Reason why $P(X = k) = 0$ but $P( k_1 \leq X \leq k_2) \neq 0$ for continous random variable $X$ This might be too easy (or almost trivial) for many out here, but i am having troubles with this. 
I know for any continous random variable $X$, $P(X = k) = 0$, however the probability that $X$ falls in $[k_1,k_2]$ is not necessarily zero, where $k_1 < k_2$.
I've been using this notion since high school and collage, but I never seem to grasp the idea. Am i being too superficial or is it okay to pause and wonder a bit here? 
Thanks for your time.
 A: There is a fundamental difference in the way probabilities are defined
for discrete and continuous random variables.
Discrete. The support of a discrete random variable is a finite or
countable set of points. By a table, a formula, or a graph the probability
for each point is specified. For example, consider the number $X$ of Heads
in $n = 20$ tosses of a fair coin, so that $X \sim Binom(20, 1/2).$
The support is the integers from 0 through 20, and the formula
$$P(X = k) = {n \choose k}(1/2)^k,$$
for $k = 0, 1, \dots, 20$ specifies the probability for each point in the
support. Thus $P(X = 15) = 0.01478577,$ as computed in R statistical software. 
dbinom(15, 20, .5)
## 0.01478577

If you want the probability of an "interval," then you need to sum the probabilities of all of the points that lie in the interval. For example,
$$P(13 \le X \le 16) = P(X=13)+P(X=14)+P(X=15)+P(X=16) = 0.1302996,$$
computed as:
sum(dbinom(13:16, 20, .5))
## 0.1302996

For discrete distributions, intervals have probabilities only in terms of the
probabilities of the individual points they contain.
Continuous. By contrast, the distribution of a continuous random variable
is defined in terms of a density function. One might say that heights (in inches) of the
men in a particular population are distributed as the random variable
$Y \sim Norm(\mu = 69, \sigma = 3.5),$ and we might represent its density
function as $f(y).$ 
Probabilities of intervals are defined in terms of the area under the density function within the particular interval in question. For example, 
$$P(57 < Y \leq 70) = \int_{57}^{70} f(y)\,dy = 0.6121481.$$
computed (by 'numerical integration') as 
diff(pnorm(c(57,60), 69, 3.5))
## 0.6121481

or approximated by standardizing and referring to a printed table of 
values of the standard normal CDF. (Numerical integration is necessary
for normal distributions because it is not possible to express a normal CDF
in 'closed' mathematical form.)
The probabilities
$P(57 \le Y \le 70)$ and $P(57 < Y < 70)$ are the same, because the
area of a boundary line is 0. 

If you ask for the probability a person chosen at random from this population is 68 inches tall, then technically you might write $P(Y = 68) = 0.$ But that indicates he is
exactly $68.000\dots$ inches tall, which is not feasible to measure.
You might say, "But that is ridiculous. I know a guy who is 68" tall."
What you may really mean is that you know someone who is between 67.5" and 68.5"
tall. Then you have an interval and you can find its probability (even if
small): $P(67.5 < Y \le 68.5) = 0.1090839.$
diff(pnorm(c(67.5,68.5), 69, 3.5))
## 0.1090839


For a continuous random variable, the way to make sense of a "point" is to
ask for the probability of the interval you actually intended. (Or if
you actually intended a point measured with perfect accuracy, then
0 probability can be taken as a reflection of the impossibility of doing that.)
