Why the probability of single element event is "zero" on continuous interval? Why the probability of a single element event is "zero" on the continuous model?, the explanation given is based on probability additivity axioms. but how? some more explanation with the source would be helpful.
cool thanks!
 A: Intuition only approach (because I'm ignorant of the corresponding math).
You have the interval $[0,1]$ and are asked to choose a number at random from this interval.
$E$ is the event that you choose the number $(1/2)$ and you are asked to compute
$p(E).$
So, as a worshipper of Archimedes (or whoever), you reason as follows:
I can estimate $p(E)$ by assuming that the interval $[0,1]$ is broken into exactly $n$ sub intervals :
$[0,\frac{1}{n}), [\frac{1}{n},\frac{2}{n}), \cdots [\frac{n-1}{n},1].$
Then, the number $(1/2)$ is in only one of these subintervals.
Therefore, $p(E) < $ probability of choosing the corresponding subinterval, which is clearly $\frac{1}{n}$.
This analysis holds, even if you allow $n$ to grow unbounded (i.e. approach $\infty$).  In such a scenario, the chance of selecting the corresponding sub-interval, which is $\frac{1}{n}$ must approach $0$.
A: Assume $\xi$ is a random variable and $p(x)$ is the density function of $\xi$.
For all $t\in R$, we have $P(\xi =t)\leq P(t\leq \xi\leq t+h)=\int_{t}^{t+h}p(x)dx$ for any $h>0$.
Therefore $0\leq P(\xi =t)\leq \lim\limits_{h\to 0^+}\int_{t}^{t+h}p(x)dx=0$, so $P(\xi =t)=0$.
A: Suppose that there is a biased random number generator that can produce numbers between $0$ and $1$. The probability density function corresponding to this generator is $f(x)=2x$:

To find the probability of falling into a particular interval $[a,b]$, you have to compute the integral between $a$ and $b$ of the probability density function. In our example, that means finding $\int_{a}^{b} 2x \, dx$:
$$
\int_{a}^{b} 2x \, dx = \left[x^2\right]_{a}^{b} = b^2 - a^2
$$
In fact, you need not use integration. The area of the trapezium shown below in below is given by
$$
\text{area} = \text{base} \times \text{average height} = (b-a) \times \frac{2a+2b}{2} = b^2 - a^2
$$

Since the probability of an event corresponds to area, we know that
$$
P(a \leq x \leq b) = b^2 - a^2 \, .
$$
The probability of a single event happening corresponds to an interval of width zero. In other words,
$$
P(x=a) = P(a \leq x \leq a) = a^2 - a^2 = 0
$$
This means that in a continuous probability distribution, the chance of a single event happening is equal to $0$.
You probably still find this answer a little unconvincing. In a discrete setting, the sum of the probabilities of all the events is equal to one. In a continuous setting, not only is this untrue, it doesn't make any sense: there is no sensible way to define the sum of uncountably many terms. One way to reconcile the differences between discrete and continuous settings is to realise that a continuous probability distribution is in a sense the 'limit' of a discrete one, as user2661923 has already mentioned. Imagine we approximate the probability density function $f(x) = 2x$ with $10$ rectangles of width $0.1$:

These rectangles give us an upper bound for the chance of falling into an interval, with
\begin{align}
P(x \in [a,a+0.1]) &\leq \text{interval width} \times \text{height of rectangle at end of interval}\\
&= 0.1 \times 2(a+0.1) = \frac{a+0.1}{5}
\end{align}
What about the general case? If we split the probability distribution function into $n$ intervals of width $1/n$, then
$$
P(x \in [a,a+1/n]) \leq 1/n \times 2(a+1/n) = \frac{2(a+1/n)}{n}
$$

Furthermore, since $a$ is a subset of $[a,a+1/n]$
$$
P(x = a) \leq P(x \in [a,a+1/n]) \, .
$$
Hence,
$$
P(x=a) \leq P(x \in [a,a+1/n]) \leq \frac{2(a+1/n)}{n} \, .
$$
For any interval of width $1/n$, the above inequality must be satisfied. If we make the interval arbitrarily small, then $n$ tends to infinity, and
$$
\frac{2(a+1/n)}{n}
$$
tends to $0$. This means for every $\varepsilon > 0$, $P(x=a) < \varepsilon$. But since probabilities must be nonnegative, we are left with one option:
$$
P(x = a) = 0 \, .
$$
A: Suppose otherwise, so that the chance of picking any element is positive, say $p$. Assuming uniform distribution, this probability is equal for all values in the interval by definition. Then for each of the infinitely many values in the interval, it is chosen with probability $p$, so the total probability would be infinite, which is absurd.
Of course, you don't need a uniform distribution either. You can easily convince yourself that summing uncountably many positive values is going to give you an infinite result.
