Chebysev inequality proof

Question

Let $(X,\Omega,\mu)$ be a measure space and let $f$ be an extended real valued measurable function defined on $X$. Proof that

$$ \mu\left(\{x\in X : |f(x)|\geq t\}\right)\leq \frac{1}{t}\int_X |f|~d\mu $$ for any t>0.

Then, conlude that the measure of the set $$\{x\in X : |f(x)|\geq t\}$$

is finite for every t

Answer 1

Define $A$ as $A=\{x\in X : |f(x)|\geq t\}$. Then we have

$$ \mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{A} ~d\mu =\int_{X} \chi_A ~d\mu $$ where $\chi_A(x)=\mathbf 1(x\in A)$. Then we have: $$ \mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{X} \chi_A ~d\mu \\ = \int_{X} \frac{t}{t}\chi_A ~d\mu \leq \int_{X} \frac{|f|}{t}\chi_A ~d\mu \\ \leq \int_{X} \frac{|f|}{t}~d\mu \\ $$

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Another inequality is as follows: $$ \mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{X} \chi_A ~d\mu \\ = \int_{X} \frac{t^2}{t^2}\chi_A ~d\mu \leq \int_{X} \frac{f^2}{t^2}\chi_A ~d\mu \\ \leq \int_{X} \frac{f^2}{t^2}~d\mu \\ $$