I expect that the following statement is true:
Let $p \in (0, \infty], f \in L^p(\mathbb{R}^n), \{f_j\} \subset L^1(\mathbb{R}^n)$ and $f_j \to f \in \mathcal{S}'(\mathbb{R}^n)$. In addition, if $\{f_j\}$ is $L^1$-bounded, then, $f \in L^1(\mathbb{R}^n)$.
I can neither prove nor disprove this claim. I would be grateful if anyone could give me some advice.
My English is not very good. Please forgive me if there are any grammatical errors or confusing parts.
Lets assume $p≥1$. $f_j\to f$ in $\mathcal S'$ means that $\int_{\Bbb R} f_j s\to\int_{\Bbb R} f s$ for all $s\in\mathcal S$, this condition can be weakened a bit, fix some $n$ and then you have:
$$\int_{[-n,n]}f_j(x) s(x)\,dx\to\int_{[-n,n]}f(x)s(x)\,dx$$ for all $s\in \mathcal S$. In particular the left-hand side is bounded above by $C\cdot \|s\|_\infty$ for $C:=\sup_{j}\|f_j\|_1$, so the right-hand side also is bounded above by this, ie:
$$\int_{[-n,n]}f(x) s(x)\,dx≤ C\,\|s\|_\infty$$
Three things:
Written out: $$\|f\lvert_{[-n,n]}\|_1 = \sup_{s\in C[-n,n], \|s\|_\infty ≤1} f[s] =\sup_{s\in \mathcal S, \|s\|_\infty ≤1}\ \int_{[-n,n]}f(x)s(x)\,dx≤ C$$ And then: $$\lim_n \int_{[-n,n]}|f(x)|\,dx ≤ C$$ giving that $f\in L^1$.