For an $L^p$ function $f:\mathbb R \to \mathbb R$ and a positive number $\delta$, we define a new function $A_\delta f$ by the step function $$ A_\delta f = \sum_{n\in\mathbb Z} \textbf{avg }^\delta_n(f) \ \chi_{I^\delta_n} \ , $$ where $I^\delta_n$ is the interval $(\ (n-1)\delta,\, n\delta\ ]$ and $ \textbf{avg }^\delta_n(f)$ is the mean value $$ \textbf{avg }^\delta_n (f) = \frac 1 \delta \int_{I^\delta_n} f. $$
Question. How to prove that $$ \lim_{\delta \to 0} \| A_\delta f - f\|_p =0 \ ?$$ (The norm is of course the $L^p$-norm.)
What I have / haven't done. I already have proved that
\begin{align} \text{the answer is YES if $f$ is of compact support. } \end{align} And what I have to do is to extend this result to the whole $L^p(\mathbb R)$ by the denseness of $C_c(\mathbb R)$ in $L^p(\mathbb R)$.
To do this I used the Minkowski inequality to get $$ \|A_\delta f - f \|_p \le \|A_\delta (f - f_k)\|_p + \| A_\delta f_k -f_k \|_p + \|f-f_k\|_p \ , $$ where $f_k$ is a sequence in $C_c(\mathbb R)$ which converges to $f$ in $L^p(\mathbb R)$. Then the second and third terms in the right-hand side become small if $\delta$ is small and $k$ large.
But I don't see why the first term $\|A_\delta (f-f_k)\|_p$ becomes small when $\delta \to 0$ and $k \to \infty$. Is this true? If so, then how can I prove it?
Thanks for any help. The answer may or may not follow my approach. You may even restrict $\delta$ to have the sequential values; that is, you may assume $\delta_n = 1/n$ or $\delta_n = 1/2^n$, etc, if that makes the thing easier.
The disjointness of $I_{n}^{\delta}$ yields that \begin{align*} \|A_{\delta}(f-f_{k})\|_{p}^{p}&=\sum_{n\in\mathbb{Z}}|\text{avg}_{n}^{\delta}(f-f_{k})|^{p}|I_{n}^{\delta}|\\ &=\delta\sum_{n\in\mathbb{Z}}|\text{avg}_{n}^{\delta}(f-f_{k})|^{p}. \end{align*} While, \begin{align*} \sum_{n\in\mathbb{Z}}\left(\dfrac{1}{\delta}\int_{I_{n}^{\delta}}|f-f_{k}|\right)^{p}\leq\sum_{n\in\mathbb{Z}}\dfrac{1}{\delta}\int_{I_{n}^{\delta}}|f-f_{k}|^{p}=\dfrac{1}{\delta}\|f-f_{k}\|_{p}^{p}, \end{align*} so $\|A_{\delta}(f-f_{k})\|_{p}^{p}\leq\|f-f_{k}\|_{p}^{p}$.