Probability of continuous random variable It is claimed that probability of continuous R.V. is 0 for ALL x in R. That is, for every point the probability is zero.
But, somehow when we sum all these zero probabilities over the entire domain $\mathbb{R}$, the total becomes equal to 1.  
Can somebody explain how 0s can add up to 1?  This defies kindergarden math.
 A: Kindergarden math does not incorporate the concept of infinity, let alone distinguishes the different orders of magnitude of infinity. (Kindergarden math does not even include negative numbers, and if you are stuck with the kindergarten math, you may never understand why you owe money on a house, let alone understand how compound interest works.)
A density of a random variable is a Radon-Nikodym derivative of the measure implied by the random variable with respect to Lesbegue measure. It does take an infinity for this thing to work out, as the finite summation of zeroes will still be zero, as you correctly noted. There are far more complicated things involved here, including measurability that relies on "good" sets of a real line being closed under countable unions, intersections and complements.
Without real analysis, these things are difficult to grasp. Are there more numbers in rational numbers than in natural numbers? (No, the cardinalities of the two sets are the same.) Are there more numbers on the real line than in natural numbers? (Yes, they are different types of infinities.) Are there more numbers on a unit square than there are on the real line? (No, the cardinalities are the same.) The proofs for each of these statements can be easily constructed... but again the kindergarten math does not incorporate the concept of a proof, just taking the teacher's word for it ;)
A: In fact, what you sum up to become 1 is not the probabilities of the variable being each of the values: you sum up (more or less) the probabilities of the variable being within an infinitely small interval around each value.
While the limit of these individual probabilities is also zero, the limit of their sum need not be (this is only so because there is an infinite number of summands, so it is not easy to create an example of this).
There is a difference between an integral and a sum.
A: Suppose I am in the business of building mathematical models for clients and a client gives me the outputs from his "random number generator" that has generated a million random numbers $x_1$, $x_2, \ldots, x_{10^6}$
all between $0$ and $1$, and all different.  He would like to have a model 
for the output of this random number generator as a random variable $X$.
First attempt: $X$ is a discrete random variable taking on values $x_1$, $x_2, \ldots, x_{10^6}$ with equal probability.  That is indeed very satisfactory.  It matches the data perfectly, the client is happy, and so am I as I pocket my fee.
But the next day, the client is back because he has run his random number
generator some more and none of the new outputs $x_{10^6+1}$, 
$x_{10^6+2}, \ldots$, match what the model is predicting, viz., the
output will be one of $x_1$, $x_2, \ldots, x_{10^6}$ (with equal probability).
So, now I have to think a bit.
Second attempt: I look at the histogram of the numbers generated
by the random number generator and it looks pretty flat all the way across
$(0,1)$.  It looks like $X$ can take on any real number value between
$0$ and $1$ and so I model $X$ as a continuous random variable.
But what value should I assign to, say, $P\{X = 0.2173333605\}$? It
appears just once in the list and thus has relative frequency $10^{-6}$
on the first million outcomes but looking at $x_{10^6+1}$, 
$x_{10^6+2}, \ldots$, I see that the relative frequency is decaying away 
towards $0$.  I look at $2\times 10^6$ outputs and the relative frequency
of the value  $0.2173333605$ has halved, while the relative frequency of
$0.92387634504$ is stuck at $0$ because it is not among the first two
million trials.
So since probabilities in the model should be similar to observed 
relative frequencies in real life, I say $P\{X = x\} = 0$ for 
every real  number $x$, $0 < x < 1$.  Wait a minute!  Where did
the probability disappear?  Well, the probability is hiding in the
intervals.  Roughly $500,000$ of $x_1, \ldots, x_{10^6}$ are in
the interval $(0, 0.5)$ while roughly $1,000,000$ of
$x_1, \ldots, x_{2\times 10^6}$ are in the interval $(0, 0.5)$.
In other words, the relative frequencies of the intervals are
pretty stable but the relative frequency of a specific number
is either $0$ or decaying away towards $0$.  Similar remarks
apply to other intervals.  So the model for
this continuous random variable $X$ is that 


*

*For any $a, b$ such that $0 \leq a < b \leq 1$, $P\{a < X < b\} = b-a$.

*For any $a, b$ such that $a \leq 0 < b \leq 1$, $P\{a < X < b\} = b$.

*For any $a, b$ such that $0 \leq a < 1 \leq b$, $P\{a < X < b\} = 1-a$.

*For any $a, b$ such that $a \leq 0 < 1 \leq b$, $P\{a < X < b\} = 1$.


More succinctly, for $a < b$, $P\{a < X < b\} = \min\{b, 1\} - \max\{a, 0\}$,
and $X$ is said to be uniformly distributed on $(0, 1)$.  Its 
probability density function is 
$$f_X(x) = \begin{cases} 1, & 0 < x < 1,\\0, & \text{otherwise}.\end{cases}$$ 
Note that the function is nonnegative and the area under the curve is $1$.  Also
$$P\{a < X < b\} = \int_a^b f_X(x) dx.$$
More generally, a continuous random variable $Y$ has a probability density
function $f_Y(y)$ that is nonnegative and has area $1$, and
$$P\{a < Y < b\} = \int_a^b f_Y(y) dy.$$
As @NickSabbe and StasK have told you, for continuous random variables,
probabilities are obtained via integrals not via sums.  Indeed, it is possible
that $f_Y(y) > 1$ for some values of $y$.  As long as
$$P\{-\infty < Y < \infty\} = \int_{-\infty}^{\infty} f_Y(y) dy = 1$$
it is perfectly OK to have $f_Y(y) > 1$ for some values of $y$.  The
density of the probability can exceed $1$, but the total probability
is $1$.  Note that $f_Y(y) > 1$ will definitely break the sums that
you want to use. 
