Find a function $\phi: \mathbb{R} \to \mathbb{R}$ such that $P[(a,\infty)] = \int_{a}^{\infty} \phi (x) dx$ for every $a \in \mathbb{R}$. Let $(\Omega, \mathcal{F}, P)$ be a probability space with $\Omega = \mathcal{R}$ the real line, $\mathcal{F}$ the Borel sets, and probability measure on the $\pi$-system of open semi-infinite positive intervals by
$$P[(a,\infty)]= e^{-a^{2}}$$. 
a) Find a function $\phi: \mathbb{R} \to \mathbb{R}$ such that $P[(a,\infty)] = \int_{a}^{\infty} \phi (x) dx$ for every $a \in \mathbb{R}$. 
My approach,
I know that $\int_{a}^{\infty} \phi (x) dx = e^{-a^{2}}$ so I was thinking I can use Fundamental Theorem of Calculus here and differentiate both sides with respect to x. 
b) Find the probability distribution $\mu_{y}(dy)$ for $Y(\omega):= \omega^{2}$.
$$P(Y(\omega)\leq y) = P(\omega^2 \leq y) = P(-\sqrt{y} \leq \omega \leq \sqrt{y}) = P[(\infty,\sqrt{y}] - P[(\infty,-\sqrt{y}] = 1 - e^{-\sqrt{y}^{2}} - 1 + e^{\sqrt{y}^{2}} = e^{y} - e^{-y}.$$
Then, we can differentiate that wrt to $y$ to get the probability distribution. Is this approach correct?

There is no probability measure $P$ such that $P(a,\infty)=e^{-a^{2}}$ for all real numbers $a$. I think this equation was supposed to hold only for $a \geq 0$ and $P(a,\infty)$ is supposed to be $1$ for $a<0$. Using the equation for all $a$ has led to a wrong answer: probabilities have to be between $0$ and $1$ but $e^{y}-e^{-y} \to \infty$ as $ y \to \infty$ so your answer cannot be right. I will let you modify the answer based on the modified formula for $P(a,\infty)$. 
The correct answer is $P(Y\leq y)=1-e^{y}$ for $y \geq 0$ and $0$ fro $y <0$.  
