P(X=a) verses $f_X(a)$ for continuous r.v. Textbook says that for continuous r.v. the probability at a specific value is zero:
$$P(X=a) = 0$$
They say that the proof of this fact is:
$$P\Big(a\Big) \le P\Big((a-\epsilon) \le x \lt a\Big)$$
(because "a" is a subset of the range $(a-\epsilon)$ to "a").  
Then if we allow epsilon to approaches zero:
$$\lim \limits_{\epsilon \to 0} P\Big((a-\epsilon) < x \lt a\Big) = \lim \limits_{\epsilon \to 0} \Big(F_X(a) - F_X(a-\epsilon)\Big)$$
therefore:
$$P(X= a) = 0$$
(note: $F_X(x) = P(X \le a)$ is the cdf of pdf $f_X(x)$)
However, a paragraph later in the textbook, it says that the pdf of continuous r.v. has the property:
$$f_{X}(x) \ge 0$$
My question is this: What's the difference between $P(X=a)$ and $f_X(a)$?  shouldn't it be : $f_{X}(a) = 0$ 
without the greater than sign?
 A: A continuous RV has no probability mass function (PMF) $p$ since a requirement for a PMF is that the sum of the PMF values is $1$. Attempting to define a PMF $p$ for a continuous RV $X$ leads to $p(a)=\int_a ^a f(x)\text{d}x=0.$ Therefore, we cannot obtain $\sum_{-\infty < a<\infty}p(a)=1$ for a continuous $X$.
Also, we cannot have $f(x)=0$ for all $x$ since (letting $S$ denote the sample space) we need $$\mathbb{P}(X\in S)=\int_{\mathbb R}f(x)\text{d}x=1.$$
The analogue of the PMF for a continuous RV $X$ is a function (PDF) $f$.
Continuous case: 


*

*$f\ge 0$,

*$\mathbb P(X \in A)=\int_A f(x)\text{d}x$,

*$\int_{-\infty}^\infty f(x)\text{d}x=1$
Discrete case: 


*

*$p\ge 0$,

*$\mathbb{P}(X\in A)=\sum_{x\in A}p(a),$

*$\sum_{-\infty<a<\infty}p(a)=1.$
A: For discrete r.v:
$$P(X=a) = f_X(a)$$
However, for continuous r.v.:
$$P(X=a) \ne f_X(a)$$
It equals:
$$P(X=a) = \lim \limits_{\epsilon \to 0} \int \limits^{a+\epsilon}_{a}f_x(v)~dv = 0$$
because:
$$P(a<X<b) = \int \limits^{b}_{a}f_x(v)~ dv = 0$$
