Compute $\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt$ Compute $\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt$. 
So if I create a number line marking a-b and a+b. If that the integral above has 5 different answers depending on where (-t,t) is located on the number line. 


*

*Of course if (-t,t) if before the mark of a-b and a+b on the number line then the integral is equal to zero. 

*If a bit of (-t,t) hangs outside of the mark of a-b but some of it is in (a-b,a+b), then the integral is equal to y-a+b.

*Similarly, if a bit of (-t,t) hangs outside of the mark of a+b but some of it is in (a-b,a+b), then the integral is equal to a+b+t-y.

*If (-t,t) is contained in (a-b,a+b) then the integral is equal to 0.


So what I have is 
$$\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt=\begin{cases}
0, & \text{if }y<a-b \\
y-a+b, & \text{if } a-b\leq y \leq ? \\
1, & \text{if }?\leq y \leq ?\\
a+b+t-y, & \text{if }?\leq y \leq a+b \\
0, & \text{if }y > a+b \\
\end{cases}$$
What would go in the "?" area?
 A: I'm assuming $\chi_{(-t,t)}$ is the characteristic function of $(-t,t)$, i.e.
\begin{equation}
 \chi_{(-t,t)}(y) =
 \begin{cases}
  1, & y \in (-t,t),   \\
  0, & y \notin (-t,t).
 \end{cases}
\end{equation}
The way you explain the cases isn't clear, though, as you state that they depend on $t$? But $t$ is the variable of integration, it ranges from $a-b$ to $a+b$, so the cases should be in terms of $y$, $a$, and $b$ (consistent with the incomplete answer you give).
Also, regarding the first case stated: if $y < a-b$, then it is still possible that $y \in (-t,t)$ for $t \in (a-b,a+b)$ (so the integral is nonzero). For instance, if $a=6$, $b=1$, and $y=-3$, then we have $\int_5^7 \chi_{(-t,t)}(-3) dt = \int_5^7 1 dt = 2$, as $-3\in (-t,t)\;\forall t \in (5,7)$. So I disagree with the first case listed.
Instead, I think there are a lot of cases to consider based on the signs of $a$ and $b$. Let me take the liberty of assuming $0 \leq b \leq a$ so that $0 \leq a-b \leq a+b$ for simplicity.
$$
\longleftarrow
\!\!\!-\!\!\!-\!\!\!\!\!\!\!
\stackrel{-(a+b)}{\bullet}
\!\!\!\!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\!\!\!\!
\stackrel{-(a-b)}{\bullet}
\!\!\!\!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!
\stackrel{0}{\bullet}
\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!
\stackrel{a-b}{\bullet}
\!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!
\stackrel{a+b}{\bullet}
\!\!-\!\!\!-\!\!\!\!\!\!\!\!\!
\longrightarrow
$$
Then I claim
\begin{equation}
 \int\limits_{a-b}^{a+b}\chi_{(-t,t)}(y)dt = 
 \begin{cases}
  0,
  &y \leq -(a+b),\\
  a+b+y,
  &-(a+b) < y \leq -(a-b),\\
  2b,
  &-(a-b) < y \leq a-b,\\
  y-(a-b),
  &a-b < y \leq a+b,\\
  0,
  &a+b < y.
 \end{cases}
\end{equation}
To justify this: The integrand is always $0$ or $1$, so we just compute the size of the interval $I = \{t\in(a-b,a+b):y\in(-t,t)\}$.
If $y \leq -(a+b)$, then $y \notin (-t,t)$ whenever $a-b<t<a+b$, so we get $I = \emptyset$.
If $-(a+b)<y\leq-(a-b)$, then $I$ is the interval from $-(a+b)$ to $y$.
If $-(a-b) < y \leq a-b$, then $y\in(-t,t)$ for every $t$ under consideration, so $I = (a-b,a+b)$.
etc.
However, we need the assumption $0 \leq b \leq a$.  For instance, if $a$ and $b$ can be anything, then the first condition becomes
\begin{equation}
 \int\limits_{a-b}^{a+b}\chi_{(-t,t)}(y)dt = 
 \begin{cases}
  0,
  &y \leq \min\{\pm(a-b),\pm(a+b)\},\\
        \vdots
    \end{cases}
\end{equation}
A: Actually, you can change it a little, $\chi_{(-t,t)} (y)=\chi_{(\max(-y,y), +\infty)} (t)=\chi_{(|y|,+\infty)} (t),$ so we change the integral to $J=\int_{a-b}^{a+b} \chi_{(|y|, +\infty)} (t) dt= \mu((|y|, +\infty)\cap (a-b, a+b ))$.  
When $|y| \leq a-b, J=2b.$ 
When $|y| \geq a+b, J=0.$
When $a-b<|y|<a+b, J=a+b-|y|$.
Then we consider the sign of $y$ and the $b-a\geq 0$ or not.
