I am trying to solve the following exercise:
Let $X$ be an absolutely continuous random variable with density $f$. Let $g$ be a non-negative non-decreasing function with $\mathbb{E}(g(X))=1$. Show that $$\mathbb{P}(X>u)\leq \mathbb{E}(g(X)\mathbf{1}_{X>u}),\ \ \forall u\in\mathbb{R}.$$
I have the following attempt: since $\mathbb{E}g(X)=1$, we have that $$1=\int_{-\infty}^{u}g(x)f(x)dx+\mathbb{E}(g(X)\mathbf{1}_{X>u})\leq g(u)\int_{-\infty}^{u}f(x)dx+\mathbb{E}(g(X)\mathbf{1}_{X>u})=g(u)\mathbb{P}(X\leq u)+\mathbb{E}(g(X)\mathbf{1}_{X>u}).$$ Write $\mathbb{P}(X\leq u)=1-\mathbb{P}(X>u)$, we have that $$g(u)\mathbb{P}(X>u)\leq g(u)-1+\mathbb{E}(g(X)\mathbf{1}_{X>u}),$$ and thus $$\mathbb{P}(X>u)\leq 1-\dfrac{1}{g(u)}+\dfrac{\mathbb{E}(g(X)\mathbf{1}_{X>u})}{g(u)}\leq 1+\dfrac{\mathbb{E}(g(X)\mathbf{1}_{X>u})}{g(u)}.$$ I do not know how to push it further.
Another attempt is to use Markov inequality. Since $X$ is absolutely continuous and $g$ is non-decreasing and nonnegative, we have $$\mathbb{P}(X>u)=\mathbb{P}(X\geq u)=\mathbb{P}(g(X)\geq g(u))\leq\dfrac{\mathbb{E}g(X)}{g(u)}=\dfrac{1}{g(u)}.$$
What have I missed?
If $g(u) \geq 1$, we have that, since $g(u) \leq g(X)$ on $\{X > u\}$, $$ \Pr(X > u) = \mathbf{E}(\mathbf{1}_{\{X > u\}}) \leq \mathbf{E}(g(u)\mathbf{1}_{\{X > u\}}) \leq \mathbf{E}(g(X)\mathbf{1}_{\{X > u\}}). $$
On the other hand, if $g(u) < 1$, we have analogously that $$ \Pr(X \leq u) = \mathbf{E}(\mathbf{1}_{\{X \leq u\}}) \geq \mathbf{E}(g(u)\mathbf{1}_{\{X \leq u\}}) \geq \mathbf{E}(g(X)\mathbf{1}_{\{X \leq u\}}), $$ so that, since $\mathbf{E} g(X) = 1$, $$ \Pr(X > u) = 1 - \Pr(X \leq u) \leq \mathbf{E}g(X) - \mathbf{E}(g(X)\mathbf{1}_{\{X \leq u\}}) = \mathbf{E}(g(X)\mathbf{1}_{\{X > u\}}). $$