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\ifsolutions
\title[Math 617 (Winter 2024): Solutions~5]{Math 617 (Winter 2024,
        Phillips): Solutions to Homework~5}
\date{14~February 2024}
\else
\title[Math 617 (Winter 2024): Homework~5]{Math 617 (Winter 2024,
        Phillips): Homework~5}
\date{7~February 2024}
\fi

\begin{document}

% \setcounter{section}{-1}

\maketitle

\begin{pbm}[Problem~4 parts (a) and~(d) in Chapter~7 of
 Rudin]\label{2017_NP_R84}
Each part of this problem counts as one normal problem.

Let $p \in [1, \I]$.
\begin{enumerate}
\item\label{2017_NP_R84_ConvInLp}
Let $f \in L^1 (\R)$ and $g \in L^p (\R)$.
Imitate the proof of Theorem~7.14 of Rudin to show
that the integral defining $(f * g) (x)$ exists for almost all $x$,
that $f * g \in L^p (\R)$,
and that $\| f * g \|_p \leq \| f \|_1 \| g \|_p$.
(For $p \in (1, \I)$, you will need to use H\"{o}lder's inequality
on carefully chosen functions
involving powers of the ones you are given.)
%  \item\label{2017_NP_R84_ConvEqNorm}
%  [Part (b) of Rudin's problem; omitted.]
%  \item\label{2017_NP_R84_NoEqNorm}
%  [Part (c) of Rudin's problem; omitted.]
\item\label{2017_NP_R84_AppId}
Prove that for every $\ep > 0$ there are nonzero $f \in L^1 (\R)$
and $g \in L^p (\R)$ such that
\[
\| f * g \|_p > (1 - \ep) \| f \|_1 \| g \|_p.
\]
\end{enumerate}
\end{pbm}

\ifsolutions

\begin{proof}[Solution to part~(\ref{2017_NP_R84_ConvInLp})]
As in the proof of Theorem 7.14 of Rudin,
we may assume that $f$ and $g$ are Borel functions,
and we consider $| f |$ and $| g |$ first.

Our first objective is to show that if $p < \I$ then
\begin{equation}\label{claim1}
\int_{- \I}^{\I}
   \left( \int_{- \I}^{\I} | f (x - t) g (t) | \, d t \right)^p \, d x
  \leq \| f \|_1^p \| g \|_p^p,
\end{equation}
in particular, that the outer integral exists.
Let $q$ be the conjugate exponent to $p$.
We apply H\"{o}lder's inequality (in Theorem~3.5 of Rudin),
in the form
\[
\int_{- \I}^{\I} h_1 (x) h_2 (x) \, d x
  \leq \left( \int_{- \I}^{\I} h_1 (x)^q \, d x \right)^{1/q}
       \left( \int_{- \I}^{\I} h_2 (x)^p \, d x \right)^{1/p}
\]
for nonnegative \mfn{s} $h_1$ and $h_2$.
We take
\[
h_1 (t) = | f (x - t) |^{1/q}
\andeqn
h_2 (t) = | f (x - t) |^{1/p} | g (t) |.
\]
This gives
\[
\begin{split}
\int_{- \I}^{\I} | f (x - t) g (t) | \, d t
 & \leq \left( \int_{- \I}^{\I} | f (x - t) | \, d t \right)^{1/q}
       \left( \int_{- \I}^{\I}
            | f (x - t) | \cdot | g (t) |^p \, d t \right)^{1/p}
                          \\
 & = \| f \|_1^{1/q}
       \left( \int_{- \I}^{\I}
            | f (x - t) | \cdot | g (t) |^p \, d t \right)^{1/p}.
\end{split}
\]
So
\begin{equation}\label{claim1a}
\left( \int_{- \I}^{\I} | f (x - t) g (t) | \, d t \right)^{p}
   \leq \| f \|_1^{p/q}
          \int_{- \I}^{\I} | f (x - t) | \cdot | g (t) |^p \, d t.
\end{equation}
Now $(x, t) \mapsto | f (x - t) g (t) |$ is Borel,
for the same reason as in the proof of Theorem~7.14 of Rudin,
and Lebesgue measure on $\R$ is $\sm$-finite,
so Fubini's Theorem for nonnegative functions implies that
\[
x \mapsto \int_{- \I}^{\I} | f (x - t) g (t) | \, d t
\]
is Borel.
Thus the left hand side of~(\ref{claim1a})
is also a Borel function of $x$.
The right hand side of~(\ref{claim1a})
is also a Borel function of $x$, by Theorem~7.8 of Rudin
applied to the $L^1$ functions $| f |$ and $| g |^p$.
Therefore, using the $L^1$ norm estimate from Theorem~7.14 of Rudin
at the second step and $\frac{p}{q} + 1 = p$ at the third step,
we get
\[
\begin{split}
\int_{- \I}^{\I}
 \left( \int_{- \I}^{\I} | f (x - t) g (t) | \, d t \right)^{p} \, d x
 & \leq \| f \|_1^{p/q}
       \int_{- \I}^{\I}
          \left( \int_{- \I}^{\I}
             | f (x - t) | \cdot | g (t) |^p \, d t \right) \, d x
                          \\
 & \leq \| f \|_1^{p/q}
       \left( \int_{- \I}^{\I} | f | \, d m \right)
       \left( \int_{- \I}^{\I} | g |^p \, d m \right)
                          \\
 & \leq \| f \|_1^p \| g \|_p^p.
\end{split}
\]
This proves the claim.

