Question about Measure Theory

Question

Let $(\Omega, U, P)$ be a measure space and X be random variable and its distribution function $F_x(x)=P(\{\omega: X(\omega)\le x\})=P(-\infty , x]$ and the function F is continuous at x.

If the event $A\in \mathscr{U}$ is a countable set in $\Bbb R$, then I need to demonstrate that $P(A)=0$

Please help me showing this. Thank you so much:)

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I know that $P(\phi )=0$, $P(\Omega )=1$, $P(A) \ge 0$ for any set $A \in \mathscr{A}$

I write $A_n$ be a collection of disjoint sets in $\mathscr{U}$

$$C_1 = A_1$$ $$C_2 = A_2 | A_1 \subset A_2$$ $$...$$ $$C_n = A_n \setminus (\bigcup_{k=1}^{n-1}A_k) \subset A_n$$

Then,

$$\bigcup_{n=1}^{\infty} C_n = \bigcup_{n=1}^{\infty} A_n$$

since A is countable, disjoint sets are also countable.

$$P(A)=0=\sum_{n\ge 1} P(C_n)$$ because point probability is zero.

This answer correct? then by using continuity of F, I will show point probability is zero.

Answer 1

Note that:

If F is continuous, then we have

$$P(X = b) = 0$$

It follows that if $A$ is a countable measurable set w/ enumeration $A=\{a_n\}_{n=1}^{\infty}$, then

$$P(X \in A) = P(X \in \bigcup_{n=1}^{\infty} \{a_n\}) = \sum_{n=1}^{\infty} P(X \in \{a_n\}) = \sum_{n=1}^{\infty} P(X = a_n)$$

Answer 2

Using the continuity of $F$, show that $\forall x \in \mathbb{R}$, $P(\{x\})=0$. Then use the definition of a measure.