# Baby Rudin, chapter 10, problem 1 - independence of order in multiple integrals.

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

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## 1 Answer

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:

1. $F$ converges to $f$, monotonically, at every point except possibly the boundary of $H$.
2. $f$ is bounded above
3. Outside of $H$, $F$ agrees with $f$.
4. 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 $(a_1,\dots,a_{k-1})$.

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Thank you for your answer! I'm a little upset because the method of proof in example 10.4 doesn't seem to help at all. – user1337 Mar 6 '13 at 16:31
@MichaelTouitou There is some resemblance between (*) and (7). Inequality (7) is based on "transversality" of intersection between line segments and the boundary (they make nonzero angle). For general convex sets this intersection is not transverse everywhere, but it is transverse when we stay away from the boundary of the projection. – user53153 Mar 6 '13 at 16:45
@user53153: I am confused by the first paragraph of your response. It appears that the k-simplex $H$ defined by Rudin is a convex polygon with an edge parallel to $x_k$, and yet Rudin proves that $f_{k-1}$ is continuous. – Doug Jun 26 '14 at 0:10
It should be mentioned that User1337 is forgetting an assumption by Rudin that $H$ has a nonempty interior. – Doug Jun 26 '14 at 0:11