2.8

Recall from Proposition 1.12 (ii) on P.65:

Suppose $f$ is integrable on $\mathbb{R}^d$. Then $\forall \epsilon > 0$ there is a $\delta$ > 0 such that: $$m(E) < \delta \hspace{0.25cm} \Rightarrow \int_E |f| < \epsilon$$

Now, simply make the observation that: $$ |F(x) - F(y)| = \Big| \int_{-\infty}^x f(t) dt - \int_{-\infty}^y f(t) dt \Big| \leq \ldots$$ $$ \ldots \leq \Big| \int_x^y f(t) dt\Big| \leq \int_x^y |f(t)|dt$$ Since we already have that $f$ is integrable, certainly, $\forall \epsilon > 0 \hspace{0.25cm} \exists \delta > 0$ such that: $$|x-y| < \delta \Rightarrow \int_x^y |f(t)|dt < \epsilon$$ Thus, we have $F(x)$ must be uniformly continuous.