Hello, twittytwitter!
A strange problem . . . I think I've solved it.
. . But my approach isn't very formal either.
Suppose that: .$\displaystyle g(x) \;=\;\begin{Bmatrix} 1 && x \leq 0 \\ 1 && x > 0 \end{Bmatrix}$
(a) Find: .$\displaystyle \lim_{R\to\infty} \int^R_{\text{}R} g(x)\,dx$
The graph of $\displaystyle y \,=\,g(x)$ looks like this, for $\displaystyle x \in [R,\,R]$ Code:


*    *1
:  R
 +    +    +  
R  :
1o    *


Since the integral represents area,
. . $\displaystyle \int^R_{\text{}R} g(x)\,dx$ is the total area of the shaded regions below. Code:


*    *1
:::A::: R
 +    +    +  
R :::B:::
1o    *


Since the area of $\displaystyle B$ is the negative of the area of $\displaystyle A$,
. . the total area is 0 (zero).
That is: .$\displaystyle \int^R_{\text{}R}g(x)\,dx \;=\;0$
Therefore: .$\displaystyle \lim_{R\to\infty}\int^R_{\text{}R} g(x)\,dx \;=\;0$
(b) Repeat (a), but: .$\displaystyle \lim_{R\to\infty} \int^{R+n}_{\text{}R}\!\! g(x)\,dx $ . with $\displaystyle n > 0.$ The graph is a variation of the one in part (a). Code:


*    *1
:::A::: R R+n
 +    +    +   *  
R :::B:::::C::
1o    *   *


Once again, the area of $\displaystyle B$ is the negative of the area of $\displaystyle A.$
. . And the area of $\displaystyle C$ is: .$\displaystyle n$
Hence: .$\displaystyle \int^{R+n}_{\text{}R}g(x)\,dx \;=\;n$
Therefore: .$\displaystyle \lim_{R\to\infty}\int^{R+n}_{\text{}R}\!\! g(x)\,dx \;=\;n$
c) Does: .$\displaystyle \int^{\infty}_{\text{}\infty} g(x)\,dx$ . converge or diverge?
As shown in part (a), it converges to 0.