Proof of Markov's and Chebyshev's Inequalities

Markov. The set $\{X\ge a\}$ is an event, since $\{X\ge a\}=\bigcap_{j\in\mathbb{N}}\{X>a-1/j\}$ and each $\{X>a-1/j\}\in\mathcal{F}$ by Measurable Function and Real-Valued Measurable Function. Pointwise,

$$a\,\mathbf{1}_{\{X\ge a\}}(\omega)\ \le\ X(\omega),$$

since the left side is $a\le X(\omega)$ on the event and $0\le X(\omega)$ off it. The left side is a nonnegative simple function with integral $a\,P(X\ge a)$; by monotonicity of the nonnegative integral (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) and Expectation, Variance, and Moments,

$$a\,P(X\ge a)\ \le\ \mathbb{E}[X],$$

and dividing by $a>0$ gives Markov's inequality (trivially valid when $\mathbb{E}[X]=\infty$).

Chebyshev. Let $\mu=\mathbb{E}[X]$ and $Y=(X-\mu)^{2}$, a nonnegative random variable (a random variable by Step 0(a) of the proof of Linearity and Monotonicity of the Lebesgue Integral and the power argument of Expectation, Variance, and Moments) with $\mathbb{E}[Y]=\operatorname{Var}(X)$, finite by hypothesis. Since $|X-\mu|\ge a$ holds exactly when $Y\ge a^{2}$, Markov's inequality applied to $Y$ with threshold $a^{2}$ gives

$$P\bigl(|X-\mu|\ge a\bigr)=P\bigl(Y\ge a^{2}\bigr)\ \le\ \frac{\mathbb{E}[Y]}{a^{2}}=\frac{\operatorname{Var}(X)}{a^{2}}.\qquad\blacksquare$$