Let $f_n : [a,b] \to \mathbb{R}$ be continuous and uniformly bounded. Define $$g_n(x) = \int_{a}^{x} f_n(s) ds, \ \ x \in [a,b], \ n \geq 1.$$ Show that $g_n(\cdot)$ is equicontinuous.
We must show that for all $\epsilon > 0$ there exists a $\delta(x_0, \epsilon)$ such that for every $n$ and for every $x_0 \in [a,b]$, $$|g_n(x) - g_n(x_0)| < \epsilon, \ \ \text{whenever} \ \ |x-x_0| < \delta.$$
$$|g_n(x) - g_n(x_0)| = | \int_{a}^{x} f_n(s) ds - \int_{a}^{x_0} f_n(s) ds | $$ $$ \leq |\int_{a}^{\delta + x_0} f_n(s) ds - \int_{a}^{x_0} f_n(s) ds| = |\int_{a}^{\delta} f_n(s) ds|.$$
Since $f_n$ is uniformly bounded, we know that $|f_n(x)| \leq M$ for all $n$. So, $$|\int_{a}^{\delta} f_n(s) ds| \leq \int_{a}^{\delta} |f_n(s)| ds \leq M(\delta-a).$$
So, for $\delta < \frac{\epsilon}{M} + a$, we have that $|g_n(x) - g_n(x_0)| < \epsilon$.
Does this look right? I honestly haven't done problems involving integrals in forever, so I am not sure if every step I did is valid? I would be very happy if someone could verify.