Integral of simple functions in standard and non-standard representation

Suppose $m$ is finite. Using the linearity of the integral, we have $$ \int \left(\sum_{k=1}^m b_k1_{F_k}(x)\right)\mu(\mathrm dx) = \sum_{k=1}^m b_k\int1_{F_k}(x)\mu(\mathrm dx) = \sum_{k=1}^mb_k\mu(F_k) $$ regardless of the shape of the collection $F_k$.

If you don't want to use linearity, note that given some finite collection $\{F_k\}_{k=1}^m$ of measurable sets there is a unique coarsest partition $\mathscr G = \{G_i\}_{i=1}^n$ such that $F_k$ are unions of some elements in $\mathscr G$. Let $g:\{1,\dots,n\}\to2^{\{1,\dots,m\}}$ be the index function uniquely defined by $$ k\in g(i)\quad\Leftrightarrow\quad G_i\subset F_k $$ for any $k\in\{1,\dots,m\}$ and any $i\in \{1,\dots,n\}$. Furthermore, note that for any $k\in \{1,\dots,m\}$ the inverse of $g$ satisfies $$ \{i:k\in g(i)\} = \{i:G_i\subset F_k\} $$ is partition of $F_k$ in $\mathscr G$. In particular, $\mu(F_k) = \sum_{i:k\in g(i)}\mu(G_i).$ Then we have: $$ \varphi(x) = \sum_{k=1}^m b_k1_{F_k}(x) = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)1_{G_i}(x) $$ where the first equality is the definition of $\varphi$ and the latter function is standard simple one. Thus $$ \int\varphi\;\mathrm d\mu = \sum_{i=1}^n\left(\sum_{k\in g(i)}b_k\right)\mu(G_i) = \sum_{k=1}^n b_k\left(\sum_{i:k\in g(i)}\mu(G_i)\right) = \sum_{k=1}^n b_k\mu(F_k) $$ where we passed from the summation over $G_i$ to the summation over $b_k$.