If you have some continuous distribution $F(x)$ and you truncated the distribution at some inclusive point $m$ i.e. $F(x | x\ge m)$ does the truncated distribution have a mass point at $m$?
Also does it matter if the point is inclusive or is it the same distribution as $F(x|x>m)$?
For a continuous distribution the truncated density is proportional to the untruncated density: $$ f(x | x > m) = \frac{f(x)}{1 - F(m)} $$ Thus it only has mass points if the untruncated distribution has any which is not the case for continuous distributions. Whether or not the equality condition in the truncation is strict or not also plays no role because it has no probability mass. The truncated distribution is the same.
If you want to obtain a distribution with a point mass at $m$ the usual approach would be to use censoring instead of truncation. Hence, a latent distribution $x^* \sim F$ would be assumed which is only observed if $x^* > m$ and $m$ otherwise, thus creating the point mass: $$ x = \left\{ \begin{array}{ll} x^* & \mbox{if } x^* > m \\ m & \mbox{if } x^* \le m \end{array} \right. $$ In this censored distribution: $P(x = m) = P(x^* \le m) = F(m)$.
