This comes from the paragraph following equation (27) on page 6 of this paper. It's not crucial to the argument — any such bound will do — but it's not clear to me why this particular bound is appropriate.
Using a discount $\gamma < 1$ corresponds to dropping the terms with $l ≫ 1/(1 − γ)$ [in the equation $\sum_{l=0}^\infty\gamma^l\chi(l;s,a)$].
Given a bound $C$ on $\chi$, those terms can be bounded by $\frac{C\gamma^l}{1-\gamma}$, but why is that negligible?