Proof of Markov's inequality

Proof

We compare $X$ with a multiple of an indicator function.

1. Let $X$ be a random variable with values in $[0, \infty)$ on a probability space $(\Omega, \mathcal{F}, P)$, and let $a > 0$.
2. By the definition of the expected value, $E[X] \in [0, \infty]$. If $E[X] = \infty$, the right side is $\infty$ and the inequality holds. So assume $E[X] < \infty$.
3. Let $A = \set{X \ge a}$, an event, and let $\mathbf{1}_A$ be its indicator function.
4. We claim $a \mathbf{1}_A \le X$ pointwise. On $A$, the left side is $a$ and $X \ge a$. Off $A$, the left side is $0$ and $X \ge 0$.
5. The integral of an indicator is the probability of its set, so $E[a \mathbf{1}_A] = a P(A)$.
6. By step 4, $X - a \mathbf{1}_A \ge 0$. The integral of a nonnegative function is nonnegative, so $E[X - a \mathbf{1}_A] \ge 0$.
7. Both $X$ and $a \mathbf{1}_A$ are integrable. By linearity of expectation and step 5, $E[X] - a P(A) \ge 0$.
8. Dividing by $a > 0$ gives $P(X \ge a) \le E[X]/a$.