Chebychev inequality generalization, essential supremum

Question

$\DeclareMathOperator{\esssup}{ess\,sup}$ I don't understand fully why would this be true. $P$ is probabilty here, and $\mathcal{E}X$ means expected value of $X$. Let $g: \mathbb{R} \to \mathbb{R}_+$ be a nondecreasing Borel function. Then we have: $$ \frac{\mathcal{E}g(X)-g(a)}{\esssup g(X)} \leq P(X\geq a) \leq \frac{\mathcal{E}g(x)}{g(a)}$$ I'm looking for a proof of the left inequality. I tried searching for it, but no results. $\esssup$ here means $\inf \limits_{E: P(E)=0}[ \sup \limits_{\omega \in \Omega \setminus E} |X(\omega)| ] $

Answer 1

Note that$\DeclareMathOperator{\esssup}{ess\,sup}$

$$\mathbb{E}g(X)-g(a) = \mathbb{E}(g(X)-g(a)) = \int_{\Omega} \big( g(X(\omega))-g(a) \big) \, d\mathbb{P}(\omega).$$

Since $g$ is non-decreasing, we have

$$g(X(\omega))-g(a) \leq 0 \qquad \text{for all} \, \, \omega \in \{X<a\}$$

and so $$\mathbb{E}g(X)-g(a) \leq \int_{\{X \geq a\}}\big(g(X(\omega))-g(a)\big) \, d\mathbb{P}(\omega) \stackrel{g \geq 0}{\leq} \int_{\{X \geq a\}} g(X(\omega)) \, d\mathbb{P}(\omega).$$

By the very definition of the essential supremum, this gives

$$\mathbb{E}g(X)-g(a) \leq \mathbb{P}(X \geq a) \esssup g(X)$$

implying the desired inequality.