R&C Analysis-Prop 1.25

Proposition 1.25:
Let $s$ and $t$ be nonnegative measurable simple functions on $X$. For $E \in R$, define
$\varphi(E)=\int_E s d\mu$
Then $\varphi$ is a measure. Also,
$\int_E (s+t) du = \int_E s d\mu + \int_E t d\mu$
Proof:
Measure should be less than $\infty$ satisfy countable additive property. Using the definition for integral of nonnegative measurable simple functios
Let $E_1,E_2,\cdots$ be disjoint members of $E$. Then,
$\varphi(E) = \sum_{i=1}^\infty \alpha_i \mu(A_i \cap E)$
$=\sum_{i=1}^\infty \alpha_i \sum_{r=1}^\infty \mu(A_i \cap E_r)=\sum_{r=1}^\infty \sum_{i=1}^\infty \alpha_i \mu(A_i \cap E_r)$
$=\sum_{r=1}^\infty \varphi(E_r)$
which shows that $\varphi$ has countable additive property.
Since $\varphi(\phi) = 0$, $\varphi$ is not identically $\infty$.

Next part of proof assumes considers $\beta_1,\beta_2,\cdots,\beta_j$ as distinct values of $t$ and lets
$B_j=\{x:f(x)=\beta_j\}$. Obviously, for the sum of simple functions, the measure is defined only $E_{ij}=A_i \cap B_j$.
Then, $\int_{E_{ij}}(s+t) d\mu = \sum_{i,j=1}^\infty (\alpha_i+\beta_j)\mu(E_{ij})$
And
$\int_{E_{ij}} s d\mu + \int_{E_{ij}}t d\mu = \sum_{i=1}^\infty \alpha_i \mu_{E_{ij}} + \sum_{j=1}^\infty \beta_j \mu_{E_{ij}}$
Couple of observations. Book seem to have missed the sums before RHS for both these equations.
Instructive to see the master stroke, Rudin delivers here.
First observation is that sum of simple functions equation now holds for $E_{ij}$ instead of $X$. Since $X$ is disjoint union of $E_{ij}(1\leq i \leq n, 1 \leq m)$ and it is already established that the integrals of simple functions yield a measure and from the facts measures can be added,, conclusion follows.