Distribution Function is absolutely continuous or singular? $$F(x) = 
\begin{cases}
0,&  x < -1\\
\frac{x}{3} + \frac13,& -1 \leq x \leq 0\\
\frac{x}{3} + \frac23,& 0 < x \leq 1\\
1,& 1 \leq x
\end{cases}$$
This $F(x)$ is a distribution function, since it is non-decreasing, right-continuous and etc...
I'm having problems to decide if it is absolutely continuous or if it is singular. 
I guess this is mixed. 
Decided that, I have to find out if I can induce a measure such that $\mu((a, b]) = F(b) - F(a)$ and explicit this measure. 
I guess this is $\frac{1}{3}$Leb in $(-1, 0)$, then it has a $\frac{1}{3}\delta_0$ in $\{0\}$ and it is $\frac{1}{3}$Leb in $(0, 1]$
 A: The distribution function is not right-continuous, as you have written it. It's clear that $F(0)=\frac13$ but $\lim_{x\to0^+}F(x)=\frac23$.
I will assume that you instead meant:
$$F(x) = \frac13(1+x)\cdot \mathsf 1_{[-1,0)}(x) + \frac13(2+x)\cdot\mathsf 1_{[0,1]}(x)+\mathsf1_{(1,\infty)}(x)  $$
(the difference being that $F(0)=\frac23$). This distribution is neither absolutely continuous nor singular. Let $\nu$ be the probability measure induced by $F$, i.e. for any $t\in\mathbb R$, $\nu((-\infty,t])=F(t)$. The Lebesgue decomposition of $\nu$ is then $\nu=\nu_c+\nu_d$ where for a Borel set $B$, \begin{align}\nu_c(B) &= \int_B f_c\ \mathsf dm,\\ \nu_d(B) &= \int_B f_d\ \mathsf d\mu.\end{align} Here $m$ denotes Lebesgue measure, $\mu$ counting measure, and $f_c$, $f_d$ the densities with respect to $m$ and $\mu$. That is,
\begin{align}
f_c &= \frac13\cdot\mathsf 1_{[-1,0)\ \cup[0,1]},\\
f_d &= \frac13\cdot\mathsf 1_{\{0\}}.
\end{align}
For contrast, a singular probability distribution $\sigma$ is one which satisfies $\sigma(N)$ for some set $N$ with zero Lebesgue measure and $\sigma(\{n\})=0$ for all $n\in N$. An example is the Cantor distribution, which has the Cantor function as its distribution function.
