Recently,I met a problem while reading Literature https://arxiv.org/abs/1906.00496#:~:text=We%20establish%20continuity%20mapping%20properties%20of%20the%20non-centered,for%20if%20is%20radial%20and%20for%20for%20general., where the article seemed to provide an incorrect proof, and I tried to solve it on my own but ran into a bottleneck. The following description of the problem is my distilled version of the proof process from the article, which can be found in page 17325, the last step of the proof of Lemma 2.3 in the article, thank you!
Condition: Let $f\in L^1(\mathbb{R}^d)$ and $\{f_k\}_{k\in\mathbb{N}}\subset L^1(\mathbb{R}^d)$ be such that $\|f_k-f\|_{L^1(\mathbb{R}^d)}\to0$ as $k\to\infty$, define the sets $X_k=\{x\in\mathbb{R}^d:f_k>0\},Y=\{x\in\mathbb{R}^d:f<0\}.$
Question: Does $|X_k\cap Y|\to 0$ as $k\to\infty$?
No. In any dimension, let $f$ be any everywhere-negative $L^1$ function, and for each $k$ define $$ f_k(x)= \begin{cases} f(x) & \|x\|\leq k\\ -f(x) & \|x\|> k \end{cases} $$ Then $\|f_k-f\|_{L^1(\mathbb R^d)}\to 0$, but $Y=\mathbb R^d$, so $$|X_k\cap Y|=|X_k|=|\{x\in\mathbb R^d\mid\|x\|>k\}|=\infty$$ for all $k$.
Remark
One can prove the following weaker statement, namely, that given the condition, we have $$\int_{X_k\cap Y}g\to 0$$ for any $g\in L^1(\mathbb R^d)$, and that is enough to fix the proof in the linked paper, using $g=2|\nabla f|$.
To prove this, for each $\delta>0$ let $Y_\delta=\{x\in\mathbb R^d\mid f(x)<-\delta\}$. Fix $\epsilon>0$, and choose $\delta>0$ such that $$\int_{Y\backslash Y_{\delta}} |g|\leq\epsilon\text{.}$$ (Such a $\delta$ exists by the dominated convergence theorem, as the functions $|g|\chi_{Y\backslash Y\delta}$ converge pointwise to $0$ with $\delta$, and are dominated by $|g|$, so we may choose a sequence $\delta_i\to 0$ for which the integrals converge.)
Now we observe that $$|X_k\cap Y_\delta|\leq \frac{\|f_k-f\|_{L^1(\mathbb R^d)}}{\delta}\to 0\text{,}$$ so that for each $M\in\mathbb N$, $$\overline{\lim_{k\to\infty}}\int_{X_k\cap Y_\delta}|g|\leq \lim_{k\to\infty}M|X_k\cap Y_\delta|+\int_{\{x\in\mathbb R^d\mid g(x)>M\}}|g| =\int_{\{x\in\mathbb R^d\mid g(x)>M\}}|g|\text{,}$$ but as $M\to\infty$ the right hand side converges to $0$ (applying the dominated convergence theorem to $|g|\chi_{\{x\in\mathbb R^d\mid g(x)>M\}}$), so that $$\int_{X_k\cap Y_\delta}|g|\to 0\text{.}$$
We then have $$\overline{\lim_{k\to\infty}}\int_{X_k\cap Y}|g|\leq \lim_{k\to\infty}\int_{X_k\cap Y_\delta}|g| +\int_{Y\backslash Y_\delta}|g|\leq \epsilon\text{.}$$ Since $\epsilon$ was arbitrary this completes the proof.