I'm dealing with this problem.
Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $\{f_n\}$ a sequence of nonnegative integrable functions. Suppose $f_n\xrightarrow{\mu} f_0$ and $\displaystyle\int_\Omega f_nd\mu\xrightarrow{n\rightarrow\infty}\int_\Omega f_0d\mu<\infty$. Prove that $\displaystyle\int_{\Omega}|f_n-f_0|d\mu\xrightarrow{n\rightarrow\infty} 0$.
If the convergence of $\{f_n\}$s to $f_0$ were almost everywhere, then the problem would become Scheffé's lemma.
By convergence in measure, for each $\epsilon>0$ and natural $n$, define : $\displaystyle E^\epsilon_n:=\{\omega\in\Omega\;;\;|f_n-f_0|(\omega)\geq\epsilon\} \;\wedge\;F_n^\epsilon:=\Omega-E^\epsilon_n$ $$\int_\Omega |f_n-f_0|d\mu=I_1+I_2\quad;\quad I_1=\int_{E^\epsilon_n} |f_n-f_0|d\mu\;\wedge\;I_2=\int_{F^\epsilon_n} |f_n-f_0|d\mu$$ We have $\mu(E_n^\epsilon)\rightarrow 0$ and $|f_n-f_0|<\epsilon$ over $F_n^\epsilon$. But I can't control the terms $|f_n-f_0|$ in $I_1$ and $\mu(F_n^\epsilon)$ in $I_2$ when the measure of space is not finite !
EDIT
My book has never mentioned "Scheffé's lemma".
Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$. Since $f_n \xrightarrow{\mu} f_0$, there exists a subsequence $\{f_{n_{k_j}}\}$ of $\{f_{n_k}\}$ such that $f_{n_{k_j}} \to f_0$ almost everywhere. Since $\int_\Omega f_{n_{k_j}}\, d\mu \to \int_\Omega f_0\, d\mu$, by Scheffé's lemma, $\int_\Omega |f_{n_{k_j}} - f_0|\, d\mu \to 0$. Since $\{f_{n_k}\}$ was arbitrary, $\int_\Omega |f_n - f_0|\, d\mu \to 0$.
Alternative: Since $f_n + |f_0| - |f_n - f_0| \ge 0$ and $\{f_n + |f_0| - |f_n - f_0|\}$ converges in measure to $2|f_0|$, Fatou's lemma gives
$$\int_\Omega 2|f_0|\, d\mu \le \varliminf \int_\Omega (f_n + |f_0| - |f_n - f_0|)\, d\mu = 2\int_\Omega |f_0|\, d\mu - \varlimsup \int_\Omega |f_n - f_0|\, d\mu.$$
Since $\int_\Omega |f_0|\, d\mu$ is finite (since $f_0 \ge 0$ a.e., so that $\int_\Omega |f_0|\, d\mu = \int_\Omega f_0 < \infty$), we deduce $\varlimsup \int_\Omega |f_n - f_0|\, d\mu \le 0$. Hence $\int_\Omega |f_n - f_0|\, d\mu \to 0$.