How to prove the probability of a specific point or set equals to 0 in a continuous distribution? Here we will consider a probability triple $(\Omega , F , P)$ when $\Omega = (0 , 1]$. We define $F$ to be the smallest $\sigma$-algebra that contains all intervals of the form of $(a , b)$, where $0 \leqslant a < b \leqslant 1)$. The probability measure $P$ is defined so that $P((a , b)) = b^2 - a^2$
1) Show that, both set $\{1\}$ and $\{0.5\}$ belongs to the $\sigma$-algebra $F$.
2) Prove that $P(\{0\}) = P(\{0.5\}) = 0$.
In this question, I totally don't know how to get on with (1).
For (2), I know that the probability of a specific point in a continuous probability equals $0$, but I don't know how to prove that.
 A: As you've noted, 2) follows from a well-known theorem:
Theorem: For a continuous probability distribution on an interval $(a,b)$ the density at any particular point is $0$.
This theorem can actually be written far more generally, which I'll comment upon after the proof. For your case, the following is sufficient:
Proof: Let $\mu$ be a continuous density function on an interval such that $\mu(\{x\})>0$ for some $x$. Then since $\mu$ is continuous, there exists an $\epsilon$ such that $\mu(\{y\})>0$ for all $y\in (x-\epsilon,x+\epsilon)$. Now just apply additivity to obtain a subset of that interval with arbitrarily large density. Since the total density needs to be one, we are done.
The key ideas of this proof are:


*

*A function that is continuous at $x$ has values close to $f(x)$ at points close to $x$

*The underlying set is dense at $x$

*Density of disjoint sets is additive


It would be a good exercise to try to formulate the most general form of this theorem that you can, by restricting the proof the the narrowest application of these principles that you can. For example, your function is continuous everywhere whereas this theorem can say something interesting about functions that are continuous at any point. This theorem also doesn't need to be specific to intervals in $\mathbb{R}$ as 2. is a much weaker assertion than "the underlying set is an interval"
