Show that, $Z$ is $\mathcal N(0,1)$ 
If $Y\sim\mathcal N(0,1)$ and let $a>0$. Let $$Z=\ \begin{cases} 
      Y&\text{if } |Y|\le a\\
      -Y &\text{if }|Y|> a\\
       \end{cases}\
$$
  Show that $Z\sim\mathcal N(0,1)$

I tried to prove it using the characteristic function
$$\phi_Z(u)=\phi_{(Y\mathbf1_{|Y|\le a}-Y\mathbf1_{|Y|>a })}(u)\overset{?}=\phi_{Y\mathbf1_{|Y|\le a}(u)}\phi_{-Y\mathbf1_{|Y|> a}}(u)$$
$$=$\phi_{Y\mathbf1_{|Y|\le a}(u)}\phi_{Y\mathbf1_{|Y|> a}}(-u)=\phi_{Y\mathbf1_{|Y|\le a}}(u)\overline{\phi_{Y\mathbf1_{|Y|> a}}(u)}$$
and since $Y$ is symmetric,which means that it is real-valued, we obtain;
$$\overline{\phi_{Y\mathbf1_{|Y|> a}}}(u)=\phi_{Y\mathbf1_{|Y|> a}}(u)$$
Hence;
$$\phi_{Y\mathbf1_{|Y|\le a}}(u)\phi_{Y\mathbf1_{|Y|> a}}(u)=\ \begin{cases} 
      E(e^{iuY})E(e^{iu0})=E(e^{iuY})&\text{if } |Y|\le a\\
      E(e^{iu0})E(e^{iuY})=E(e^{iuY})&\text{if }|Y|> a\\
       \end{cases}\
$$
$\textbf{My Question is:}$
Is the first step allowed, are $Y\mathbf1_{|Y|\le a}$ and $-Y\mathbf1_{|Y|>a }$ independent ?
 A: 
$\textbf{My Question is:}$ (...) are $Y\mathbf1_{|Y|\le a}$ and $-Y\mathbf1_{|Y|>a }$ independent ?

Of course not. Let $U=Y\mathbf1_{|Y|\le a}$ and $V=-Y\mathbf1_{|Y|>a }$, then $[U=0]=[|Y|\gt a]\cup[Y=0]$, $[V=0]=[|Y|\leqslant a]$, and $[U=V=0]=[Y=0]$. What does all this tell you about the possibility that $(U,V)$ is independent?
An extended version of the result at the beginning of the post is the following. 

Let $\theta:[0,\infty)\to\{-1,1\}$ denote any measurable function and $Y=SX$ where $(X,S)$ is independent with $X\geqslant0$ almost surely and $P(S=1)=P(S=-1)=\frac12$. 
Then $\theta(|Y|)\cdot Y$ is distributed like $Y$.

Which in turn follows from the factoid below.

Assume that $P(S=1)=P(S=-1)=\frac12$ and that $T$ is independent of $S$ with $|T|=1$ almost surely.
Then $ST$ is distributed like $S$.

A: Let $h$ be an arbitrary bounded Borel function then
$$\eqalign{
\mathbb{E}(h(Z))&=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+
\underbrace{\frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(-y)e^{-y^2/2}dy}_{y\leftarrow-y}\cr
&=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+
\frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(y)e^{-y^2/2}dy\cr
&=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}h(y)e^{-y^2/2}dy\cr
&=\mathbb{E}(h(Y))
}
$$
It follows that $Z$ and $Y$ have the same distribution, since $h$ is arbitrary, and we are done.
