After checking some olde posts about the Lebesgue measure and how dos it play its role when I random variable is normally distributed I found this post. My question is, what happens when a random variable $X$ follows a normal distribution $N(\mu,\sigma^2)$ instead of a standard normal?
Suppose that $(\Omega,\mathcal{F},\mathbb{P})$ is a standard probability space and $x:(\Omega, \mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is a random variable normaly distributed where $x\sim N(\mu,\sigma^2)$. How is the lebesgue measure defined in here?
My attempt, correct me if I am mistaken. If $F$ is the CDF of the random variable $x$, namely
$$F(y)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^ye^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}d\Lambda(x)$$
$F$ is strictly increasing and hence one-to-one and continuous. The range of $F$ is $(0,1)$ with $F(-\infty)=0$ and $F(+\infty)=1$. The inverse of $F$ is $x:(0,1)\to\mathbb{R}$ which is also increasing, one-to-one and continuous with $x(0)=-\infty$ and $x(1)=+\infty$. Since $x_{|(0,1)}$ is continuous, it is measurable on the Lebesgue probability space. Observe that
$$\Lambda\left(\{\omega\in[0,1]:x(\omega)\leq y\}\right)=\Lambda\left(x^{-1}((-\infty,y])\right)=F(y)$$
Thanks in advance!