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

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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.