Bounded sequence in $\mathcal{L}^p$ implies $\mathcal{L}^p$ convergence if sequence converges a.e

Question

I'm working on $\mathcal{L}^p(E,\mathcal{A},\mu)$ space and I would like to prove the following:

Let $p\in [1,\infty[$ and $\mu(E)< \infty$, suppose also that

i) $f_n \to f, \mu$ a.e

ii) there is $r>p$, such that $\sup_n \int|f_n|^rd\mu < \infty $,

then $f_n \to f$ in $\mathcal{L}^p$.

Using Jensen's inequality, I know that assumption number ii) together with $\mu(E)<\infty$ implies that $||f_n||_p<M $ for $M\in \mathbb{R}^+$. I am tempted to say that since the sequence is bounded there must be a convergent subsequence and that by the completeness of the space it is in contained in $\mathcal{L}^p$. Is it correct? If so, where should I go from there? Some help would be appreciated

Answer 1

Taking $p=2$, $E=[0,1]$ with Lebesgue measure, $$f_n=n^{1/2}\chi_{[0,1/n]}$$ shows that this does not follow from knowing just that $||f_n||_p\le M$. The compactness you invoke doesn't exist.

Answer 2

You can estimate $\displaystyle \int_E |f_n - f|^p \, d\mu$ by letting $\epsilon > 0$ and writing $$ \int_E |f_n - f|^p \, d\mu = \int_{E \cap |f_n - f| < \epsilon} |f_n - f|^p \, d\mu + \int_{E \cap |f_n - f| \ge \epsilon} |f_n - f|^p \, d\mu.$$ The first integral satisfies the simple bound $$ \int_{E \cap |f_n - f| < \epsilon} |f_n - f|^p \, d\mu \le \epsilon^p \mu(E).$$ The second integral can be estimated using Holder's inequality: for $r > p$ you have $$\int_{E \cap |f_n - f| \ge \epsilon} |f_n - f|^p \, d\mu \le \mu(\{ E \cap |f_n - f| \ge \epsilon\})^{1-p/r} \left( \int_E |f_n - f|^r \, d\mu \right)^{p/r}.$$

Use $f_n \to f$ in two ways: first, since $\mu(E) < \infty$ you have $$ \lim_{n \to \infty} \mu(\{ E \cap |f_n - f| \ge \epsilon\}) = 0$$ and second, Fatou's lemma gives you $$ \sup_n \int_E |f_n - f|^r \, d\mu \le 2^r \sup_n \left(\int_E |f_n|^r \, d\mu + \int_E |f|^r \, d\mu \right) \le 2^{r+1}M$$ where $M = \sup \displaystyle \int_E |f_n|^r \, d\mu < \infty$.

You should be able to piece together the limit from here.