Let $f \in L^1 (\mathbb R)$. Suppose $g\_n \in L^1 (\mathbb R)$ are a sequence of positive functions.


Define, for each $n$, the function $f\_n$ by


$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$


**Question:** Is it true that if $g\_n \to 0$ in weak $L^1$ norm, then $f\_n \to f$ in $L^1$ strong?


*Remark:* Implicit here is that one has to prove the $f\_n$ are all eventually in $L^1$.



Convergence of integral averages in $L^1$
The answer is negative even if $f$ is supported in $[0,1]$ and $g\_n \to 0$ uniformly in $[0,1]$. The reason is that $f \in L^1$ does not suffice for the Hardy-Littlewood maximal function to be in $L^1$, so choosing $g\_n$ to capture that maximal function will yield a counterexample. It is still useful to see an explicit counterexample. I will add that next.


For $x \in (0,1]$, let $$f(x)=\frac1{x \cdot \log^2(2/x)}\,,$$ with $f(x)=0$ for all $x \notin (0,1]$. Then $f$ is in $L^1[0,1]$.
Define $g\_n(x)= \min\{x,1/n\}$ for $x \in (0,1]$, with $g\_n(x)=\frac1{n(x^2+1)}$ for $x \notin [0,1]$. Then


$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy$$
satisfies
$$f\_n(x)=\frac{1}{2x\log(2/x)} \quad \text{for all} \quad x \in (0,1/n)\,,$$
so $f\_n \notin L^1[0,1]$.


**Remark** If we assume that $f\log\_+(f) \in L^1$ then the answer to the original question is positive, because in that case The Hardy-Littlewood maximal function is in $L^1$, so one can appeal to dominated convergence. See, e.g., Theorem 3.4 in [this paper](https://www.jstor.org/stable/2314249).


