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.

$\textbf{i)First part}$ 
$\Pr(Z\in A)=\Pr(Y\in A, |Y|\le a)+\Pr(-Y\in A, |Y|> a)$
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;
$\Pr(X\in(-a,a);Y\in(-a,a))=\Pr(X\in(-a,a))\neq\Pr(X\in(-a,a))\Pr(Y\in(-a,a))$
