$a-b=\int I(a\leq x)-I(b\leq x)dx$?

Question

Is it true that for any $a,b\in \mathbb{R}$, we have that $a-b=\int_{-\infty}^{\infty}\left( I(a\leq x)-I(b\leq x)\right)dx$?

I see that the difference of indicator functions is only nonzero when $a\leq x < b$ or $b\leq x <a$. Moreover, $ I(a\leq x)-I(b\leq x)$ equals $1$ in the former case and equals $-1$ in the latter. So, depending whether $a<b$ or not, the above integral is $b-a$ or $a-b$, respectively.

So, is this equality wrong?

*the original statement is for positive real numbers $a,b$, but I believe it does not help much. **I'm hesitant to conclude it is wrong because this result was used in an article.

Thanks in advance!

Answer 1

The identity is always false. If you change the left side to $b-a$ then it is always true.