I am solving an exercise of real analysis:
Given that $f_n $ and $f$ is lebesgue integrable on $E$,$f_n\to f$ a.e. when $n\to\infty$.
If $$\lim_{n\to\infty}\int_Ef_n(x)\mathrm{d}x=\int_Ef(x)\mathrm{d}x, \lim_{n\to\infty}\int_E|f_n(x)|\mathrm{d}x=\int_E|f(x)|\mathrm{d}x$$ then prove that $$\lim_{n\to\infty}\int_E|f_n(x)-f(x)|\mathrm{d}x=0.$$ My approach is as follows：
Since$|f_n(x)-f(x)|\leq|f_n(x)|+|f(x)|$, and $$\lim_{n\to\infty}\int_E|f_n(x)|+|f(x)|\mathrm{d}x=2\int_E|f(x)|\mathrm{d}x,$$ by the generalized dominated convergence theorem, $$\lim_{n\to\infty}\int_E|f_n(x)-f(x)|\mathrm{d}x=\int_E\lim_{n\to\infty}|f_n(x)-f(x)|\mathrm{d}x=\int_E|\lim_{n\to\infty}f_n(x)-f(x)|\mathrm{d}x=0.$$ I thought the answer is just like this, until I found the answer of this question is so long that I think my solution may has some wrong.
So, where am I wrong? Is this exercise not really obvious?
update:
Knowing that Scheffé's lemma is much stronger than this exercise, I confirm that my proof is correct and this stupid exercise is not really good for understand the property of $L^1$ convergence.