For $p = \I$ the analog of~(\ref{claim1}) is that
\begin{equation}\label{claim2}
\sup_{x \in \R} \int_{- \I}^{\I} | f (x - t) g (t) | \, d t
  \leq \| f \|_1 \| g \|_{\I}.
\end{equation}
This is obvious.

It is now immediate that
$\int_{- \I}^{\I} | f (x - t) g (t) | \, d t < \I$
for almost every $x$ (for every $x$ if $p = \I$).
Therefore
$\int_{- \I}^{\I} f (x - t) g (t) \, d t$
exists for almost every $x$ (for every $x$ if $p = \I$).

We next claim that the function
\[
h (x) = \int_{- \I}^{\I} f (x - t) g (t) \, d t
\]
is \mb.
For $n \in \N$ define
\[
g_n (x) = \left\{ \begin{array}{ll}
 g (x) & \hspace{3em} {\mbox{$|x| \leq n$ and $| g (x) | \leq n$}} \\
     0 & \hspace{3em} {\mbox{otherwise.}}
    \end{array} \right.
\]
Then $g_n \in L^1 (\R)$,
so Theorem~7.8 of Rudin implies that
\[
h_n (x) = \int_{- \I}^{\I} f (x - t) g_n (t) \, d t
\]
is \mb.
For every $x$ for which
$\int_{- \I}^{\I} | f (x - t) g (t) | \, d t < \I$,
the Dominated Convergence Theorem implies that
$\limi{n} h_n (x) = h (x)$.
Thus, $h$ is the pointwise almost everywhere limit of \mfn{s},
hence \mb.
This proves the claim.

Now for $p < \I$ we get, using~(\ref{claim1}) at the second step,
\[
\int_{- \I}^{\I} | h | \, d m
\leq \int_{- \I}^{\I}
 \left( \int_{- \I}^{\I} | f (x - t) g (t) | \, d t \right)^{p} \, d x
\leq \| f \|_1^p \| g \|_p^p.
\]
whence
\[
\| f * g \|_p = \| h \|_p \leq \| f \|_1 \| g \|_p.
\]
For $p = \I$ we get, using~(\ref{claim2}) instead,
\[
\| f * g \|_{\I}
 \leq \sup_{x \in \R} | h (x) |
 \leq \sup_{x \in \R} \int_{- \I}^{\I} | f (x - t) g (t) | \, d t
 \leq \| f \|_1 \| g \|_{\I}.
\]
This completes the proof.
\end{proof}

The first two solutions to part~(\ref{2017_NP_R84_AppId})
are written using functions normalized to have
$\| f \|_1 = \| g \|_p = 1$.
This is neater in a sense, but not really needed,
and not normalizing would simplify the notation a little.

\begin{proof}[Solution to part~(\ref{2017_NP_R84_AppId})]
We first assume $p < \I$.
Choose $\rh \in (0, 1)$
so small that $\rh < 1 - (1 - \ep)^p$.
Define
\[
f = \left( \frac{1}{2 \rh} \right) \ch_{[- \rh, \rh]}
\andeqn
g = \left( \frac{1}{2^{1/p}} \right) \ch_{[- 1, 1]}.
\]
Then one checks directly that $\| f \|_1 = \| g \|_p = 1$.
Let $m$ be Lebesgue measure on~$\R$.
We have
\[
\begin{split}
(f * g) (x)
& = \int_{- \I}^{\I} f (y) g (x - y) \, d y
  = \left( \frac{1}{2 \rh \cdot 2^{1/p}} \right)
         \int_{- \rh}^{\rh} \ch_{[- 1, 1]} (x - y) \, d y
\\
& = \left( \frac{1}{2 \rh \cdot 2^{1/p}} \right)
         \int_{- \rh}^{\rh} \ch_{[x - 1, \, x + 1]} (y) \, d y
\\
& = \left( \frac{1}{2 \rh \cdot 2^{1/p}} \right)
         m \big( [- \rh, \rh] \cap [x - 1, \, x + 1] \big)
\\
& = \begin{cases}
   0 & x \leq - 1 - \rh
        \\
   \frac{x + 1 + \rh}{2 \rh \cdot 2^{1/p}}
           & - 1 - \rh < x \leq - 1 + \rh
       \\
   \frac{2 \rh}{2 \rh \cdot 2^{1/p}} & - 1 + \rh < x \leq 1 - \rh
        \\
   \frac{1 + \rh - x}{2 \rh \cdot 2^{1/p}} & 1 - \rh < x \leq 1 + \rh
       \\
   0 & 1 + \rh < x.
\end{cases}
\end{split}
\]
Considering only the interval $[- 1 + \rh, \, 1 - \rh]$,
we get
\[
\| f * g \|_p^p
 \geq [ 1 - \rh - (- 1 + \rh)] (2^{- 1/p})^p
 = 1 - \rh
 > (1 - \ep)^p,
\]
so
$\| f * g \|_p > 1 - \ep = (1 - \ep) \| f \|_1 \| g \|_p$,
as desired.. 

