In the proof to proposition 4.2 of 'The Riemann Integral' it is stated that the net of simple functions converges uniformly for continuous functions. This question aims to prove this in a general setting.
Given a compact metric space: $X$
together with the Borel algebra: $\mathcal{B}(X)$
and a Banach space: $E$
Consider a continuous function: $f:X\to E$
and measurable (tagged) partitions: $\forall A\in\pi:\quad A\in\mathcal{B}(X)$
together with simple functions: $f_\pi:=\sum_{A\in\pi} f(a)\mathbb{1}_{A}\quad a\in A$
Then the net of simple functions converges uniformly: $$\lim_{\pi}\|f_\pi-f\|_\infty=0$$ where the order on partitions is given by: $$\pi'\geq\pi:\iff \forall A'\in\pi'\exists A\in\pi: A'\subseteq A$$
Can somebody help me proving this?
By local continuity there is a delta for any point: $$d(x,z)<\delta(z)\implies d(f(x),f(z))<\frac{1}{2}\epsilon$$ By compactness there is a finite cover: $$X=\bigcup_{i=1}^N B_{\delta(z_i)}(z_i)$$ Choose the partition: $$\pi_\epsilon:\quad A_n:=B_{\delta(z_n)}(z_n)\setminus\bigcup_{i=1}^{n-1} B_{\delta(z_i)}(z_i)$$ Then for any subpartition: $$\pi\geq\pi_\epsilon:\quad \|s_\pi(x)-f(x)\|\leq\|s_\pi(x)-f(z_{i_0})\|+\|f(x)-f(z_{i_0})\|<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon$$
Note that uniform continuity was not required!