Editing Chebyshev's inequality (section)

===Measure-theoretic statement===
Let <math>(X,\,\Sigma,\,\mu)</math> be a [[measure space]], and let ''f'' be an [[extended real number line|extended real]]-valued [[measurable function]] defined on ''X''. Then for any real number <math>t > 0</math> and <math>0 < p < \infty</math>,

:<math>\mu(\{x\in X\,:\,\,|f(x)|\geq t\}) \leq {1\over t^p} \int_{X} |f|^p \, d\mu.</math>

More generally, if ''g'' is an extended real-valued measurable function, nonnegative and nondecreasing, with <math>g(t) \neq 0</math> then: {{citation needed|date=May 2012}}

:<math>\mu(\{x\in X\,:\,\,f(x)\geq t\}) \leq {1\over g(t)} \int_X g\circ f\, d\mu.</math>

This statement follows from the [[Markov inequality]], <math>
\mu(\{x\in X:|F(x)|\geq \varepsilon\}) \leq\frac1\varepsilon \int_X|F|d\mu
</math>, with <math>F=g\circ f</math> and <math>\varepsilon=g(t)</math>, since in this case <math>\mu(\{x\in X\,:\,\,g\circ f(x)\geq g(t)\}) \geq \mu(\{x\in X\,:\,\,f(x)\geq t\}) </math>.
The previous statement then follows by defining <math>g(x)</math> as <math>|x|^p</math> if <math>x\ge t</math> and <math>0</math> otherwise.