If $p = \I$,
take $f = \ch_{[0, 1]}$ and $g = 1$.
Then
\[
\| f \|_1 = \| g \|_{\I} = 1
\andeqn
f * g = 1,
\]
so
$\| f * g \|_{\I} = 1 > (1 - \ep) \| f \|_1 \| g \|_{\I}$.
\end{proof}


\begin{proof}[Alternate solution to part~(\ref{2017_NP_R84_AppId})]
We first assume $p < \I$.
Instead of choosing $f$ to be a multiple
of the characteristic function
of a short interval,
as in the first solution,
we choose $g$ to be a multiple
of the characteristic function
of a long interval.

Choose $M \in (0, \I)$
so large
that $\frac{1}{M} < 1 - (1 - \ep)^p$.
Define
\[
f = \left( \frac{1}{2} \right) \ch_{[- 1, 1]}
\andeqn
g = \left( \frac{1}{(2 M)^{1/p}} \right) \ch_{[- M, M]}.
\]
Then one checks directly that $\| f \|_1 = \| g \|_p = 1$.
Let $m$ be Lebesgue measure on~$\R$.
We have
\[
\begin{split}
(f * g) (x)
& = \int_{- \I}^{\I} f (y) g (x - y) \, d y
  = \left( \frac{1}{2 \cdot (2 M)^{1/p}} \right)
         \int_{- 1}^{1} \ch_{[- M, M]} (x - y) \, d y
\\
& = \left( \frac{1}{2 \cdot (2 M)^{1/p}} \right)
         \int_{- 1}^{1} \ch_{[x - M, \, x + M]} (y) \, d y
\\
& = \left( \frac{1}{2 \cdot (2 M)^{1/p}} \right)
         m \big( [- 1, 1] \cap [x - M, \, x + M] \big)
\\
& = \begin{cases}
   0 & x \leq - M - 1
        \\
   \frac{x + M + 1}{2 \cdot (2 M)^{1/p}}
           & - M - 1 < x \leq - M + 1
       \\
   \frac{2}{2 \cdot (2 M)^{1/p}} & - M + 1 < x \leq M - 1
        \\
   \frac{M + 1 - x}{2 \cdot (2 M)^{1/p}} & M - 1 < x \leq M + 1
       \\
   0 & M + 1 < x.
\end{cases}
\end{split}
\]
Considering only the interval $[- M + 1, \, M - 1]$,
we get
\[
\| f * g \|_p^p
 \geq [ M - 1 - (- M + 1)] ((2 M)^{- 1/p})^p
 = 1 - \frac{1}{M}
 > (1 - \ep)^p,
\]
so
$\| f * g \|_p > 1 - \ep = (1 - \ep) \| f \|_1 \| g \|_p$,
as desired.. 

If $p = \I$,
take $f = \ch_{[0, 1]}$ and $g = 1$.
Then
\[
\| f \|_1 = \| g \|_{\I} = 1
\andeqn
f * g = 1,
\]
so
$\| f * g \|_{\I} = 1 > (1 - \ep) \| f \|_1 \| g \|_{\I}$.
\end{proof}

\begin{proof}[Third solution to part~(\ref{2017_NP_R84_AppId})]
In this solution, we prove something stronger, and much more
interesting, namely that for {\emph{every}} $g \in L^p (\R)$
(for every $g \in C_0 (\R)$ if $p = \I$)
there exists a sequence $(f_n)_{n \in \N}$ in $L^1 (\R)$
such that $\| f_n \|_1 = 1$ for all $n$,
and such that $\limi{n} \| f_n * g \|_p = \| g \|_p$.
In fact, the sequence $(f_n)_{n \in \N}$ will depend on neither $p$
nor the choice of the function $g$, and we will even get
\[
% \begin{equation}\label{AppId}
\limi{n} \| f_n * g - g \|_p = 0.
% \end{equation}
\]

Choose any nonnegative \cfn{} $f \colon \R \to [0, \I)$
such that $\supp (f) \S [-1, 1]$
and $\int_{- \I}^{\I} f \, d m = 1$.
Define $f_n (x) = n f (x / n)$.

