@robjohn: Let $f$ be a measurable function from $(\Omega, \mathcal{F}, \mu)$ to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. The Markov inequality is

$$\int_\Omega |f| \, d\mu
\geq \epsilon \mu( |f| \geq \epsilon ) $$

for any $\epsilon >0$.

My understanding of the proof without the tail formula of expectation is

$$\int_\Omega |f| \, d\mu
\geq \int_{\{ |f| \geq \epsilon \}} \epsilon \, d \mu
= \epsilon \mu( |f| \geq \epsilon ) $$

I guess your understanding of the proof based on the tail formula of expectation is: