Point estimator When the point estimator under consideration has a pdf, then 
$P[T=\tau(\theta)]=0 $ 
where $\tau(.)$ is some function of parameter $\theta$
$T$ is an estimator of $\tau(\theta)$
But I did many exercises to find point estimators of the parameters of density functions.
For example the two point estimators of the mean, $\mu$ and the variance $\sigma^2$ of a normal density are $\bar X$ and $\frac{n}{n-1}S^2$  respectively.
But for the following statement, are those probabilities necessarily zero? 
"When the point estimator under consideration has a pdf, then 
$P[T=\tau(\theta)]=0$"
 A: Although the probability that $\bar{X} = \mu$ is equal to $0$, this doesn't make using $\bar{X}$ necessarily illogical.
Consider $E(\bar{X})$.
$$
E(\bar{X}) = E \Big( \frac{1}{N} \sum_{i=1}^N x_i \Big) =  \frac{1}{N} \sum_{i=1}^N E(x_i) = \frac{1}{N} \sum_{i=1}^N \mu = \mu
$$
So $E(\bar{X}) = \mu$. It's unbiased.
What about the variance of $\bar{X}$? Let's call that $V(\bar{X})$
$$
V(\bar{X}) = V \Big( \frac{1}{N} \sum_{i=1}^N x_i \Big) = \frac{1}{N^2}  \sum_{i=1}^N V(x_i) + 0 = \frac{1}{N^2}  \sum_{i=1}^N \sigma^2  = \frac{\sigma^2}{N}
$$
(I shortened that derivation quite a bit. In the third step, the $+0$ term represents a large number of covariance terms between $x_i$ and $x_j$ where $j \neq i$, which are equal to zero under iid sampling.)
So, the expected value of $\bar{X}$ is equal to $\mu$, while its variance is inversely proportional to $N$. You can imaging the distribution of $\bar{X}$ as a gaussian, centered around $\mu$. As you add more and more data ($N$ gets bigger) the variance of this gaussian shrinks, and it gets more and more probable that $\bar{X}$ is "close" to $\mu$. That's the motivation - it may never be equal to it, but it hopefully should not be too wrong. 
In fact, it can be proven that $\bar{X}$ is the minimum variance estimator for $\mu$. There's no other statistic to estimate $\mu$ you can find using your dataset with a smaller variance - that is as tightly constrained to be "near" $\mu$ as $\bar{X}$. Google "Cramer-rao lower bound" to see why.
