &= \left| \int_{\mathbb{R}^d} f(x - y) g(y) dy \right| \\ &\le \int_{\mathbb{R}^d} |f(x - y) - f_\varepsilon(x - y)| |g(y)| dy + \int_{\mathbb{R}^d} |f_\varepsilon(x - y)| |g(y) - g_\varepsilon(y)| dy + \int_{\mathbb{R}^d} |f_\varepsilon(x - y) g_\varepsilon(y)| dy \\ &\le \| f - f_\varepsilon \|_1 \| g \|_1 + \| f_\varepsilon \|_1 \| g - g_\varepsilon \|_1 + \int_{\mathbb{R}^d} |f_\varepsilon(x - y) g_\varepsilon(y)| dy. \end{align*}\]

Since a simple function is integrable and has compact support (Prop 2.21 (d)),
the last term \(\int |f_\varepsilon(x - y) g_\varepsilon(y)| dy \to 0\) as \(|x| \to \infty\) (by density argument).

Thus,
\[ \lim_{|x| \to \infty} (f * g)(x) = 0. \]

②

\(|x - y| \ge |x| - |y|\), so \(\lim_{|x| \to \infty} |x - y| = \infty\).
The simple function \(f_\varepsilon\) is supported on a set of finite measure,
\(\lim_{|x| \to \infty} f_\varepsilon(x - y) = 0\).
So,
\[ \lim_{|x| \to \infty} \int_{\mathbb{R}^d} |f_\varepsilon(x - y) g_\varepsilon(y)| dy = 0. \]

③

Then,
\[ \lim_{|x| \to \infty} |(f * g)(x)| \le \| f - f_\varepsilon \|_1 \| g \|_1 + \| f_\varepsilon \|_1 \| g - g_\varepsilon \|_1 < 2 \varepsilon, \]
and this holds for all \(\varepsilon > 0\). Thus,
\[ \lim_{|x| \to \infty} (f * g)(x) = 0. \]