Okay so I have a measurable function $f$ and a set $E_{n,i}$ $$E_{n,i}=\left\{ x:\frac{i-1}{2^{n}} \leq f(x)<\frac{i}{2^{n}}\right\}$$ where $i=1,...,n2^{n}$ and $n=1,2,...$
Then I have another set $A_{n}$ given by $$ A_{n}=\{x:f(x)\geq n\} $$ and then I have a sequence of functions $s_{n}$ given by $$ s_{n}=\sum^{n2^{n}}_{i=1} \frac{i-1}{2^{n}} \mathbb{1}_{E_{n,i}} +n\mathbb{1}_{A_{n}} $$ So I'm slightly unsure about whether my approach to evaluate this is correct. My instincts tell me to first fix $n$ and then work from there. Say $n=4$ then we would have $$ s_{4}=\sum^{4(2^{4})}_{i=1} \frac{i-1}{2^{4}} \mathbb{1}_{E_{4,i}} +4\mathbb{1}_{A_{4}}$$ which then gives us $$ s_{4}=\sum^{64}_{i=1} \frac{i-1}{16} \mathbb{1}_{E_{4,i}} +4\mathbb{1}_{A_{4}}$$ and I feel pretty confident that is correct but I have no one to verify this. Obviously it would be pretty ridiculous to try and do this by hand but just to get the idea of a few of the sets $E_{4,i}$ I have $$E_{4,1}=\left\{ x:0 \leq f(x)<\frac{1}{16}\right\}$$ and then $$ E_{4,2}=\left\{ x:\frac{1}{16} \leq f(x)<\frac{2}{16}\right\}$$ and $$E_{4,3}=\left\{ x:\frac{2}{16} \leq f(x)<\frac{3}{16}\right\}$$ and finally $$E_{4,64}=\left\{ x:\frac{63}{16} \leq f(x)<\frac{64}{16}\right\}$$ so clearly $$ \bigcup^{64}_{i=1} E_{4,i}=\{x:f(x)<4\} $$ which makes sense since it seems that the whole point of the sets $E_{n,i}$ is to cover $[0,n)$ in the sense that $$ \bigcup^{n2^{n}}_{i=1} E_{n,i} \subseteq \{x:f(x)<n\} $$ Now here is the part I think I have messed up on: $$\int s_{4} d\mu=126 \, \mu(\{x:f(x)<4\})+4 \, \mu(\{x:f(x)\geq 4\})$$ where $\mu$ is the Lebesgue measure and the 126 comes from $$ \sum^{64}_{i=1} \frac{i-1}{16}=126$$ at least according to MuPad and the additive property of the Lebesgue measure for disjoint sets implies that $$ \sum^{64}_{i=1} \mu(E_{4,i})=\mu\left(\bigcup^{64}_{i=1} E_{4,i} \right) $$ and we can treat the sum $$ \sum^{64}_{i=1} \frac{i-1}{16} $$ as a constant with respect to the integration $$\int s_{4} d\mu$$ which allows us to write $$ \int s_{4} d\mu=\sum^{64}_{i=1} \frac{i-1}{16} \int \sum^{64}_{i=1} \mathbb{1}_{E_{4,i}} \, d\mu+4\int \mathbb{1}_{A_{4}} d\mu $$ and I think that last part in treating the summation as a constant might be wrong.
Your understanding is correct all the way up until the last equation, which should be: $$ \int s_{4} d\mu=\sum^{64}_{i=1} \frac{i-1}{16} \int \mathbb{1}_{E_{4,i}} \, d\mu+4\int \mathbb{1}_{A_{4}} d\mu\tag1 $$ (you should have just one summation.) To arrive at (1) all you're doing is applying the theorem on linearity of integration: $$ \int (af+bg)\,d\mu = a\int f\,d\mu + b\int g\,d\mu $$ and applying it to a large but finite sum of functions. In this case the only functions in sight are the $\mathbb{1}_{E_{4,i}}$ (64 of these) and $\mathbb{1}_{A_4}$; everything else is a scalar coefficient.