I'm trying to see if the following holds:
Let $a\in\mathbb{R}$ and $b\in\mathbb{R}$, then it is always possible to write:
$$( a - b )=\int_{-\infty}^{\infty}\left(I_{\{b\leqslant x\}} - I_{ \{ a \leqslant x\}} \right)dx$$
My steps are as follows:
\begin{align} \int_{-\infty}^{\infty}\left(I_{\{b\leqslant x\}} - I_{ \{ a \leqslant x\}} \right)dx = & \int_{-\infty}^{\infty}I_{\{b\leqslant x\}} dx - \int_{-\infty}^{\infty}I_{ \{ a \leqslant x\}} dx \\ = & \int_{b}^{\infty}dx - \int_{a}^{\infty} dx \\ (\textrm{replacing infinite limits with finite values})=&\lim_{c\to\infty} \left[ \int_{b}^{c}dx - \int_{a}^{c} dx \right]\\ =&\lim_{c\to\infty} \left[ (c-b)- (c-a) \right]\\ =&\lim_{c\to\infty} \left[ (a-b) \right]\\ =&(a-b) \end{align}
I'm not sure whether my steps are valid; any pointers/insights would be appreciated. Many thanks in advance.
Edited using suggestions:
Assuming $b\leq a$ \begin{align} \int_{-\infty}^{\infty}\left(I_{\{b\leqslant x\}} - I_{ \{ a \leqslant x\}} \right)dx = & \int_{-\infty}^{\infty}\left(I_{ \{b\leqslant x \leqslant a \}} \right)dx \\ = & \int_{b}^{a}dx \\ =&(a-b) \end{align}
Your argument is not valid. The expression $\int_b^{\infty} dx-\int_a^{\infty} dx$ is not defined because both the integrals are $\infty$. To avoid this problem verify that $I_{\{b\leq x\}} -I_{\{a\leq x\}} =I_{\{b\leq x <a\}}$ and then integrate. Remark: I have assumed that $b \leq a$. Because of the symmetry in the equation there is no loss of generality in assuming this. [ If you interchange $a$ and $b$ you get the same equation multiplied by $(-1)$].