Given the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and the r.v. $X$ defined on it, the (pushforward) measure $\mu_X$ is defined as $\mu_X(B) := PX^{-1}(B), \forall B \in \mathcal{B}(\mathbb{R})$, and the distribution function $F_X(x)$ is defined as $F_X(x) := \mu_X((-\infty, x])$. $\mu_X$ is nonatomic if $$\forall A_1 \in \mathcal{B}(\mathbb{R}): \mu_X(A_1)>0 \rightarrow \exists A_2\in \mathcal{B}(\mathbb{R}) \quad \& \quad A_2\subset A_1: \mu_X(A_2)>0.$$
If such a set $A_2$ does not exists, $A_1$ is an atom of $\mu_X$.
There is also this fact that $F_X$ is continuous iff $\mu_X(\{x\}) = 0, \forall x \in \mathbb{R}.$
To prove "if $F_X$ is continuous $\rightarrow$ $\mu_X$ is nonatomic", this is how I proceed:
Suppose not: there exists $A \in \mathcal{B}(\mathbb{R})$, an atom of $\mu_X$ with $c := \mu_X(A) > 0$. Restricting $\mu_X$ to $A$, we can consider the measurable space $(A, \mathcal{B}(\mathbb{R})_A)$. Let $F(x) = \mu_X((-\infty, x]) \in [0, c]$. $F$ is a distribution function and hence is non-decreasing, $F(x) \downarrow 0$ as $x \rightarrow -\infty$, and $F(x) \uparrow c$ as $x \rightarrow \infty$. Continuity of F implies $F(x) \in (0, c)$ for some $x \in \mathbb{R}$, then the set $B = (-\infty, x]$ gives a contradiction to the fact that $A$ is an atom.
Questions:
1- Is there anything about this proof not rigorous enough that needs modification?
2- How can I prove the converse of the claim?
Any other proof strategy is appreciated as well.
Best Answer
If $\mu_X$ is nonatomic, for every $x \in \mathbb R$ we have $$\eqalign{0 &= \mu_X(\{x\}) = \lim_{n \to \infty} \mu_X((x-1/n, x+1/n]) \cr &= \lim_{n \to \infty} (F(x+1/n) - F(x-1/n))}$$ Since $F$ is nondecreasing, given $\epsilon > 0$ we have $|F(x+t) - F(x)| < \epsilon$ for $|t| < 1/n$ if $n$ is large enough that $F(x+1/n) - F(x-1/n) < \epsilon$.