Convergence in measure implies $L^1$-convergence assuming $||f_n||_{L^p}$ is unif. bounded Let $(X,\mu)$ be a finite measure space ($\mu(X)<\infty$) and $f_n, f$ measurable functions such that $f_n\to f$ in measure, plus $||f_n||_{L^p}<C$ for all $n$, for a fixed $p>1$, and for a fixed $C>0$. How can we conclude that $f_n\to f$ in $L^1$?
 A: Solution 1: (Thanks to saz in the comments, and Matthew Tyler). Assume au contraire, then after passing to a subsequence we have  $||f_n-f||_{L^1}>\epsilon\forall n$. Since $f_n\to f$ in measure, after passing to a (sub)subsequence we have $f_n\to f$ a.e. Then $|f_n|^p\to |f|^p$ a.e. so by Fatou's Lemma, 
$$ \int_X |f|^p \le \liminf_{n\to\infty} \int_X |f|^p\le C^p \hspace{.5cm}\Rightarrow \hspace{.5cm}||f||_{L^p}\le C,$$
so $f\in L^p$, and $||f_n-f||_{L^p}\le ||f_n||_{L^p}+||f||_{L^p}\le 2C$ are also uniformly bounded. Replacing the $f_n$ with $f_n-f$, we have $||f_n||_{L^1}>\epsilon\forall n$ but $f_n\to 0$ a.e. (and in measure too), plus $\{||f_n||_{L^p}\}$ is uniformly bounded, say by $D$. Then for any $\delta>0$ we have by Holder, where $\frac{1}{p}+\frac{1}{q}=1$:
$$ \epsilon \le \int_X |f_n| = \int_{\{|f_n|\le\delta\}}|f_n| + \int_{\{|f_n|>\delta\}}|f_n| \le \delta\mu(X) | ||f_n\chi_{\{|f_n|>\delta\}}||_{L^1(X)}\\ \le \delta\mu(X) + ||f_n||_{L^p(X)} ||\chi_{\{|f_n|>\delta\}}||_{L^q(X)} \le\delta\mu(X) + D\mu(\{|f_n|>\delta\})^{1/q}.$$
Convergence in measure implies $\mu(\{|f_n|>\delta\})\to 0$ as $n\to\infty$, and so the above inequality gives $\epsilon\le\delta\mu(X)$ for any $\delta>0$, absurd!

Solution 2: (by Jose27 in the comments). Notice $f\in L^p$ as above. Set $g=f_n-f, \alpha>0$ arbitrary and $\mu_a:=\mu(\{|g|>\alpha\})$. Jensen's Inequality gives 
$$ \bigg|\frac{1}{\mu_\alpha}\int_{\{|g|>\alpha\}} |g|\bigg|^p \le \frac{1}{\mu_\alpha}\int_{\{|g|>\alpha\}} |g|^p \hspace{.5cm}\Rightarrow\hspace{.5cm} \int_{|g|>\alpha} |g| \le \mu_\alpha^{1-1/p}\left(\int_{\{|g|>\alpha\}} |g|^p\right)^{1/p} =\mu_\alpha^{\frac{p-1}{p}}||g||_{L^p(\{|g|>\alpha\})}.$$
Notice fixing $\alpha$, $\mu_\alpha\to 0$ as $n\to\infty$, by convergence in measure. Hence,
$$||f_n-f||_{L^1} \le \int_{\{|g|\le\alpha\}} |g| + \int_{\{|g|>\alpha\}}|g| \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}||g||_{L^p(\{|g|>\alpha\})}\\ \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}(||f_n||_{L^p(X)} + ||f||_{L^p(X)}) \le \alpha\mu(X) +\mu_\alpha^{\frac{p-1}{p}}(C + ||f||_{L^p(X)}),$$
so letting $n\to\infty$ we see $\limsup_{n\to\infty} ||f_n-f||_{L^1(X)} \le \alpha\mu(X)$. Since this holds for any $\alpha>0$ we see the limsup is $0$, as desired.
