Suppose $\Omega=[0,1]$ and $f$ is continuous, monotonic, and of bounded variation on $\Omega$ with $M:= \text{Var}_\Omega(f).$
Let $\| \cdot\|:=\| \cdot\|_{L_\infty(\Omega)}.$ Let $T=\{0=t_0,...,t_n=1\}$ be the uniform partition of the range of $f,a_k$ is the average of the function values at the endpoints on $[f^{-1}(t_{k-1}),f^{-1}(t_k)],$ and $S_n(x):=a_k,x\in[f^{-1}(t_{k-1}),f^{-1}(t_k)),k=1,...,n.$ Then there is a theorem says that$$\|f-S_n\|\le M/2n.$$
The above theorem uses the uniform partition of the range. Now if we use a random partition of the range of $f$ and assume the partition points $t_i \sim U(0,1), $ where $i=1,..,n-1.$ Can we get the result $E(\| f-S_n\|)\le C/n$ for some constant $C$? How to prove or disprove it? I have trouble expressing the expectation explicitly.
I showed that the expectation of the length of each subinterval is $1/n.$ But I am not sure if the result is useful.