Markov's Inequality

Theorem
Given a measure space $$\displaystyle(X, \Sigma, \mu)$$ and an $\displaystyle A$-measurable function $$\displaystyle f$$ where $$A \in \Sigma$$, then:


 * $$\mu(\{x\in A : |f(x)| \geq t\}) \leq \frac{1}{t}\int_{A} |f| \mathrm d\mu$$ for any positive $$t \in \R$$.

Proof
Pick any t and define $$B = \{x\in A : |f(x)| \geq t\}$$.

Let $$\displaystyle 1_{B}$$ 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, $$t 1_{B}(x) = t\cdot 1 = t \leq |f(x)|$$.

In the second case, $$t 1_{B}(x) = t\cdot 0 = 0 \leq |f(x)|$$.

Hence $$\forall x \in A$$, $$t1_{B}(x) \leq |f(x)|$$.

By the monotonicity of the Lebesgue integral, $$\int_{A} t1_{B} \mathrm d\mu \leq \int_{A} |f|\mathrm d\mu$$.

But by the linearity of the Lebesgue integral, $$\displaystyle\int_{A} t1_{B}\mathrm d\mu = t\int_{A} 1_{B}\mathrm d\mu = t\mu (B)$$.

Hence, dividing through by $$t$$, we get $$\mu(B)\leq \frac{1}{t}\int_{A} |f| \mathrm d\mu$$.

Markov's Inequality in Probability
Given a probability space $$(\Omega, \Sigma, \Pr)$$, Markov's inequality asserts that for a random variable $$X$$, $$\Pr(|X| \geq t) \leq \frac{\mathrm E(|X|)}{t}$$ for any $$t > 0$$.

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