I'm trying to do the following problem, and I'm getting hung up on one part. Here is the problem:
Let $f: X \longrightarrow \mathfrak{X}$ be a continuous map from a compact Hausdorff space $X$ into a Banach space $\mathfrak{X}$, with $\mu$ a Radon measure on $X$. Consider elements of the form: $$I_{\lambda}(f) = \sum_{k=1}^n f(s_k)\mu(E_k)$$ where the $E_k$'s are Borel sets partitioning $X$ and $s_k \in E_k \subset \{s \in X \: | \: \lvert f(s)-f(s_k) \rVert \leq \epsilon \}$ for $\epsilon > 0$. With $\lambda = \{E_1,\dots,E_n,\epsilon\}$, prove that $(I_{\lambda}(f))_{\lambda \in \Lambda}$ is a convergent net. We denote the limit by: $$\int_{X} f(s) \: d\mu(s).$$
My main issue I'm having is what the ordering needs to be on the set $\Lambda$. I figured at first that I would want something like $\lambda \leq \mu$ if and only if $\mu$ contains a refinement of the partition in $\lambda$, and $\epsilon_{\mu} \leq \epsilon_{\lambda}$. However, this doesn't seem to work, as the smaller the $\epsilon$ I stipulate, this seems to directly affect how many $E_k$'s I would need in my partition since $E_k \subset \{s \in X \: | \: \lVert f(s)-f(s_k) \rVert \leq \epsilon\}$. Hence if $\epsilon > 0$ was really small, I should expect there to be many more $E_k$'s as $\mu(\{s \in X \: | \: \lVert f(s)-f(s_k) \rVert \leq \epsilon\}) \longrightarrow 0$ as $\epsilon \rightarrow 0^{+}$.
Thus, my point is that I shouldn't be able to just change $\epsilon$ independently of changing the $E_1,\dots,E_n$. But this goes against what my ordering is supposedly okay with.
Is there another ordering I'm supposed to be using?
Your proposed ordering works fine. To be precise, we should say that $\Lambda$ consists of triples $(\{E_1,\dots,E_n\},\epsilon,\{s_1,\dots,s_n\})$ with each $s_k\in E_k$, and $(\{E_1,\dots,E_n\},\epsilon,\{s_1,\dots,s_n\})\leq (\{F_1,\dots,F_m\},\eta,\{t_1,\dots,t_m\})$ if $\eta\leq \epsilon$ and each $F_l$ is contained in $E_k$ for some $k$.
Then given $\epsilon>0$, choose a partition $\lambda_0=(\{E_1,\dots,E_n\},\epsilon_0,\{s_1,\dots,s_n\})$ with $\epsilon_0<\frac{\epsilon}{\mu(X)}$.
(To see that such a partition exists, note that at each $x\in X$, we have by continuity a neighborhood $U$ where $\|f(y)-f(x)\|<\epsilon/2$ for $y\in U$. By compactness there is a finite cover $U_1,\dots,U_n$ of $X$ of such neighborhoods, and so we may take $E_1=U_1$, $E_k=U_k\backslash\bigcup_{j<k}E_j$ as our partition, and may choose any points in $E_k$ as our $s_k$.)
Then for any partition $\lambda=(\{F_1,\dots,F_m\},\eta,\{t_1,\dots,t_m\})$ with $\lambda\geq \lambda_0$, we have each $F_l$ contained in some $E_{k_l}$, so we can estimate
\begin{align*} \|I_\lambda(f)-I_{\lambda_0}(f)\| &=\left\|\sum_{k=1}^nf(s_k)\mu(E_k) -\sum_{l=1}^m f(t_l)\mu(F_l) \right\| \\ &= \left\|\sum_{l=1}^mf(s_{k_l})\mu(F_l) -\sum_{l=1}^m f(t_l)\mu(F_l) \right\|\\ &= \left\|\sum_{l=1}^m(f(s_{k_l})-f(t_l))\mu(F_l) \right\|\\ &\leq \sum_{l=1}^m\left\|(f(s_{k_l})-f(t_l)) \right\|\mu(F_l)\\ &\leq \epsilon_0\sum_{l=1}^m\mu(F_l) = \epsilon_0\mu(X)<\epsilon\text{.} \end{align*}
It quickly follows that the net is a Cauchy net, hence by completeness of $\mathfrak X$ the net converges.