If $A$ and $f$ are bounded, then $f$ is integrable in the extended sense (?) [Spivak]

I have a problem with one of the theorems in Spivak's Calculus on Manifolds. I will give some background first:

An open cover $\mathcal{O}$ of an open set $A \subset \mathbb{R}^n$ is admissible if each $U \in \mathcal{O}$ is contained in $A$. If $\Phi$ is subordinate to $\mathcal{O}$, $f:A \to \mathbb{R}$ is bounded in some open set around each point of $A$, and $\{x:f \text{ is discontinuous at } x \}$ has measure $0$, then each $\int_A \varphi \cdot |f|$ exists. We define $f$ to be integrable (in the extended sense) if $\sum_{\varphi \in \Phi} \int_A \varphi \cdot |f|$ converges (the proof of Theorem 3-11 shows that the $\varphi$'s may be arranged in a sequence). This implies convergence of $\sum_{\varphi \in \Phi} \int_A \varphi \cdot f$, which we define to be $\int_A f$. These definitions do not depend on $\mathcal{O}$ or $\Phi$ (but see Problem 3-38)

Then comes part (2) theorem 3-12:

If $A$ and $f$ are bounded, then $f$ is integrable in the extended sense.

The proof is as follows:

If $A$ is contained in the closed rectangle $B$ and $|f(x)| \leq M$ for $x \in A$, and $F \subseteq \Phi$ is finite, then

$$\sum_{\varphi \in F} \int_A \varphi \cdot |f| \leq \sum_{\varphi \in F} M \int_A \varphi=M \int_A \sum_{\varphi \in F} \varphi \leq Mv(B), \tag{1}$$

since $\sum_{\varphi \in F} \varphi \leq 1$.

My difficulties:

Say $f=1$, $A=\mathbb{Q} \cap [0,1]$ and take $B=[0,1]$. Since $A$ is compact we can find a finite partition of unity $\Phi$. Taking $F=\Phi$ in (1) means we have to deal with the expression

$$\int_A \sum_{\varphi \in F} \varphi=\int_A 1,$$

which doesn't seem to exist, since $A$ is not Jordan-measurable.

Is this really a problem? If so, can it be resolved?

Thanks!

EDIT:

Obviously, $A$ in my example is not compact. However, I still don't understand why the integrals $\int_A \varphi \cdot |f|$ exist, as $A$ is an arbitrary bounded set.

Furthermore, the theorem implies that $\int_A 1$ exists where $A=\mathbb{Q} \cap [0,1]$. How can this be?

• This $A$ is not open ("An open cover $\mathcal O$ of an open set $A$"), nor is it compact (it is not closed because it does not contain all its limit points, i.e. there are rational sequences converging to irrational limits). – BaronVT Apr 20 '15 at 18:29
• @BaronVT Thanks, I've edited my question now. – user1337 Apr 20 '15 at 18:35
• It still seems to me that Spivak is requiring $A$ to be open, which $\mathbb Q \cap [0,1]$ is not. – BaronVT Apr 20 '15 at 18:57