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Let $\Omega\subset \mathbb{R}^n$ be a bounded domain and $H:=\{x\in\Omega:x_1=0\}$ a hyperplane in $\Omega$, and $\{f_k\}$ a sequence of absolutely integrable functions on $\Omega$ with $L^1$ norm less than $1$. Assume that $x_1\cdot f_k \to f_0$ in $L^1$ sense. as $k\to\infty$.

Question: Is $\frac{f_0}{x_1}$ absolutely integrable? If so, is the $L^1$ norm of it less than $1$? (Note that $1/x$ is not integrable near the origin in $\mathbb{R}^1$.)

Clues: Note that, $f_k \to f_0/{x_1}$ in $L^1$ sense as $k\to\infty$ does not hold. For example, take $\Omega=(-1,1)$, $f_k=k/2$ on $(-1/k,1/k)$ and $=0$ somewhere else. Then $||f_k||_{L^1}=1$, and $f_0=0$

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The answer is YES. Convergence in $L^{1}$ implies a.e. convergence of some subsequence. If $x_1f_{k_j} \to f_0$ a.e then Fatou's Lemma gives $\int |\frac {f_0} x| \leq \lim \inf \int |f_{k_j}| \leq 1$.

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