Integral of characteristic function Let $\alpha >0$ and let $x \in \mathbb{R}$. Show that $$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \chi_{[-\alpha,\alpha]}(x-t).\chi_{[-\alpha,\alpha]}(t)\;dt= \frac{1}{\sqrt{2} \pi}\bigg(m([-\alpha,\alpha] \cap [x-\alpha,x+\alpha] \bigg).$$  Here $\chi_{[-\alpha,\alpha]}(x)$ is the characteristic function.
I have no idea i to show this, i always ended up with $\frac{1}{\sqrt{2 \pi}} m(A)$. Any help and hints would help thanks!!! 
 A: Here  $x$ is fixed. $x-t \in [-\alpha, \alpha]$ and $t \in [-\alpha, \alpha]$ iff $x-\alpha \leq t \leq x+\alpha$ and $-\alpha \leq t \leq \alpha$ iff $t \in [-\alpha, \alpha] \cap [x-\alpha,x+ \alpha]$. Hence the integral becomes the measure of $[-\alpha, \alpha] \cap [x-\alpha,x+ \alpha]$.
A: We have 
$$\chi_{[-\alpha,\alpha]}(x-t).\chi_{[-\alpha,\alpha]}(t)=1 \iff t \in [-\alpha,\alpha] \cap [x-\alpha,x+\alpha]$$
and
$$\chi_{[-\alpha,\alpha]}(x-t).\chi_{[-\alpha,\alpha]}(t)=0 \iff t \notin [-\alpha,\alpha] \cap [x-\alpha,x+\alpha].$$
With $A=[-\alpha,\alpha] \cap [x-\alpha,x+\alpha]$ we get therefore
$$\int_{-\infty}^{\infty} \chi_{[-\alpha,\alpha]}(x-t).\chi_{[-\alpha,\alpha]}(t)\;dt=m(A).$$
A: $1_{[-a,a]}(x-t)=1 \Longleftrightarrow x-t=b \in [-a,a] \Longleftrightarrow t \in [-a-x,a-x] \Longleftrightarrow 1_{[x-a,x+1]}(t)=1$
So the integral is equal to $$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}1_{[-a,a]}(t)1_{[x-a,x+a]}(t)dt=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}1_{[-a,a]\cap[x-a,x+a]}(t)dt$$
A: Note that $x-t \in [-\alpha,\alpha] \iff t\in [x-\alpha,x+\alpha]$. Thus:
$$\chi_{[-\alpha,\alpha]}(x-t)\chi_{[-\alpha,\alpha]}(t) = \chi_{[-\alpha,\alpha]\cap[x-\alpha,x+\alpha]}(t)$$ and: 
$$\int_{-\infty}^{\infty}\chi_{[-\alpha,\alpha]}(x-t)\chi_{[-\alpha,\alpha]}(t)dt = \int_{-\infty}^{\infty}\chi_{[-\alpha,\alpha]\cap[x-\alpha,x+\alpha]}(t)dt = m([-\alpha,\alpha]\cap[x-\alpha,x+\alpha])$$
