Tuesday, March 12, 2019

integration - If $mu(|f_n|^p)$ is bounded and $f_nto f$ in measure then $f_nto f$ in $L^1$




Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of real measurable functions s.t.,




(a) The sequence $\displaystyle(\int |f_n|^p\ \mathsf d\mu)_{n\in\Bbb{N}}$ is bounded.



(b) The sequence $(f_n)_{n\in\Bbb{N}}$ converges in measure to a measurable function $f$.



Use Hölder's inequality to show that $$\displaystyle\lim\limits_{n\to\infty}\int|f_n-f|\ \mathsf d\mu=0.$$




So far I've shown that $f$ is $p$-integrable and that, for all $n\in\Bbb{N}$ and for all real number $\alpha>0$



$$\displaystyle \int |f_n-f|\ \mathsf d\mu\leq \alpha\mu(\Omega)+\int_{ \{|f_n-f|>\alpha\}}|f_n-f|\ \mathsf d\mu$$




I'm trying to use Hölder on the last term of above inequality to show that this part goes to $0$. But I don't know how to do it.



A hint would be more appreciated then the whole answer.


Answer



By Holder's inequality,
$$\int_{|f_n-f|>\alpha} |f_n - f| d\mu \leq \int_{|f_n-f|>\alpha} |f_n| d\mu + \int_{|f_n-f|>\alpha} |f| d \mu \leq$$
$$ \mu(\{\omega\in\Omega:|f_n(\omega)-f(\omega)|>\alpha\})^{1/q} \left(||f_n||_p+||f||_p\right)$$



As $n\to\infty$, $\mu(\{\omega\in\Omega:|f_n(\omega)-f(\omega)|>\alpha\})\to 0$, and $\left(||f_n||_p+||f||_p\right)<\infty$.




Then $$\mu(\{\omega\in\Omega:|f_n(\omega)-f(\omega)|>\alpha\})^{1/q} \left(||f_n||_p+||f||_p\right)\xrightarrow{n\to\infty}0$$


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