First suppose that $g \in C_{\mathrm{c}} (X)$.
Choose $M$ such that $\supp (g) \S [-M, \, M]$.
If $p \neq \I$, set
\[
\ep_0 = \frac{\ep}{2^{1/p} (M + 1)^{1/p}}.
\]
If $p = \I$, set $\ep_0 = \frac{1}{2} \ep$.
Use uniform continuity of $g$ to choose $\dt > 0$
such that whenever $s, t \in \R$ satisfy $| s - t | < \dt$,
then $| g (s) - g (t) | < \ep_0$.
\Wolog{} $\dt < 1$.
Let $n > 1 / \dt$, so that $\supp (f_n) \S [- \dt, \, \dt]$.
A change of variables allows us to rewrite the convolution as
\[
(f_n * g) (x) = \int_{- \I}^{\I} f_n (t) g (x - t) \, d t.
\]
Also, from $\int_{- \I}^{\I} f \, d m = 1$ we get
$\int_{- \I}^{\I} f_n \, d m = 1$, whence
\[
\begin{split}
| (f_n * g) (x) - g (x) |
& = \left| \int_{- \I}^{\I} f_n (t) g (x - t) \, d t
     - \int_{- \I}^{\I} f_n (t) g (x) \, d t \right|           \\
& \leq \int_{- \I}^{\I} f_n (t) | g (x - t) - g (x) | \, d t.
\end{split}
\]
The integrand on the right is zero whenever
$| t | \geq \dt$, and also whenever $| x | \geq M + \dt$.
For $| t | < \dt$ we have $| g (x - t) - g (x) | < \ep_0$.
From $\int_{- \I}^{\I} f_n \, d m = 1$ we therefore get
\[
| (f_n * g) (x) - g (x) | \leq \ep_0
\]
for all $x$.
When $p = \I$, this tells us that
$\| f_n * g - g \|_{\I} \leq \ep_0 < \ep$.
So assume $p \neq \I$.
We have already observed that
$(f_n * g) (x) = g (x) = 0$ for $| x | \geq M + \dt$.
Therefore
\[
\begin{split}
\| f_n * g - g \|_p
& \leq \left( \int_{- \I}^{\I}
           | (f_n * g) (x) - g (x) |^p \, d x \right)^{1/p}   \\
& < [2 (M + \dt) \ep_0^p]^{1/p}
  < 2^{1/p} (M + 1)^{1/p} \ep_0
  = \ep.
\end{split}
\]

Now we consider an arbitrary function $g \in L^p (\R)$
($g \in C_0 (\R)$ if $p = \I$).
Let $\ep > 0$.
Choose $h \in C_{\mathrm{c}} (X)$ such that
$\| h - g \|_p < \frac{1}{3} \ep$.
Choose $N$ so large that $\| f_n * h - h \|_p < \frac{1}{3} \ep$
for all $n \geq N$.
One checks that $f_n * g - f_n * h = f_n * (g - h)$.
Using the result of Part~(a) and $\| f_n \|_1 = 1$ for all $n$,
we then get
\[
\begin{split}
\| f_n * g - g \|_p
 & \leq \| f_n * (g - h) \|_p + \| f_n * h - h \|_p
                + \| h - g \|_p       \\
 & \leq \| f_n \|_1 \| g - h \|_p + \| f_n * h - h \|_p
                + \| h - g \|_p       \\
 & < \tfrac{1}{3} \ep + \tfrac{1}{3} \ep + \tfrac{1}{3} \ep
 = \ep
\end{split}
\]
for all $n \geq N$, as desired.
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Problem~6 in Chapter~7 of Rudin]\label{2017_NP_R806}
This problem counts as $1{.}5$ ordinary problems.
Do not use anything about polar coordinates from previous courses.

Let
\[
S^{d - 1} = \{ x \in \R^d \colon \| x \| = 1 \}
\]
be the unit sphere in $\R^d$.
Show that every $x \in \R^d \SM \{ 0 \}$ has a unique
representation $x = r z$ with $r \in (0, \I)$ and $z \in S^{d - 1}$.
Thus, $\R^d \SM \{ 0 \}$ may be regarded as the Cartesian product
$(0, \I) \times S^{d - 1}$.

