Proof of Markov's and Chebyshev's Inequalities

Markov's inequality. By The Lebesgue Integral and Null Sets (vi) with $\mu=P$, $f=X$ and $t=a$, the set $\{X\ge a\}$ is an event and

$$a\,P(X\ge a)\le\int_{\Omega}X\,dP=\mathbb{E}X$$

(Expectation (i); the product on the left is the real product, since $P(X\ge a)\le P(\Omega)=1$ is a real number in $[0,1]$ by Basic Properties of a Measure (iii) and Addition and Multiplication on [0,∞] (ii)).

Chebyshev's inequality. Since $X\in\mathcal{L}^{1}(P)$ (Lᵖ Inclusions and Constants on a Finite Measure Space (i), Elementary Properties of the p-Seminorm (i)), $\mathbb{E}X$ is a real number, and $Y=(X-\mathbb{E}X)^{2}$ is a nonnegative random variable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)), hence an element of $\mathcal{M}^{+}(\Omega,\mathcal{F})$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)). For real $t$, $|t|\ge a$ holds exactly when $t^{2}\ge a^{2}$, as $a>0$; hence

$$\{|X-\mathbb{E}X|\ge a\}=\{Y\ge a^{2}\}.$$

As $X-\mathbb{E}X\in\mathcal{L}^{2}(P)$ (Lᵖ Inclusions and Constants on a Finite Measure Space (ii), Minkowski's Inequality (i)), $Y$ is integrable (Hölder's Inequality (ii)), so its expectation as an element of $\mathcal{M}^{+}(\Omega,\mathcal{F})$ is the real number $\mathbb{E}\bigl[(X-\mathbb{E}X)^{2}\bigr]=\operatorname{Var}X$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi), Variance and Covariance (i), Variance and Covariance (ii)). By (i) with the level $a^{2}>0$,

$$a^{2}\,P\bigl(|X-\mathbb{E}X|\ge a\bigr)=a^{2}\,P(Y\ge a^{2})\le\mathbb{E}Y=\operatorname{Var}X,$$

and dividing by $a^{2}>0$ gives $P(|X-\mathbb{E}X|\ge a)\le\operatorname{Var}X/a^{2}$.