TITLE: Baby Rudin, chapter 10, problem 1 - independence of order in multiple integrals.
QUESTION [8 upvotes]: In problem 1, Rudin asks for a generalization of example 10.4, which states that if $f$ is continuous in the standard k-simplex $Q^k$, then the integral $\int_{Q^k} f$ exists, and that the order of the $n$ separate integrations is immaterial. He proves that by approximating $f$ by a sequence of functions which are continuous in a k-cell containing $Q^k$.
In problem 1 we need to show that the same hold, if $Q^k$ is replaced by any compact and convex $H \subseteq \mathbb R^k$. There's a hint, suggesting that one should approximate $f$ by continuous functions in $\mathbb R^k$ with support $\subseteq H$.
For fixed $\delta>0,$I've introduced the function $\varphi_\delta(t)
=\begin{cases}
\frac{1}{\delta}t & 0 \le t \le \delta \\
1 & t>\delta
\end{cases}$ 
and approximating $f$ by $F(x)=f(x) \varphi_\delta(\rho_{H^c}(x))$, where $\rho_E(x)=\inf \{ d(x,y) : y \in E\}$ is the distance between $x$ and $E$.
I can't follow his proof, Namely the inequality (7) on page 247 is problematic for me.
Thanks in advance
EDIT: Lebesgue's theory is not allowed

REPLY [4 votes]: Inequality (7) is false for general $H$. And for a good reason. Recall that in Rudin's notation 
$$f_{k-1}(x_1,\dots,x_{k-1})=\int_{} f(x_1,\dots,x_k)\,dx_k$$
where I omit the limits of integration (they correspond to the size of some cube containing $H$). 
In Example 10.4 the function $f_{k-1}$ turned out to be continuous. But for general convex sets $H$ it 
is not. For example, consider some convex polygon in the plane: if one of its edges happens 
to be parallel to the axis $x_k$, and if $f\equiv 1$, the function $f_{k-1}$ will be discontinuous. 
Consequently, there is no hope to approximate it uniformly by $F_{k-1}$, which is what (7) was for. 
We must set a more modest goal. As $\delta\to 0$, the functions $F$ converge to $f$ in the following sense: 

$F$ converges to  $f$, monotonically, at every point except possibly the boundary of $H$.
$f$ is bounded above
Outside of $H$, $F$ agrees with $f$. 
On every compact set $K$ contained in the interior of $H$, the convergence $F\to f$ is uniform. 

Lemma. If $F$, $f$ and $H$ satisfy (1,2,3,4), then so do $F_{k-1}$, $f_{k-1}$ and $H_{k-1}$, where
$H_{k-1}$ is the orthogonal projection of $H$ along the $x_k$ axis. 
Assume the lemma for now. Apply it  first to $F$, $f$ and $H$, then to $F_{k-1}$, $f_{k-1}$ and $H_{k-1}$, etc. 
Eventually we arrive at $H_1$ being a closed interval, $F_1$ converging to $f_1$ in the sense described above. 
It is an easy exercise in one-dimensional integration to prove that
$$\int F_1\to \int f_1$$
Now we are done, because the integrals $\int F_1$ do not depend on the numbering of coordinates (this
was proved early in Chapter 10).  
It remains to prove the lemma, which is not super difficult but is quite boring... a good illustration of 
why people do not want to use Riemann integral in higher dimensions. It helps to draw pictures.
By a standard compactness argument, it suffices to prove that every point of the interior of $H_{n-1}$ has a
neighborhood on which convergence is uniform. Pick such a point 
$(a_1,\dots,a_{k-1})$. Consider the set $$T=\{t\in\mathbb R: (a_1,\dots,a_{k-1},t)\in H\}$$ which is a bounded interval.
Introduce the function
  $$d(t)=\operatorname{dist}((a_1,\dots,a_{k-1},t),\partial H)$$
Show that $d$ is concave (because $H$ is convex). We have $d(t_0)>0$ for some $t_0$. 
By concavity this implies a linear bound of the form 
$$d(t)>c \operatorname{dist}(t,\partial T)\tag{*}$$ where $\partial T$
is the boundary of the interval $T$. This gives you a version of (7): 
the length of the set $\{t \in T: d(t)<\delta\}$ is at most $c^{-1}\delta$. Moreover, since $d$ is uniformly (Lipschitz)
continuous, an
estimate  similar to (*) holds for segments parallel to $(a_1,\dots,a_{k-1},t)$ and near it. This gives a
uniform bound on $f_{k-1}-F_{k-1}$ in a neighborhood of $(a_1,\dots,a_{k-1})$.