$P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu$

Hm, how to proceed? We can only attempt to replace the $1$ with the $X$ to get
some non-trivial bound on $X$. But we know that $X \geq A$. so we should perhaps
first introduce the $A$:
$P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = 1/A \int_{\{X \geq A\}} A d \mu$

Now we are naturally led to see that this is always less than $X$:
$\begin{aligned}
&P(X \geq A) \equiv = \int_{\{X \geq A\}} 1 d \mu = \\
& 1/A \int{\{X \geq A\}} A d \mu < 1/A \int_{\{X \geq A\}} X d \mu = 1/A \mathbb{E}[X]
\end{aligned}$

This completes marov's inequality:
$P(X \geq A) \leq \mathbb{E}[X]/A$

So we are "smearing" the indicator $1$ over the domain $\{X \geq A\}$ and attempting
to get a bound.