Let $m_d$ be Lebesgue measure on $\R^d$.
Define a measure $\sm_{d - 1}$ on $S^{d - 1}$ by
\[
\sm_{d - 1} (E)
 = d \cdot m_d ( \{ r z \colon {\mbox{$z \in E$ and $0 < r < 1$}} \} )
\]
for every Borel set $E \S S^{d - 1}$.
Prove that for every nonnegative Borel function
$f \colon \R^d \to [0, \I]$ we have
\begin{equation}\label{IntPC}
\int_{\R^d} f \, d m_d
 = \int_0^{\I}
    r^{d - 1} \left( \int_{S^{d - 1}} f (r z) \, d \sm_{d - 1} (z)
       \right) \, d r.
\end{equation}
Check that this coincides with familiar results when $d = 2$ and
when $d = 3$.
\end{pbm}

\begin{proof}[Hint]
Check that the formula is true when $f$ is the characteristic
function of a set of the form
\[
\bigl\{ r z \colon {\mbox{$z \in E$ and $r_1 < r < r_2$}} \bigr\}
\]
for a Borel set $E \S S^{d - 1}$ and $0 \leq r_1 < r_2 \leq \I$.
Pass from these to characteristic functions of Borel sets in $\R^d$.
\phantom\qedhere
\end{proof}

\ifsolutions

\begin{proof}[Solution]
(This solution is a little sketchy at one point, and omits
the check of the last sentence when $d = 3$.)

To do the first paragraph properly, we really want to show that
$h (r, z) = r z$ is a homeomorphism
$h \colon (0, \I) \times S^{d - 1} \to \R^d \SM \{ 0 \}$.
To prove this, we simply write down the \ct{} inverse
$x \mapsto \bigl( \| x \|, \, (1 / \| x \|) x \bigr)$.

% \vspace{1ex}

Next, we verify the formula~(\ref{IntPC}).
First, let $E \S S^{d - 1}$ be Borel,
suppose $0 \leq r_1 < r_2 \leq \I$, and set
\[
R = \{ r z \colon {\mbox{$z \in E$ and $r_1 \leq r < r_2$}} \}
 \cap ( \R^d \SM \{ 0 \} ).
\]
(This set differs slightly from the set in the suggestion,
but is easier to work with.)
We verify~(\ref{IntPC}) for $f = \ch_R$.
Set
\[
S = \{ r z \colon {\mbox{$z \in E$ and $0 < r < 1$}} \}.
\]
Using $m_d (r F) = r^d m_d (F)$ for every Borel set $F$
(Theorem~2.20(e) of Rudin, together with the identification
of the factor appearing there as $| \det (T) |$),
from $r_1 S \S r_2 S$ and $R = r_2 S \SM r_1 S$ we get
\[
k \cdot m_d (R) = r_2^d \sm_{d - 1} (E) -  r_1^d \sm_{d - 1} (E).
\]
It is immediate to check that this is consistent with
the right hand side of~(\ref{IntPC}).

We next verify that~(\ref{IntPC}) holds for the characteristic
function of an open subset $U \S \R^d \SM \{ 0 \}$.
Choose a sequence of finite partitions ${\mathcal{P}}_n$
for $n \in \N$
of $S^{d - 1}$ into Borel sets such that
${\mathcal{P}}_{n + 1}$ refines ${\mathcal{P}}_n$ for every $n$,
and such that the diameter of each set in ${\mathcal{P}}_n$
is at most $2^{-n}$.
The collections
\[
{\mathcal{Q}}_n
 = \left\{ \left[ \frac{k}{2^n}, \, \frac{k + 1}{2^n} \right)
            \cap (0, \I)
     \colon {\mbox{$k \in \Z$ and $k \geq 0$}} \right\}
\]
are partitions of $(0, \I)$ with the same properties.
Then (some detail omitted)
$h^{-1} (V) \S (0, \I) \times S^{d - 1}$ is the
increasing union of the sets
%
\begin{equation}\label{Eq_4211_FnDfn}
F_n = \bigcup \{ A \times B
        \colon {\mbox{$A \in {\mathcal{P}}_n$,
                     $B \in {\mathcal{Q}}_n$,
                     and $A \times B \S h^{-1} (V)$}} \}.
\end{equation}
%
Thus $U$ is the increasing union of the sets $h (F_n)$.
For each $n \in \N$,
the set $h (F_n)$ is a disjoint union of sets $h (A \times B)$
with $A \in {\mathcal{P}}_n$ and $B \in {\mathcal{Q}}_n$,
and we proved above that~(\ref{IntPC}) holds for the characteristic
functions of such sets.
Therefore~(\ref{IntPC}) holds for $\ch_{F_n}$ by the
version of the Monotone Convergence Theorem for series
of nonnegative functions.
Thus~(\ref{IntPC}) holds for $\ch_U$ by the usual version
of the Monotone Convergence Theorem.

