# essential supremum of a measurable function as an integral

This question arises from a paper I read.

Let $$f$$ be a measurable function on $$\mathbb{R}^n$$. The essential supremum of $$f$$ is defined by,

$$\operatorname{ess}\sup _{x \in \mathbb{R}^n} f(x) = \inf \{t : f(x) \leq t \text{ for almost all } x \in \mathbb{R}^n\}$$.

Now, let $$0 < \lambda <1$$ and let $$f,g \in L^1 (\mathbb{R})$$ be non-negative. Let,

$$s(x) = \operatorname{ess}\sup f\left(\dfrac{x-y}{1- \lambda}\right)^{1- \lambda} g\left(\dfrac{y}{\lambda}\right)^ \lambda$$ (over all $$y \in \mathbb{R}^n$$) .

Then, $$s$$ can be written as,

$$s(x) = \sup_{\phi \in D} \int_{\mathbb{R^n}} f\left(\dfrac{x-y}{1- \lambda}\right)^{1- \lambda} g\left(\dfrac{y}{\lambda}\right)^ \lambda \phi(y) dy$$,

Where $$D$$ is a countable dense subset of the unit ball of $$L^1 (\mathbb{R}^n)$$.

Can someone explain how $$s$$ is written in terms of an integral. In the paper, the author has trivialized this but I don't get the idea behind it. I assume this is some sort of a standard representation of the essential supremum of a measurable function on $$\mathbb{R}^n$$.

If you take $$\phi$$ in the unit ball of $$L^1(\mathbb{R}^n)$$ in the supremum, then this equality is a consequence of the isometric isomorphism between the dual of $$L^1$$ and $$L^\infty$$. That you can pass to a dense subset is a standard argument: The supremum of a continuous function on a dense subset is equal to the supremum on the entire space.