# Continuity of a random variable and of its cumulative distribution function

Could you help me to understand if the following statements are correct?

Let $X$ be a random variable with cdf $F$ (so $F$ is, by definition, non-decreasing, right continuous, $\lim_{x\rightarrow-\infty}F(x)=0$, $\lim_{x\rightarrow\infty}F(x)=1$)

1) $X$ is a continuous random variable if and only if $F$ is a continuous functions

2) $F$ is a continuous function strictly monotone (increasing) if and only if $F^{-1}$ exists

3) $X$ is a continuous random variable and $F$ strictly monotone function (increasing) if and only if $F^{-1}$ exists

The first statement: $\implies$: $X$ being a continuous random variable means that it has a density. Just write $F$ as an integral of this density to see that $F$ is continuous. $\impliedby$: If $F$ is continuous, then it is absolutely continuous with respect to the Lebesgue measure so then by the Radon-Nikodyn theorem $X$ has a density.
The second statement: This has nothing to do with probability. $\implies$ is obvious I hope. For $\impliedby$ note that $F$ has an inverse if and only if it is bijective. Since $F$ is nondecreasing, it must be strictly increasing. A strictly increasing (or decreasing) bijective function has to be continuous. See this for a proof.
Statement 1) is false -- usually "$$X$$ is a continuous random variable" means "there is an integrable function $$f$$ (called the density of $$X$$) such that $$\mathbb{P}(X \leq t) = \int_{-\infty}^t f(x)\,dx$$". Since $$f$$ is integrable, this expression (which gives the cdf) is continuous. But the converse need not be true. The Cantor function is the classic example of a continuous cdf which is not absolutely continuous (i.e. cannot be expressed as a cumulative integral of some other function).