It now follows from the Dominated Convergence Theorem
that~(\ref{IntPC}) holds for the characteristic
function of a bounded $G_{\dt}$-set $G \S \R^d \SM \{ 0 \}$.
Every bounded Borel set $F \S \R^d \SM \{ 0 \}$ is contained in a
bounded $G_{\dt}$-set $G \S \R^d \SM \{ 0 \}$
such that $m_d (G \SM F) = 0$.
If $m_d (F) = 0$,
applying the first sentence to $G$, we see that
both sides of~(\ref{IntPC}) for $\ch_G$ are zero.
By monotonicity, both sides of~(\ref{IntPC}) for $\ch_F$ are zero.
Thus, (\ref{IntPC})~holds for bounded Borel sets of measure zero.
Letting $F$ be an arbitrary bounded Borel set
and letting $G$ be as before,
we find that (\ref{IntPC})~holds for $\ch_G$ and $\ch_{G \SM F}$.
Since everything is finite, we subtract, showing that
(\ref{IntPC})~holds for $\ch_F$.

Now let $F \S \R^d \SM \{ 0 \}$ be an arbitrary Borel set.
Applying the Monotone Convergence Theorem to the characteristic
functions of $F_n = F \cap B_n (0)$,
we see that (\ref{IntPC})~holds for $\ch_F$.
Writing an arbitrary nonnegative Borel function as the
pointwise increasing limit of simple nonnegative Borel functions
and applying the Monotone Convergence Theorem once again,
we see that (\ref{IntPC})~holds for arbitrary
nonnegative Borel functions on $\R^d \SM \{ 0 \}$.
Since $m_d ( \{ 0 \} ) = 0$,
we may replace $\R^d \SM \{ 0 \}$ by $\R^d$.

\vspace{1ex}

It remains to check that this gives the usual results
in dimensions~$2$ and~$3$.
We give the solution only in dimension~$2$;
dimension~$3$ is handled the same way.

Recall the usual formula:
\[
\int_{\R^2} f \, d m_2
 = \int_0^{\I} r \left( \int_0^{2 \pi}
         f \bigl( r \cos (\te), \, r \sin (\te) \bigr) \, d \te
       \right) \, d r.
\]
We thus want to show that if $f$ is a nonnegative Borel function on
$S^1 \S \R^2$, then
\[
\int_{S^1} f \, d \sm_1
 = \int_0^{2 \pi} f \bigl( \cos (\te), \,  \sin (\te) \bigr) \, d \te.
\]
Define $g \colon B_1 (0) \SM \{ 0 \} \to [0, \I]$
by
\[
g (x, y)
 = f \left( \frac{x}{\sqrt{x^2 + y^2}}, \, \frac{y}{\sqrt{x^2 + y^2}}
       \right).
\]
That is, $g (w) = f ( (1 / \| w \| ) w)$.
Using the definition of $\sm_1$ at the first step,
and Fubini and the change of variables formula
for the function
$k \colon (0, 1) \times (0, 2 \pi) \to B_1 (0) \SM \{ 0 \}$
given by $k (r, \te) = \bigl( r \cos (\te), \, r \sin (\te) \bigr)$
at the second step, we get
\[
\begin{split}
\int_{S^1} f \, d \sm_1
& = 2 \int_{B_1 (0) \SM \{ 0 \}} g \, d m_2
  = 2 \int_0^1 r \left( \int_0^{2 \pi}
         f \bigl( \cos (\te), \,  \sin (\te) \bigr) \, d \te \right)
                    \, d r      \\
& = \int_0^{2 \pi} f \bigl( \cos (\te), \,  \sin (\te) \bigr) \, d \te,
\end{split}
\]
as desired.
\end{proof}

It is essential for the argument that the union in~(\ref{Eq_4211_FnDfn})
be disjoint, even when only finitely many sets are involved
(such as when $U$ is bounded).
The previous step doesn't help with the union of
two sets of the form used when they are not disjoint.

\begin{proof}[Alternate solution (sketch)]
We describe a different method to
verify the formula~(\ref{IntPC}) for the characteristic functions
of arbitrary open sets.
From this point on, the rest of the proof is the same.

First, let $E \S S^{d - 1}$ be Borel,
suppose $0 \leq r_1 < r_2 \leq \I$, and set
\[
R (r_1, r_2, E)
 = \{ r z \colon {\mbox{$z \in E$ and $r_1 \leq r < r_2$}} \}
    \cap ( \R^d \SM \{ 0 \} ).
\]
One verifies~(\ref{IntPC}) for $f = \ch_{R (r_1, r_2, E)}$
as in the first solution.
Next, again with $0 \leq r_1 < r_2 \leq \I$, set
\[
S (r_1, r_2, E)
  = \{ r z \colon {\mbox{$z \in E$ and $r_1 < r < r_2$}} \}.
\]
Then $S (r_1, r_2, E)$ is an increasing union
\[
S (r_1, r_2, E)
 = \bigcup_{n = 1}^{\I}
           R \left( r_1 + \tfrac{1}{n}, \, r_2, \, E \right).
\]
Applying the Monotone Convergence Theorem
on both sides of~(\ref{IntPC}),
with $f_n = \ch_{R \left( r_1 + 1/n, \, r_2, \, E \right)}$,
we see that~(\ref{IntPC}) holds for $f = \ch_{S (r_1, r_2, E)}$.
In particular, it holds for the characteristic function of
any open set
of the form $S (r_1, r_2, E)$ with $E \S S^{d - 1}$ open.

