The function $f: [0,1]\rightarrow [0,1]$ and constant $k \in \mathbb{R}$ are given. We have the following maximization problem:
$$ \max _{g: [0,1]\rightarrow \mathbb{R}}\int_0^{1}f(x) g(x) d x\\ \text{s.t. } g \in [0,1], g \text{ non-decreasing},\\ \int^1_0g(x)dx\leq k. $$
Now the set $\{g \in [0,1], g \text{ non-decreasing}\}$ is compact and convex. We know from existing results that its extreme points are indicator functions of the form $\mathbf{1}(x \geq x^*)$ for some $x^* \in [0,1]$.
Therefore we can write, any feasible $g(x)$ as:
$$g(x)=\int_0^{1}\mathbf{1}(x\geq x^*)d\mu(x^*)$$,
where $\mu$ is some distribution over $[0,1]$, i.e. over the thresholds of the indicator functions in the support of $g$.
My question is, is it notationally correct to just replace the above expression of $g(x)$ in the original problem? I'm jittery because of infinite convex combinations being involved.
In other words, is it notationally correct to write the original problem as follows?
Let us denote the set of distributions over $[0,1]$ by $\Delta[0,1]$.
$$ \max _{\mu \in \Delta [0,1]}\int_0^{1}f(x) \left(\int_0^{1}\mathbf{1}(x\geq x^*)d\mu(x^*)\right) d x\\ \text{s.t. } g \in [0,1], g \text{ non-decreasing},\\ \int^1_0g(x)\left(\int_0^{1}\mathbf{1}(x\geq x^*)d\mu(x^*)\right)dx\leq k. $$
This reflects the fact that when we are optimizing over feasible $g$'s, we are equivalently optimizing over the set of distributions over $[0,1]$ (distributions over the threshold $x^*$), such that the linear constraint $\int^1_0g(x)dx\leq k$ is satisfied.