Let $f \in BV(\mathbb{R}) \cap L^1(\mathbb{R}),$ consider the piecewise constant function defined by \begin{eqnarray} f^{\delta}(x):= \frac{1}{\delta}\int\limits_{k\delta}^{(k+1)\delta}f(y)dy \quad \text{for } x\in [{k\delta},{(k+1)\delta}) \end{eqnarray} Then clearly $f^{\delta} \in L^{1}(\mathbb{R})$ and $f^{\delta} \rightarrow f$ pointwise a.e.in $\mathbb{R}.$
Now, my doubt is "Do we have $\|f^{\delta}-f\|_{L^1(\mathbb{R})}\leq C \delta$ for some $C$ independent of $\delta?$" If so how to prove it?
This is true.
We can first split up $\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}$ as follows: $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} = \int_{\mathbb{R}} |f^{\delta}(x) - f(x)| \mathrm{d} x = \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta} |f^{\delta}(x) - f(x)| \mathrm{d} x. $$ Partitioning $\mathbb{R}$ into those intervals enables us to rewrite $f^{\delta}$ using its definition. Thus, $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} = \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta} \bigg|\frac{1}{\delta}\int_{k\delta}^{(k + 1)\delta} f(y) \mathrm{d} y - f(x)\bigg| \mathrm{d} x. $$ Moving $f(x)$ inside the $y$-integral and using the triangle inequality gives us $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} \leq \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta} \frac{1}{\delta}\int_{k\delta}^{(k + 1)\delta} |f(y) - f(x)| \mathrm{d} y \mathrm{d} x. $$ The right-hand side only increases if we include all $y$ such that $|x - y| \leq \delta$, so we have $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} \leq \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta} \frac{1}{\delta}\int_{x - \delta}^{x + \delta} |f(y) - f(x)| \mathrm{d} y \mathrm{d} x. $$ Reassembling the outside integrals yields $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} \leq \int_{\mathbb{R}} \frac{1}{\delta}\int_{x - \delta}^{x + \delta} |f(y) - f(x)| \mathrm{d} y \mathrm{d} x. $$ Setting $y = x + h$ and changing the order of integration, we see that $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} \leq \frac{1}{\delta}\int_{-\delta}^{\delta} \int_{\mathbb{R}} |f(x + h) - f(x)| \mathrm{d} x \mathrm{d} h. $$ Analagously to Proposition 9.3 in Brezis' Functional Analysis, Sobolev Spaces, and Partial Differential Equations, we have that if $f \in BV(\mathbb{R})$, then $$ \|\tau_{h}f - f\|_{L^{1}(\mathbb{R})} \leq |h|\|\nabla f\|, $$ where $\tau_{h}f(x) = f(x + h)$ and $\|\nabla f\|$ denotes the total variation of $f$. Using this, $$ \|f^{\delta} - f\|_{L^{1}(\mathbb{R})} \leq \frac{1}{\delta}\int_{-\delta}^{\delta} |h| \|\nabla f\| \mathrm{d} h = \|\nabla f\|\delta. $$