Is this a distribution function? Let $F : \mathbb R \rightarrow [0,1]$ such as $F$ is right continuous, non-decreasing, $\lim_\limits{x \rightarrow -\infty} F(x) = 0$ and $\lim_\limits{x \rightarrow \infty} F(x) = 1$
Is $F$ necessarily a distribution function ? Meaning, is there a measure $\mu$ on $(\mathbb R,\mathcal B (\mathbb R))$ such that $F(x)= \mu(]-\infty, x])$ ?
If yes, does that extend to $\mathbb R^n$ ?
 A: Define $\mu (a,b]) =F(b)-F(a)$ and extend this to finite disjoint  unions of half-closed intervals $(a,b]$ by additivity . You get a measure on an algebra which generates the Borel sigma algebra. Caratheodory extension theorem tells you that $\mu$ can be extended to the Borel sigma algebra. This is all standard text book material and you can fins it in Rudin's RCA, for example. In the case of $\mathbb R^{n}$ the condition on $F$ are more complicated. You can find the detials in Billingsle'y Probability and Measure.
A: Yes, those four properties are necessary and sufficient for $F$ to be a distribution function.
Given $F$ satisfying those properties, define $X(\omega) = \sup\{y : F(y) <\omega\}$ for $\omega \in (0,1)$. Then $X$ is a random variable on $(\Omega, \mathcal{F},P)$ where $\Omega = (0,1)$, $\mathcal{F}$ is the set of Borel subsets of $(0,1)$, and $P$ is Lebesgue measure. One can show that $\{\omega : X(\omega) \leq x\} = \{\omega : \omega \leq F(x)\}$, and thus $P(X \leq x) = F(x)$. Hence $F$ is a distribution function. This construction is taken directly from Durrett, and you can read the proof in Theorem 1.2.2.
