Unimodality and continuity for probability distribution From Wikipedia about the conditions for the Vysochanskij–Petunin inequality

The sole restriction on the distribution is that it be unimodal and
  have finite variance. (This implies that it is a continuous
  probability distribution except at the mode, which may have a non-zero
  probability.)

Does it mean that unimodality and finite variance imply the distribution is continuous except at the mode? Why is that? 
Thanks!

Your interpretation is correct.  
If you regard some distributions on the integers as unimodal in the natural sense (such as the binomial and Poisson distributions) then the distribution with $$\Pr(X=0)=\frac{1}{2k^2}$$ $$\Pr(X=1)=1-\frac{1}{k^2}$$ $$\Pr(X=2)=\frac{1}{2k^2}$$ has a mean of $1$, a variance of $\frac{1}{k^2}$, and is unimodal when $k \gt \sqrt{\frac32}$. But it is bound by the Chebyshev inequality not the Vysochanskij–Petunin inequality.  
