There are two different object you might be interested in:
- the truncated normal distribution, which is a Gaussian whose range is restricted to lie within $[a,b]$. This is a density, so it's continuous (no need for epsilon balls).
- a random variable $Z$ that results from drawing a Gaussian random variable $X$ and then "thresholding" it, setting $Z=a$ if $X<a$ and to $Z=b$ if $X>b$, to $Z=X$ if $a\leq X \leq b$.
I'm guessing it's the latter object you're interested in. As you've noted, the probability of getting a sample on the border is non-zero: $P(Z=a) = \Phi((a-\mu)/\sigma)$ and $P(Z=b)=\Phi((\mu-b)/\sigma)$, where $\Phi$ is the standard Gaussian cdf, $\mu$ is the mean, and $\sigma$ the standard deviation.
However, there is zero probability of getting any particular sample within the range $(a,b)$, but the probability density within this range is simply that of the original Gaussian. (The probability of getting a sample from somewhere within this range is simply $1-P(Z=a)-P(Z=b)$.
The resulting object is therefore defined by "point masses" at $a$ and $b$ and a continuous density between $a$ and $b$.
in (-1,1) the probability of a single event is 0
do you mean the probability of getting exactly a given value is infinitesimally small or that it is impossible to get any value > -1 and < 1 (i.e. the probability density function is zero) ? From the answer below I suspect you mean the former... $\endgroup$ – Andre Holzner Jul 5 '11 at 21:32