Question: Let $f$ be a nonnegative integrable function. Show that the function $F$ defined by $$ F(x) = \int_{-\infty}^{x} f $$ is continuous.
In other words, show for all $\epsilon > 0$, there exists a $\delta > 0$ such that $$ | y - x | < \delta \quad \implies \quad \left| \int_{-\infty}^y f - \int_{-\infty}^x f \right| < \epsilon $$ for all $x,y \in \mathbb{R}$.
Issues: So I'm working out of Royden & Fitzpatrick (4th ed) and this question follows a chapter on Uniform Integrability (The Vitali Convergence Theorem). I was thinking I could formulate it using: $$\left| \int_{-\infty}^y f - \int_{-\infty}^x f \right| = \left| \int_{-\infty}^{-\infty} f - \int_{y}^{x} f \right| = \left| \int_{x}^{y} f \right| < \epsilon $$ but I'm not sure if this works or not, and how I would go from here.