Stein 2.9 (Chebyshev's inequality)

(Chebyshev's inequality)

Let $f\in{L^1(\mathbb{R}^d)}$, $f \geq 0$, $\alpha > 0$ and $E_\alpha = \{x | f(x) > \alpha \}$.

Then, $$m(E_\alpha) \leq \frac{1}{\alpha} \int f $$

$\textit{Proof}$ : $$E_\alpha = \{x | f(x) > \alpha \} = \{ x | \frac{f(x)}{\alpha} > 1 \}$$ Integration can be considered a way of measuring a set, so we can write $$m(E_\alpha) = \int \chi_{E_\alpha} = \int_{E_\alpha} 1 $$ By the monotinicity and linearity of the Lebesgue integral, $$\leq \int_{E_\alpha} \frac{f(x)}{\alpha} = \frac{1}{\alpha}\int_{E_\alpha}f \leq \frac{1}{\alpha} \int f \quad \quad \blacksquare $$