Now we claim that~(\ref{IntPC}) holds for any finite union
of sets of the form $S (r_1, r_2, E)$.
This is proved by induction on the size of the union.
We know it holds for one such set.
So assume it holds for any union of $n$ such sets,
and let $S_1, S_2, \ldots, S_{n + 1}$ be $n + 1$ such sets.
Set
\[
E = \bigcup_{k = 1}^{n + 1} S_k, \,\,\,\,\,\,
F = \bigcup_{k = 1}^n S_k
\andeqn
G = F \cap S_{n + 1} = \bigcup_{k = 1}^n (S_k \cap S_{n + 1}).
\]
Each set $S_k \cap S_{n + 1}$ is easily checked to be again of
the form we are considering.
So the formula~(\ref{IntPC}) holds for $\ch_F$ and $\ch_G$
by the induction hypothesis,
and for $\ch_{S_{n + 1}}$ by the case $n = 1$.
Since $\ch_E = \ch_F + \ch_{S_{n + 1}} - \ch_G$
and everything is finite,
we conclude that the formula~(\ref{IntPC}) holds for $\ch_E$.
This completes the induction.

Now apply the Monotone Convergence Theorem
to show that the formula~(\ref{IntPC}) holds for $\ch_E$
whenever $E$ is a countable union
of sets of the form $S (r_1, r_2, E)$.

Next, one proves using easy point set topology arguments
that every open subset of $\R^d \SM \{ 0 \}$ is a countable union of
sets of the form $S (r_1, r_2, E)$ with $E$ open.
It follows that~(\ref{IntPC}) holds for $\ch_U$
for every open set $U \S \R^d \SM \{ 0 \}$.
The conclusion for arbitrary open sets $U \S \R^d$
follows, because $\{ 0 \}$ has measure zero.

The proof is now completed as in the first solution.
\end{proof}

\begin{proof}[Second alternate solution (brief sketch)]
We could also verify~(\ref{IntPC}) for characteristic functions
of Borel sets by showing that the collection
${\mathcal{C}}$ of sets $E$
such that it holds for $\ch_E$ is a monotone class
containing all finite disjoint unions of sets $h (A \times B)$
with $A \S (0, \I)$ and $B \S S^{d - 1}$ Borel.
Using Theorem~7.3 of Rudin, it is easy to show that every
such class contains all Borel subsets of $\R^d \SM \{ 0 \}$.
The proof that ${\mathcal{C}}$ is a monotone class
is essentially the same as the proof of Theorem~7.6 of Rudin.
\end{proof}
\fi


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{pbm}[Taken from some edition of Rudin, but not in the one
  I am working from]\label{2017_NP_R812}
This problem counts as $1{.}5$ ordinary problems.
(There are a number of estimates to do.
Be sure to prove that the hypotheses of the theorems you use are
really satisfied.)

Use Fubini's Theorem and the relation
\[
\frac{1}{x} = \int_0^{\I} e^{- x t} \, d t
\]
for $x > 0$ to prove that
\[
\limi{a} \int_0^a \frac{\sin (x)}{x} \, d x = \frac{\pi}{2}.
\]
\end{pbm}

\begin{rmk}\label{R_2017_NP_R812_NotInt}
The function $f (x) = \frac{\sin (x)}{x}$
is not Lebesgue integrable on $(0, \I)$,
because
\[
\int_0^{\I} \left| \frac{\sin (x)}{x} \right| \, d x = \I.
\]
The problem asserts the existence of the improper Riemann integral,
not of the Lebesgue integral.
\end{rmk}

