Markov's Inequality

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f$ be an $A$-measurable function where $A \in \Sigma$.

Then:
 * $\ds \map \mu {\set {x \in A: \size {\map f x} \ge t} } \le \frac 1 t \int_A \size f \rd \mu$

for any positive $t \in \R$.

Proof
Pick any $t$ and define:
 * $B = \set {x \in A: \size {\map f x} \ge t}$

Let $\chi_B$ denote the indicator function of $B$ on $A$.

For any $x \in A$, either $x \in B$ or $x \notin B$.

In the first case:
 * $t \map {\chi_B} x = t \cdot 1 = t \le \size {\map f x}$

In the second case:
 * $t \map {\chi_B} x = t \cdot 0 = 0 \le \size {\map f x}$

Hence:
 * $\forall x \in A: t \map {\chi_B} x \le \size {\map f x}$

By Integral of Integrable Function is Monotone:
 * $\ds \int_A t \chi_B \rd \mu \le \int_A \size f \rd \mu$

But by Integral of Integrable Function is Linear:
 * $\ds \int_A t \chi_B \rd \mu = t \int_A \chi_B \rd \mu = t \map \mu B$

Dividing through by $t$:
 * $\ds \map \mu B \le \frac 1 t \int_A \size f \rd \mu$

Markov's Inequality in Probability
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Markov's inequality asserts that for a random variable $X$:
 * $\map \Pr {\size X \ge t} \le \dfrac {\expect {\size X} } t$

for any $t > 0$.

It can then be used to derive the probabilistic form of Chebyshev's Inequality.