Proposition 4.23 in Royden & Fitzpatrick I'm studying Proposition 4.23 (given as follows) in Royden & Fitzpatrick's Real Analysis.

$\textbf{Proposition}$ Let $f$ be a measurable function on $E$. If $f$
is integrable over $E$,  then for each $\varepsilon > 0$, there is a
$\delta > 0$ for which
\begin{equation} \text{if $A \subset E$ is
measurable and $m(A) < \delta$, then $\int_{A} \lvert f \rvert <\varepsilon$} \tag{26} 
\end{equation}
Conversely, if $m(E) < \infty$,
if for each $\varepsilon > 0$, there is a $\delta > 0$ for which (26)
holds, then $f$ is integrable over $E$.

For the "converse" part, the authors say that

Let $\delta_0 > 0$ respond to the $\varepsilon = 1$ challenge.
Since $m(E) < \infty$, according to the preceding lemma, $E$
can be expressed as the disjoint union of a finite collection of
measurable subsets $\{E_k\}_{k=1}^{N}$, each of which has measure
less then $\delta_0$. Therefore,
\begin{equation}
\sum_{k=1}^{N} \int_{E_k}  f  < N 
\end{equation}
By the additivity over domains of integration it follows that if $h$ is a nonnegative measurable function of finite support and $0 \leqslant h \leqslant f$ on $E$, then $\int_{E}h < N$. Therefore $f$ is integrable.

This argument seems a little bit weird to me because I think it is unnecessary to introduce $h$. Since
\begin{equation*}
\int_{E} \lvert f \rvert = \sum_{k=1}^{N} \int_{E_k}  \lvert f \rvert < N 
\end{equation*}
we can conclude directly that $f$ is integrable. Do I miss something?
 A: The excerpt you have taken from Royden's proof of the converse is additionally confusing because the argument is made for a nonnegative function $f$ and this is not a hypothesis.  Of course, what is not shown is the statement that the theorem is proved by establishing it separately for the positive and negative parts of $f$, and it can be assumed that $f \geqslant 0$ without loss of generality.
So to prove Lebesgue integrability of a nonnegative function $f$ over $E$ it suffices to show that $\int_E f \neq  +\infty$. (Recall that the Lebesgue integral of a nonnegative measurable function is always defined although it may  be infinite and "Lebesgue integrability"  means that the integral is finite.)
You can proceed as Royden has done by invoking the definition of the Lebesgue integral  of a nonnegative function as a supremum over integrals of bounded measurable functions of finite support.
There is another way, which you seem to be  arriving at.  Since you are trying to prove that $\int_Ef$ is a finite real number, it does not work as a valid proof is to simply write $\int_E f = \sum \ldots$ without justification, thereby initiating a circular argument. However, you can make this rigorous by applying  Royden's Theorem 11 (Additivity over Domains of Integration).
Since$f$ is nonnegative,  $\int_{E_k} f < 1 < +\infty$ for $k=1,2,\ldots, N$and $E = \cup_{k=1}^N E_k$ is a disjoint union, it follows from Theorem 11 that
$$\int_E f = \sum_{k=1}^N \int_{E_k} f  < N < +\infty$$
Of course, the proof of Theorem 11 relies on a similar approach to what Royden uses in completing the proof of this proposition. In the end quoting Theorem 11 is not providing a "better" proof.
