Variance $= 0$, show that $X=\mu$ with probability one If the variance of $X$ is zero, show that $X=\mu$ with probability one.
Using Chebychev's inequality that is, 
\begin{equation*}
P(|X-\mu|\geq k\sigma)\leq\frac{1}{k^2},
\end{equation*}
I just let $\sigma=0$,thus, $P(|X-\mu|\geq 0)$, as our absolute value is always greater than or equal to one, this probability equals one.. Does this look correct?
 A: No, it does not look correct, because $P(|X - \mu| \ge 0)$ is not the same as $P(X = \mu)$, and because the Chebyshev inequality does not apply when $\sigma = 0$.  For this problem, it would make sense to show that $P(X\neq \mu) = 0$. Now the event $(X\neq \mu)$ is the increasing union of the events $(|X - \mu| \ge \frac{1}{n})$, for $n \in \Bbb N$. Thus $P(X\neq \mu) = \lim_{n\to \infty} P(|X - \mu| \ge \frac{1}{n})$. By Markov's inequality, 
$$P\left(|X - \mu| \ge \frac{1}{n}\right)  = P\left(|X - \mu|^2 \ge \frac{1}{n^2}\right) \le n^2 \operatorname{Var}(X) = 0$$ 
for all $n \in \Bbb N$. Therefore $\lim_{n\to \infty} P(|X - \mu| \ge \frac{1}{n}) = 0$ and consequently $P(X\neq \mu) = 0$.
A: A more intuitive argument is just noting that for any non-negative random variable $X$ we have $\mathbb{E}(X) = \mathbb{E}(X \, ; X > 0) = 0$ if and only if $\mathbb{P}(X > 0) = 0$. (The expectation of $X$ over the set where it's positive can only be zero if the set itself is negligible.)  Since $\mathbb{V}(X) = \mathbb{E}[(X - \mu)^2]$, then $\mathbb{V}(X) = 0$ implies $\mathbb{P}((X - \mu)^2>0) = 0$ or equivalently $X = \mu$ almost surely.
A: The proof of Chebyshev's inequality uses the fact that $\sigma>0$. 
See: https://en.wikipedia.org/wiki/Chebyshev%27s_inequality.
And see the proof af the Markov inequality which is the basis of the proof above.
See: https://en.wikipedia.org/wiki/Markov%27s_inequality.
