Suppose $f \in C([t_0,t_1])$ and $g:\mathbb{R} \to \mathbb{R}$ where
$$g(t) = \begin{cases}f(t), \,\,\, t_0 \leq t \leq t_1\\ 0, \quad\quad x \notin [t_0,t_1]\end{cases}$$
Can it be shown:
$$\int_{t_0}^{t_1} \left|\frac{1}{2 \delta}\int_{t-\delta}^{t+\delta}g(s) \, ds \right|\, dt \leq \int_{t_0}^{t_1}|g(t)|\, dt$$
This came up in a signal processing application. It is obvious that
$$\int_{t_0}^{t_1} \left|\frac{1}{2 \delta}\int_{t-\delta}^{t+\delta}g(s) \, ds \right|\, dt \leq \int_{t_0}^{t_1} \frac{1}{2 \delta}\int_{t-\delta}^{t+\delta}|g(s)| \, ds \, dt $$
but can't get any farther.
This is an example of a type of problem related to Lebesgue's differentiation theorem which shows up in a standard course on measure theory. Let $G(x)$ denote the anti-derivative of $|g(x)|$, that is $$G(x)=\int\limits_{-\infty}^x |g(t)| dt$$.
Then $G(x)=0$ for $x\leq t_0$ and $$G(x)=\int\limits_{-\infty}^{t_{1}} |g(t)| dt \hspace{2 mm}\text{ for } x\geq t_1$$. By triangle inequality,
$$\int_{t_0}^{t_1} \left|\frac{1}{2 \delta}\int_{t-\delta}^{t+\delta}g(s) \, ds \right|\, dt \leq \int_{t_0}^{t_1} \frac{1}{2 \delta}\int_{t-\delta}^{t+\delta}|g(s)| \, ds \, dt =\int_{t_0}^{t_1} \frac{G(t+\delta)-G(t-\delta)}{2\delta} \, dt $$
The final step is to use the form of $G(x)$ to simplify
$$\int_{t_0}^{t_1}\frac{G(t+\delta)-G(t-\delta)}{2\delta} \, dt $$