Proof of Chebyshev Inequality I was going through the proof of the Chebyshev Inequality here .
And I seem to be facing some trouble in the approximation stage. I can't seem to follow how $\epsilon$ has been approximated to $(t-\mu)$.
 A: It is an inequality. The text in that document breaks up the flow slightly.
It should read like this:
$\int_{-\infty}^{\mu-\epsilon}(t-\mu)^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}(t-\mu)^2f_X(t)dt \ge \int_{-\infty}^{\mu-\epsilon}\epsilon^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}\epsilon^2f_X(t)dt$.
They did not replace $(t-\mu)^2$ with $\epsilon^2$. Rather, they exploited the inequality $(t-\mu)^2 \ge \epsilon^2$ to achieve the inequality I typed above.
A: Over the two semi infinite intervals of integration we have 1) in the first region t<μ-ϵ and 2)in the second region t>μ+ϵ.  Both regions were cleverly chosen so the ϵ$^2$<(t-μ)$^2$.  So the inequality is maintained with ϵ$^2$ replacing (t-μ)$^2$ and the rest should be easy for you.
A: Notice that for $t\in (-\infty,\mu-\epsilon]\cup[\mu+\epsilon,\infty)$ that $(t-\mu)^2 \ge \epsilon^2$. 
A: $\epsilon$ is not being approximated by $t - \mu$. What is happening is that for $t$ outside the interval $(\mu - \epsilon, \mu + \epsilon)$, $\epsilon^2 \leq (t - \mu)^2$ and hence the two integrals can be underestimated. 
The reasoning "since $t \leq \mu - \epsilon \Rightarrow \epsilon \leq | t - \mu | \Rightarrow \epsilon^2 \leq (t - \mu)^2$" on the page you refer to, applies to the rewriting of the left integral, c.q., $t$ on the left of the interval $(\mu - \epsilon, \mu + \epsilon)$. A similar reasoning applies to the right integral, c.q., the right of that interval.
