Markov Inequality proof (measure theory)

Question

I am trying to prove Markov's Inequality in measure theory as:

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be a non-negative function which satisfies $g(x)>0$ se $x>0$, and not descendant in $[0,\infty)$. Let also be a measurable function in measure space $(X,\mathcal{X},\mu)$. Then, for all $a>0$:

$\mu\{x \in X:|f(x)|\ge a\}\le\frac{\int{g\circ f d\mu }}{g(a)}$.

I am familiar with the usual proof that involves letting $g$ be

$g(f(x)) = \begin{cases}a,~~~~~~~~~~ \text{in case}~f(x) \ge a \\ 0,~~~~~~~~~~ \text{in case}~f(x) < a \end{cases}$.

Then, I guess it would go like:

$\int_{X}{f}d\mu \ge \int_{X}{g(f(x))d\mu}=\mu\{x \in X:|f(x)|\ge a\}$

But apparantly, this is not the case. I believe it is quite not accurate as $f$ in this case is not strictly positive.

I have spent hours and I cannot find the answer. Can someone please help me with this step I am missing in the middle?

Thank you very much.

Answer 1

We have $ \mu( x \in X : |f(x)| \geq a) = \int_{ |f(x)| \geq a} d\mu(x) \leq \int_{ |f(x)| \geq a} \dfrac{g(x)}{g(a)} d\mu(x) \leq \dfrac{1}{g(a)}\int g \circ x d\mu(x) $, where we used $g(x)/g(a) \geq 1$ by condition on $g$

Answer 2

I think it should be $g \circ |f|$ rather than $g \circ f$.

In that case, as $g$ is non decreasing, $|f| \geq a \Leftrightarrow g(|f|) \geq g(a)$. now you can apply Markov's inequality.