In Markov inequality proof, why is $\int_a^\infty xp(x) \, dx \ge \int_a^\infty ap(x) \, dx$

Question

Markov inequality,

$$\Pr(X \ge a) \le \frac{E[x]}{a}$$

Proof

$$\begin{aligned} E(X) &= \int_0^\infty xp(x)\,dx = \int_0^a xp(x)\,dx + \int_a^\infty xp(x)\,dx \\ &\ge \int_a^\infty xp(x)\,dx \ge \int_a^\infty ap(x) \, dx \end{aligned}$$

Why is the last inequality true?
Why is it ok to change $x$ to $a$, and we know it is less than or equal to?
How did they figure it out?
Thanks!

Answer 1

For every $x\ge a$, $$ xp(x) \ge ap(x) $$ and using positivity of $\int$ and $p(x)$: $$ \int_a^\infty xp(x) \,dx \ge \int_a^\infty ap(x) \,dx $$