Prove a random variable has normal distribution 
Let $X$ be a standard normal random variable and $a>0$ a constant.
  Define:
$Y = \begin{cases}
      \phantom{-}X & \text{if $\,|X| < a$}; \\
     -X & \text{otherwise}.
\end{cases}$
Show that $Y$ is a standard normal distribution and the vector $(X,Y)$
  is not two-dimensional normal distributed.

I tried approaching this using the expectation saying for $g$ a measurable function
\begin{align*}
E[g(Y)] 
  &= \int_{\mathbb{R}} g(y)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx  \\
  &= \int_{-\infty}^{-a} g(x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx + \int_{-a}^a g(-x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx + \int_a^{\infty} g(x) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx
\end{align*}
as $x^2 = (-x)^2$ I could reason that $Y$ must have normal distribution, but this just doesn't seem clean enough.
As to the second question I thought I could express $(X,Y)$ as a function of $X$ and then:
$E[g(X,Y)] = E[g(h(X))] = E[f(X)]$ where $f = g \circ h$ a measurable function at which point I was confused by the dimensions.
 A: The distribution of a real-valued random variable $Y$ is determined by its cdf $F := \mathbb{P}(Y \leq y)$ (because sets of the form $\{(-\infty, y]\}$ generate the Borel $\sigma$-algebra on $\mathbb{R}$). Let $N$ denote a standard normal random variable, $Y = -X$ and $y \in \mathbb{R}$; there are three cases: (i) $y \leq -a$, (ii) $-a < y \leq a$, and (iii) $a < y$. Consider case (ii):
\begin{align*}
P(Y \leq y) &= P(Y \leq -a) + P(-a < Y \leq y) \\
&= P(-X \leq -a) + P(-a < X \leq y) \\
&= P(X \geq a) + P(-a < N \leq y) \\
&= P(N \geq a) + P(-a < N \leq y) \\
&= P(N \leq -a) + P(-a < N \leq y) \\
&= P(N \leq y).
\end{align*}
The first equality follows by disjointness and finite additivity; the second because $\{\omega : Y \leq -a\} = \{ \omega : -X  \leq -a\}$ and $\{-a < Y \leq y\} = \{-a < X \leq y\}$ for $-a < y \leq a$. The second-to-last line follows by symmetry of $N$. 
The case (i) is easy and (iii) is symmetric. 
For the second question, have you drawn the support of the measure on $\mathbb{R}^2$? 
A: $\textbf{i)First part}$ 
$\Pr(Z\in A)=\Pr(Y\in A, |Y|\le a)+\Pr(-Y\in A, |Y|> a)$
Since $\displaystyle\Pr(-Y\in A, |Y|> a)=\Pr(-(-Y)\in A, |-Y|> a)=\Pr(Y\in A, |Y|> a)$
Hence 
$\Pr(Z\in A)=\Pr(Y\in A, |Y|\le a)+\Pr(Y\in A, |Y|> a)=\Pr(Y\in A)$
$\textbf{ii)Second part}$ 
Recall that $Cov(X,Y)=E(XY)-E(X)E(Y)$
and we say that two random variables $X,Y$ are uncorrelated if $Cov(X,Y)=0$
this is equivalent to $E(XY)=E(X)E(Y)$, this is of course true if $X$ and $Y$ are independent, but the converse is wrong, however it is true for the Multivariate Normal (in this case Bivariate Normal)
$X,Y$ are standard normal $\Rightarrow E(X)=E(Y)=0$, set $f(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$
and $\displaystyle E(XY)=\int_{-a}^{a}x^2f(x)-\int_{-\infty}^{-a}x^2f(x)-\int_{a}^{\infty}x^2f(x)=1-4\int_{a}^{\infty}x^2f(x)$
Now you can choose $a>0$ such that the expression above is zero (http://inci.ca/ujb6c1_2vp) 
which means that $X$ and $Y$ are independent, but this is false since;
$\Pr(X\in(-a,a);Y\in(-a,a))=\Pr(X\in(-a,a))\neq\Pr(X\in(-a,a))\Pr(Y\in(-a,a))$
