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I'm trying to prove the exercise:

Let $\{f_n\}$ be a sequence of real-valued Lebesgue measurable functions on $\mathbb{R}$. Assume that $f_n\rightarrow f$ Lebesgue almost everywhere as $n\rightarrow\infty$, f is also real-valued Lebesgue measurable function, also, $||xf_n(x)||_{L^1(\mathbb{R})}\le1$ and $||f_n(x)||_{L^2((\mathbb{R}))}\le1$.

How can I prove $\{f_n\},f\in L^1(\mathbb{R})$ and that $$||f_n-f||_{L^1(\mathbb{R})}\rightarrow0,\quad as \;n\rightarrow\infty?$$

I think the proof might be very similar to proof of Dominated Convergence Theorem, but I don't know how to dominate $f_n$, and I feel two conditions about $xf_n$ and $f_n$ are a bit strange. Can you give me some hints?

Zijie Tang
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    Take. a look at Vitali's convergence theorem – Mittens Jan 06 '25 at 16:04
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    The conditions in your problem imply that given $\varepsilon>0$, (1) there is a set $E$ such that $\int{\mathbb{R}\setminus E}|f_n|<\varepsilon$ for all $n$, and (2) there is $\delta>0$ such that $\int_A|f_n|<\varepsilon$ for all $n$ whenever $m(A)<\delta$. You can then directly apply Vitali's convergence theorem (here is a proof of Vitali's convergence theorem in a general setting) – Mittens Jan 06 '25 at 16:36

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Hint: Use the condition that $||xf_n(x)||_{L^1({\mathbb R})} \leq 1$ to show that it suffices to prove the result on any bounded interval $[-N,N]$. Then on $[-N,N]$ use Egorov's theorem in combination with the fact that on such an interval the $L^1$ norm is bounded by a constant times $N^{{1/2}}$ times the $L^2$ norm. (The latter part is also related to the Vitali convergence theorem that Mittens mentioned in the comment.)

Zarrax
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  • OP did not show any effort, not even the simple fact that $f_n \in L^{1}$. So this question is missing context and it should be closed. Please read this: https://math.meta.stackexchange.com/questions/33508/enforcement-of-quality-standards?cb=1 – Kavi Rama Murthy Jan 07 '25 at 11:23
  • Most of the questions I answer get closed for some reason or another. If that's the type of site you want this to be, that's your right, but I'm not going to look for excuses to not give hints. – Zarrax Jan 07 '25 at 15:10