Can we always integrate continuous function on compact set, by using riemann integral?

I'm currently studying Baby Rudin chapter 10, which is about integration over sets in $$\mathbb{R}^n$$. For continuous function $$f:\mathbb{R}^k \to \mathbb{R}$$, Rudin defines Riemann integral of $$f$$ over $$k$$-cell as follows:

Let $$k$$-cell $$I^k$$ as a set of $$x=(x_1 , x_2 , \cdots, x_k )$$ such that $$a_i \le x_i \le b_i$$.

Define $$f_k =f$$, and recursively define $$\begin{eqnarray} f_i (x_1 , \cdots , x_i ) = \int_{a_{i+1}}^{b_{i+1}}{f_{i+1} (x_1 , \cdots , x_i , t)dt} \end{eqnarray}$$

Now, Riemann integral of $$f$$ over $$I^k$$ is defined as a real number $$f_0$$.

It can be easily proved that this is well-defined, using uniform continuity.

In Definition 10.3, if $$f$$ is a continuous function with compact support, Rudin defines integral of $$f$$ over $$R^k$$ as $$\int_{R^k} f = \int_{I^k} f$$, where $$I^k$$ is $$k$$-cell which contains support of $$f$$.

And finally, Exercise 10.1 is the following:

Let $$H$$ be a compact convex set in $$\mathbb{R}^k$$, with nonempty interior. Let $$f \in C(H)$$, put $$f(x)=0$$ in the complement of $$H$$, and define $$\int_H f$$ as in Definition 10.3. Prove that $$\int_H f$$ is independent of the order in which the $$k$$ integrations are carried out.

Here, I'm a bit suspicious about the existence of $$\int_H f$$. To define $$\int_{I^k} f$$, ($$H \subset I^k$$), $$f_i$$ must be integrable over $$[a_i , b_i ]$$. ($$f_i , a_i , b_i$$ are defined at first yellow box) I wanted to prove this, but I failed. And I am even thinking it might not be true. Below is my progress, not written kindly.

I wanted to prove by induction in $$k$$. We need,

(i) projection of compact convex set is compact convex.

(ii) $$f_i$$ is (uniformly) continuous.

(i) can be proved easily. (ii) is a main problem, since I found counterexample. (which might be false :) )

If $$H$$ is $$\{ (x,y,z) | 0 \le z \le 1 , (x-1+z)^2 +y^2 \le (1-z)^2 \}$$, an inclined cone, and $$f=1$$, then integration over $$z$$-axis gives non-continuous function (discontinuity at $$(0,0)$$).

Of course, in above case, further integration can be well defined. But anyways, my proof can't be proceeded further.

Can anyone help me prove this, or find any counterexamples?

Thank you for reading my question. Hope to get satisfying answer here. Of course, any small ideas are also welcome.

• @Masacroso Thank you for your comment, but I'm quite confused with your comment. Assuming that you are saying that $f_i$ are always continuous, (so that condition (ii) in my post is true) I want you to check my counterexample once again. I think, if I integrate $f=1$ over $z$-axis, $(0,0)$ has value $1$, while other points on circle $(x-1)^2 +y^2 =1$ has value $0$, so that it($f_2$) is discontinuous at $(0,0)$. Sorry if I misunderstood your comment. (p.s. I can prove $f_{k-1}$ is continuous on its domain's interior, but nothing else.) – sansae Oct 9 at 17:44
• @Masacroso - sansae is correct in this. The $f_k$ are not even necessarily continuous, much less uniformly continuous. $f$ is chosen continuous on $H$, but nothing is said about the behavior of $f$ on the boundary of $H$. It need not be $0$ there. And if it isn't, then the extension to a cell will be discontinuous at the boundary. – Paul Sinclair Oct 9 at 23:37
• sanae, But you are mistaken to require the $f_i$ to be unitformly continuous. $f$ is uniformly continuous on $H$, and $0$ outside. You can prove well-definedness from that. You do not need $f_i$ to be uniformly continuous on the whole cell. Note that this definition follows the same path as Rudin's, but doesn't depend on it. – Paul Sinclair Oct 9 at 23:41
• @PaulSinclair Thanks for your hopeful comment. I also thought about proving $f_i$ uniformly continuous on interior of $H$, but I failed. Can you please give me some hint? Thanks again. – sansae Oct 10 at 1:12
• $H$ is compact, $f$ is continuous on $H$ (not just the interior of $H$). A continuous function on a compact set is automatically uniformly continuous. For each $x \in H$, the set $U_x = \{y \in H : |f(y) - f(x)| < \epsilon/2\}$ is open in $H$ and the collection of $U_x$ for all $x$ covers $H$. Because $H$ is compact, that collection has a finite subcover. – Paul Sinclair Oct 10 at 16:22