Inspired by this question, suppose that we have a function $f\in\mathrm{BV}([0,1])\cap L^1([0,1])$, take a measurable partition $\mathcal P^N$ of $[0,1]=\bigcup_{i=1}^NI_i$, and obtain a new function $f^N$ obtained by averaging over the sets $I_i$: when $x\in I_i$, $$f^N(x)=\frac1{|I_i|}\int_{I_i}f(y)\ \mathrm dy.$$ As in the linked question, we have $f^N\in L^1[0,1]$, and if the mesh of the partition tends to $0$ when $N\to\infty$, we can argue that $f^N\to f$ a.e. as $N\to\infty$.
I was wondering whether it is possible in this case to prove that $$\|f^N-f\|_{L^1[0,1]}\leq\|f\|_{TV}\mathrm{mesh}(\mathcal P^N),$$ though the approach of the linked question didn't seem to work here directly. I hope someone can help.
EDIT: as Alex Ravsky's counterexample shows, we cannot just take $\mathcal P^N$ to be any measurable partition. Instead, let's assume that we have partitions of the form $0=t_0<t_1<\cdots<t_{n-1}<t_n=1$ of the unit interval in disjoint, consecutive intervals $[t_{k-1}, t_k)$. In this case, I'm still wondering if the result holds. A similar bound is not hard to prove if we assume that the size of the smallest and largest set in the $N$th partition do not differ by more than a factor $C$, independent of $N$. However, I'm particularly interested in not making that assumption.
Again, I hope the calculations below are correct, althoug that I did not check all details. I am going do this later.
Let $[f]=\{g:[0,1]:\to\mathbb R: g=f \mbox{ almost everywhere on} [0,1]\}$.
Let $g\in [f]$ be any function.
Then $g^N(x)=f^N(x)$ for any $x\in [0,1]$.
Moreover, if $x\in I_i$ then $\inf_{x'\in I_i} g(x')\le g^N(x)\le \sup_{x''\in I_i} g(x'')$.
It follows that $|g^N(x)-g(x)|\le \Delta_i$, where $\Delta_i=\sup_{x',x''\in I_i}|g(x'')-g(x')|$.
Next, $\sum_{i=1}^{N}\Delta_i\le\operatorname{var} g,$ where $$\operatorname{var} g=\sup\{\sum_{i=0}^{N-1} |g(x_{i+1})-g(x_i)|: 0=x_0 < x_1 < \ldots < x_N=1\}.$$
Then we obtain that $$\|f^N-f\|_{L[0,1]}=\int_{[0,1]}|f^N(x)-f(x)|\mathrm dx=$$ $$\int_{[0,1]}|g^N(x)-g(x)|\mathrm dx=\sum_{i=1}^{N}\int_{I_i}|g^N(x)-g(x)|\mathrm dx\le $$ $$\sum_{i=1}^{N}\int_{I_i}\Delta_i\mathrm dx=\sum_{i=1}^{N}|I_i|\Delta_i\le \sum_{i=1}^{N}\operatorname{mesh}(\mathcal P^N)\Delta_i\le$$ $$\operatorname{mesh}(\mathcal P^N)\operatorname{var}{g}.$$
Thus if $\|f\|_{TV}=\inf_{g\in [f]} \operatorname{var} g$ then $\|f^N-f\|_{L[0,1]}\le \operatorname{mesh}(\mathcal P^N) \|f\|_{TV}.$