\ifsolutions

\begin{proof}[Solution to Problem~\ref{2017_NP_R812}]
We first claim that if $a > 0$ then
\[
\int_0^a \frac{\sin (x)}{x} \, d x
 = \int_0^{\I} \left(
         \int_0^a  e^{- x t} \sin (x) \, d x \right) \, d t.
\]
Define $f \colon [0, \I) \times (0, a] \to \R$
by $f (t, x) = e^{- x t} \sin (x)$.
Then $f$ is \ct, hence \mb.
We have $| f (t, x) | \leq x e^{- x t}$ for all $t, x \in [0, \I)$,
and an application of Fubini's Theorem for nonnegative
functions
% (ignoring $x = 0$ because that only happens
% on a set of measure zero)
shows that
\[
\int_{[0, \I) \times [0, a]} x e^{- x t} d (m \times m) (t, x)
  = \int_0^a \left(
         \int_0^{\I} x e^{- x t} \, d t \right) \, d x
  = \int_0^a 1 \, d x
  < \I.
\]
It follows that $f$ is \igb{} on $[0, \I) \times (0, a]$.
(Remembering to do this step is a substantial part of the problem.)
Now we may apply Fubini's Theorem for \igb{}  functions
to get
\[
\int_0^{\I} \left(
         \int_0^a  e^{- x t} \sin (x) \, d x \right) \, d t
  = \int_0^a \left(
         \int_0^{\I}  e^{- x t} \sin (x) \, d t \right) \, d x
  = \int_0^a x^{- 1} \sin (x) \, d x.
\]
This proves the claim.

We next claim that
\[
\int_0^a  e^{- x t} \sin (x) \, d x
 = \frac{1}{1 + t^2}
       - \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2}.
\]
This is most easily done by writing
$\sin (x) = \frac{1}{2 i} ( e^{ i x} - e^{- i x} )$,
combining the products of exponentials,
and doing some algebra after the integration is done.
It can also be done by two integrations by parts:
one gets an equation for the integral, which can be solved.
Details are omitted.

Now we claim that
\[
\limi{a} \int_0^{\I}
      \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \, d t
 = 0.
\]
To prove this, we estimate, for $t \geq 0$:
\[
\left| \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \right|
  \leq \frac{e^{- a t} ( | \cos (a) | + t | \sin (a) | )}{1 + t^2}
  \leq  \frac{e^{- a t} ( 1 + t )}{1 + t^2}.
\]
Separate estimates on $[0, 1]$ and on $[1, \I)$ show that
\[
\frac{1 + t}{1 + t^2} \leq 2
\]
for all $t \geq 0$.
Therefore
\[
\left| \int_0^{\I}
      \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \, d t
              \right|
  \leq \int_0^{\I} 2 e^{- a t} \, d t
  = \frac{2}{a},
\]
from which the claim follows.

Now we prove the result.
We have
\[
\begin{split}
\limi{a} \int_0^a \frac{\sin (x)}{x} \, d x
& = \limi{a} \int_0^{\I} \left(
         \int_0^a  e^{- x t} \sin (x) \, d x \right) \, d t  \\
& = \int_0^{\I} \frac{1}{1 + t^2} \, d t
       - \limi{a} \int_0^{\I}
          \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \, d t
                \\
& = \int_0^{\I} \frac{1}{1 + t^2} \, d t = \frac{\pi}{2}.
\end{split}
\]
This completes the proof.
\end{proof}

\begin{rmk}\label{2017_NP_R812_R}
One can also use the Dominated Convergence Theorem to
prove the third claim, that
\[
\limi{a} \int_0^{\I}
      \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \, d t
 = 0.
\]
One must, of course, prove that for every sequence
$(a_n)_{n \in \N}$ in $[0, \I)$ such that $\limi{n} a_n = \I$,
one has
\[
\limi{n} \int_0^{\I}
   \frac{e^{- a_n t} ( \cos (a_n) + t \sin (a_n) )}{1 + t^2} \, d t
 = 0.
\]
One part, pointwise convergence of the integrand to zero
almost everywhere, is easy.
The other part requires
a nonnegative \igb{}  function $g \colon [0, \I) \to \R$
such that
\[
\left| \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \right|
  \leq g (t)
\]
for all $a \in [0, \I)$ and $t \in [0, \I)$.
In fact, one easily sees that it is enough to find $g$ and some $N > 0$
such that this estimate holds
for all $a \in [N, \I)$ and $t \in [0, \I)$.
We sketch two possibilities.

First, the proof above shows that
\[
\left| \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \right|
  \leq  2 e^{- a t}
\]
for all $a \in [0, \I)$ and $t \in [0, \I)$.
Therefore the estimate holds for $a \geq 1$ with
$g (t) = 2 e^{- t}$.

Second, the proof above shows that
\[
\left| \frac{e^{- a t} ( \cos (a) + t \sin (a) )}{1 + t^2} \right|
  \leq  \frac{e^{- a t} ( 1 + t )}{1 + t^2}
  \leq  \frac{1 + t e^{- a t}}{1 + t^2}
\]
for all $a \in [0, \I)$ and $t \in [0, \I)$.
If $a \geq 1$ then $t e^{- a t} \leq t e^{-t} \leq 1$,
so the estimate holds for $a \geq 1$ with
$g (t) = \frac{2}{1 + t^2}$.
\end{rmk}
\fi

\end{document}
