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Let $f_n \; \colon (0,1) \to \mathbb{R} $ a sequence of Lebesgue integrable functions which converges almost everywhere in $(0,1)$ to $0$. Prove or Disprove: if there exists $p \in (1,+ \infty)$ such that $(f_n)$ is bounded in $L^p(0,1)$,then $\lim_{ n \to \infty} \int_0^1|f_n(x)| \, dx =0$.

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I think the property is true and Egorov's theorem will help here. Fix $\varepsilon >0$. Since $f_n \to 0$ a.e., there is a set $E \subset (0,1)$ such that $\lvert E\rvert < \varepsilon$ and $f_n \to 0$ uniformly on $(0,1)\setminus E$. Now fix $\delta > 0$ and $N$ large enough that $\lvert f_n \rvert < \delta$ on $(0,1)\setminus E$ whenever $n \ge N$. Then for $n \ge N$, \begin{align*} \int^1_0 \lvert f_n(x)\rvert dx &= \int_{(0,1)\setminus E} \lvert f_n(x)\rvert dx + \int_E \lvert f_n(x)\rvert dx \\ &\le \delta \int_{(0,1)\setminus E}dx + \lvert E \rvert^{1/q}\|f_n\|_{L^p(0,1)}\\ &\le \delta + \varepsilon^{1/q}M \end{align*} where $M$ is the bound on $f_n$ in $L^p(0,1)$, and $q$ is the dual exponent to $p$. Since $\varepsilon$ and $\delta$ are arbitrary, this shows that $\int^1_0 \lvert f_n(x)\rvert dx \to 0.$

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For future readers: this answer is a failed attempt at a counterargument. I’m leaving it here instead of deleting it since I think my faults here can be instructive. Read the comments below for more detail.

Here’s a modification to what I posted earlier: let $f_n:(0,1)\to\mathbb{R}$ be defined by

$$ f_n(x) =\begin{cases} n\ln(n), & \text{if} \;\; x\in\left(0,\frac{1}{n}\right)\\ 0, &\text{otherwise} \end{cases}. $$

We see that the $f_n\to0$ point wise, however we find that

$$ \int_0^1 |f_n|dx = \frac{1}{n}\cdot n\ln(n) = \ln(n). $$

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  • $\begingroup$ hmmm, interesting. I posted what I think is a valid proof, so either this sequence isn't bounded in $L^p$ for any $p > 1$, or I made a mistake. EDIT: This function sequence is not bounded in $L^p$ since $\|f_n\|^p_{L^p} = n^{p-1}(\ln(n))^p$ $\endgroup$
    – user307205
    Apr 11, 2020 at 22:17
  • $\begingroup$ @User8128 I believe these functions are $L^p$ bounded because they are essentially step functions. I believe we can show that $||f_n||_p= n^{\frac{p-1}{p}}\ln(n)$. $\endgroup$
    – Mnifldz
    Apr 11, 2020 at 22:21
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    $\begingroup$ @Mnifldz Each function is in $L^p$, but the sequence is not bounded in $L^p$ since $\|f_n\|_{L^p} \to \infty$ $\endgroup$
    – user307205
    Apr 11, 2020 at 22:21
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    $\begingroup$ @GiuseppeTenaglia The question seems to be this: assume that $f \to 0$ a.e and $(f_n)$ is bounded in $L^p$ for a fixed $p \in (1,\infty)$. Prove or disprove that $\|f\|_{L^1} \to 0$. If that is indeed the question, then my above answer proves that the property is true. $\endgroup$
    – user307205
    Apr 11, 2020 at 22:26
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    $\begingroup$ @Mnifldz no problem! Glad we could resolve this $\endgroup$
    – user307205
    Apr 11, 2020 at 22:26

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