Chebyshev Inequality

Suppose $f$ is an integrable function defined on a measure space (for example, a region in ${\mathbb{R}}^n$), and $p>0$. For any constant $\lambda >0$, Chebyshev’s inequality (also known as Markov’s inequality in this general context) states that the measure (area/volume) of the set of points where the absolute value of the function is greater than or equal to $\lambda$ (i.e., the level set) satisfies: $|\{x:|f(x)|\ge \lambda \}|\le \frac{1}{{\lambda }^p}\int |f(x)|{}^{p}dx$ Here, $|E|$ denotes the Lebesgue measure of the set $E$.

Proof

On the set $S=\{x:|f(x)|\ge \lambda \}$, it is clear that $|f(x)|{}^p\ge {\lambda }^p$. Therefore: $\int |f(x)|{}^{p}dx\ge {\int }_S|f(x)|{}^{p}dx\ge {\int }_S{\lambda }^{p}dx={\lambda }^p|S|$ Dividing both sides by ${\lambda }^p$ yields the inequality