Is it true that any continuous probability distribution gives probability zero to any single element in the sample space? Let $p$ be some probability distribution with a density $f$.
$p$ is defined over $\Omega$.
Is it true that for any $\omega \in \Omega$, $p(\omega) = 0$?
If not, what are the "minimal conditions" under which $p(\omega) = 0$ for any $\omega$ (the space $\Omega$ is actually a subset of $\mathbb{R}^d$ for some $d$)?
 A: The OP asks for a precise mathematical answer to an imprecise question. For example, the term "probability distribution" has a very precise meaning in terms of the measure transfered by some random object from some probability space to its image, but here you have only one probability space, and no random object. I will try to give an answer in his original setting.
Suppose that you have a measurable space $(\Omega,\mathscr{F}$), where $\Omega=\mathbb{R}$ and $\mathscr{F}$ is the class of Borel subsets of $\mathbb{R}$. The extension to higher dimensions adds nothing conceptually.
You have a probability measure $P$ over $(\Omega,\mathscr{F}$), and you say that $P$ has a density $f$. What does that mean exactly? It means that you have a measure (not necessarily a probability measure) $\lambda$ over $(\Omega,\mathscr{F}$) and
$$
  P(B) = \int_B f(\omega)\,d\lambda(\omega) \, ,
$$
for every $B\in\mathscr{F}$, where the nonnegative $f$ is such that $\int_\mathbb{R}f\,d\lambda=1$.
The measure $\lambda$ dominates $P$, in the sense that for every $B\in\mathscr{F}$, if $\lambda(B)=0$ then $P(B)=0$.
Note that for each $\omega\in\Omega$, the singleton $\{\omega\}\in\mathscr{F}$, so it is legal to talk about $P(\{\omega\})$, and that is the substance of the point raised by the comment made by whuber.
Now, to answer your question "Is it true that $P(\{\omega\})=0$ for every $\omega\in\Omega$?", consider two cases.


*

*Let $\lambda$ be Lebesgue measure. In this case, the answer is clearly "Yes", because $\lambda(\{\omega\})=0$ for every $\omega\in\Omega$, and $\lambda$ dominates $P$.

*Let $\lambda$ be such that $\lambda(\mathbb{R})=2$, and $\lambda(\{0\})=\lambda(\{1\})=1$. Let a function $f$ be defined by $f(0)=1/4$, $f(1)=3/4$, and define $f(\omega)$ arbitrarily for $\omega\notin \{0,1\}$. Now, define a probability measure $P$ over $(\Omega,\mathscr{F})$ by
$$
  P(B) = \int_B f(\omega)\,d\lambda(\omega) \, .
$$
This $P$ is a genuine probability measure, it has density $f$ with respect to $\lambda$, but
$$
  P(\{0\}) = \int_{\{0\}} f(\omega)\,d\lambda(\omega) = f(0) \cdot \lambda(\{0\}) = \frac{1}{4} \, ,
$$
and
$$
  P(\{1\}) = \int_{\{1\}} f(\omega)\,d\lambda(\omega) = f(1) \cdot \lambda(\{1\}) = \frac{3}{4} \, ,
$$
Hence, in this case the answer is "No".
Short answer: it depends on the dominating measure.
A: I'd like to share the answer to this question with you in very simple words. Because I also wondered in my early days on why R gives a value when I input a single value in any density function of a continuous variable, say for example in dnorm. As I know for a particular value the probability should be zero. 
Here an important thing to be noticed is that if you manually put the value of the random variable, say, $X$, in the form of the probability density function (p.d.f.) of a continuous random variable, then you do not get the probability. What you get is called simply the 'ordinate'. If you consider a X-Y plane, then this value is the $Y$ coordinate value against your value for $X$ coordinate. 
Following is a practical example of why a particular point gives probability zero in a continuous probability distribution: 
While for a discrete distribution an event with probability zero is impossible (e.g. rolling 3½ on a standard die is impossible, and has probability zero), this is not so in the case of a continuous random variable. For example, if one measures the width of an oak leaf, the result of 3½ cm is possible, however it has probability zero because there are uncountably many other potential values even between 3 cm and 4 cm. Each of these individual outcomes has probability zero, yet the probability that the outcome will fall into the interval (3 cm, 4 cm) is nonzero. This apparent paradox is resolved by the fact that the probability that $X$ attains some value within an infinite set, such as an interval, cannot be found by naively adding the probabilities for individual values. Formally, each value has an infinitesimally small probability, which statistically is equivalent to zero.
