Prove that if $X$ is standard normal then $|X|$ and $1_{X>0}$ are independent 
Given $X$ is standard Normal distributed ($X\sim N(0,1)$). Show that $|X|$ and $1_{X>0}$ are independent.

My attempt $|X| = X$ for $X>0$, and $|X| = -X$ for $X\leq 0$. Each of these possibilities have $p=\frac{1}{2}$ as $X$ ~ $N(0,1)$. Same probabilities work for $1_{X>0} = 1$ for $X>0$ and $0$ for $X<0$. But now, we need to find $P(|X|\cap 1_{X>0})$. This intersection turns out to be $1$ if $X>0$, and $0$ if $X\leq 0$. 
We have: $P(1) = P(0)=\frac{1}{2}$ (by definition of $1_{X>0}$). But $\frac{1}{2}$ is not equal to $P(X)P(1)$ for $X>0$ or $X\leq 0$, isn't it? 
My question I think I messed up something here on the product of the two probabilities. Could someone please help correct it?
 A: Proving two random variables are independent is a bit different than proving two events are independent. Quoting Wikipedia, 

Two random variables $X$ and $Y$ are independent if and only if for every $a$ and $b$, the two events $\{X \leq a\}$ and $\{Y \leq b\}$ are independent.

And, again, quoting Wikipedia, 

Two events $A$ and $B$ are independent if their joint probability equals the product of their probabilities, $P(A \cap B) = P(A) P(B)$. 

So, now, using these definitions, to answer your question we need to show that for every $a$ and $b$, $P(|X| \leq a \cap 1_{X > 0} \leq b) = P(|X| \leq a) P(1_{X > 0} \leq b)$.
See if you can transfer your work done so far to these definitions. 
A: We must show $$P(|X| \leq a, \mathbf{1}_{X > 0} \leq b) = P(|X| \leq a)P(\mathbf{1}_{X > 0} \leq b).$$ 
I'll sketch the case $0 \leq b <1$. The other cases should be easy.
Start by just thinking about what the events mean. The event $\{|X| \leq a, \mathbf{1}_{X > 0} \leq b)\}$ is equivalent to $\{|X| \leq a, \mathbf{1}_{X>0} = 0  \} = \{ -a \leq X \leq a, X \leq 0 \} = \{-a \leq X \leq 0\}$. 
Now we can calculate the probabilities using the density $\phi$ of the standard normal.
$$ P(|X| \leq a, \mathbf{1}_{X > 0} \leq b) = P(-a \leq X \leq 0) = \int_{-a}^{0}\phi(x)dx = 1/2 \int_{-a}^{a}\phi(x)dx   $$
Notice that we've used the symmetry of $\phi$. On the other hand we also have 
$$P(|X| \leq a)P(\mathbf{1}_{X > 0} \leq b) = \int_{-a}^{a} \phi(x)dx \int_{-\infty}^{0} \phi(x)dx = 1/2 \int_{-a}^{a}\phi(x)dx,$$
where we use the fact that $\int_{-\infty}^{0} \phi(x)dx = 1/2$.
