I'm trying to compute the CDF $F_n$ corresponding to the following data: $\Omega=[-1,2]$, $\mathcal A = \mathcal B\cap\Omega$, $P:\Omega\to\Bbb R$
\begin{equation}
P(A)=\left\{
\begin{aligned}
& \frac{m(A)}6, & 0\not\in A \\
& \frac{m(A)+3}6, & 0\in A
\end{aligned}
\right.
\end{equation}
where $m$ is Lebesgue's measure; and considering the following random variable:
\begin{equation}
X_n(\omega)=\left\{
\begin{aligned}
& 2, & -1\le\omega < \frac1n-1 \\
& n\omega-1, & \frac1n-1\le \omega \le \frac1n \\
& \frac{w+1}{nw}, & \frac1n<\omega\le2
\end{aligned}
\right.
\end{equation}
In particular, I'm having trouble trying to find the expression of $F_n(x)$ for $x\in\left[\frac3{2n},1+\frac1n\right]$:
The side on the left is easy, but the part on the right has given me headaches. I can't seem to find the interval in which $\omega$ lives, or at least this is what I have:
\begin{equation} X_n^{-1}\left(\left[\frac3{2n},x\right]\right) = \left[\frac1{nx-1},2\right] \end{equation}
But I get the following CDF:
\begin{equation} F_n(x)=\left\{ \begin{aligned} &\frac{n+x}{6n}, & -n\ge x < -1\\ & \frac{4n+x}{6n}, & -1\ge x < 0 \\ & \frac23, & 0\le x < \frac3{2n}\\ & \frac16\left(2+\frac1{1-nx}\right), & \frac3{2n}\le x < 1+\frac1n \\ & 1-\frac1{6n}, & 1+\frac1n\le x < 2 \\ & 1, & x\ge 2 \end{aligned} \right. \end{equation}
And in the interval I just got before, the function doesn't seem to do well against the others.
Does anyone know where my mistake is?
You've computed $$P\left(X_n^{-1}\left(\left[\frac3{2n},x\right]\right)\right)$$ when you actually want $$F_n(x):=P\left(X_n^{-1}\left(-\infty,x\right]\right)=P\left(X_n^{-1}\left(-\infty, \frac3{2n}\right]\right)+P\left(X_n^{-1}\left(\frac3{2n},x\right]\right).$$ So to repair your calculation, just add $$P\left(X_n^{-1}\left(-\infty, \frac3{2n}\right]\right)=:F_n\left(\frac 3{2n}\right)=\frac23$$ which you've already computed. This will render your CDF non-decreasing.
Note that $P\left(X_n^{-1}\left[\frac3{2n},x\right]\right)$ is the same as $P\left(X_n^{-1}\left(\frac3{2n},x\right]\right)$, since there's no point mass at $\frac3{2n}$.
PS: Please fix the definition of $X_n(\omega)$ in the third case $\frac1n<\omega\le 2$. You've copied over the 2nd case from the definition of $P(A)$; the value should be $\frac1n(\frac 1\omega+1)$.