$f_n \to f$ almost everywhere and $\int |f_n| \to \int |f|$ implies $\int |f_n - f| \to 0$?

Question

Suppose $f_n$ and $f$ are integrable, $f_n \to f$ almost everywhere, and $\int |f_n| \to \int |f|$. Does it necessarily follow that$$\int |f_n - f| \to 0?$$

Answer 1

Hint:

I can suppose , for ever $n=1,2,\cdots$ $$\|f_{n+1}-f_{n}\|_1\le 2^{-n} M$$ Set $$h_n=\sum_{k=1}^{n}|f_k-f_{k-1}|=|f_1|+\sum_{k=2}^{n}|f_k-f_{k-1}|$$ therefore $$\int_{\Omega}h_nd\mu=\|f_1\|_1+\sum_{k=2}^{n}\|f_k-f_{k-1}\|_1$$ $$\int_{\Omega}h_nd\mu=\|f_1\|+\sum_{k=1}^{n-1}\|f_{k+1}-f_{k}\|_1$$ $$\int_{\Omega}h_nd\mu\le\|f_1\|+M\sum_{k=1}^{n-1}2^{-k}<M+\|f_1\|_1$$ we can say there is Integrable function $h$such that ${{h}_{n}}\uparrow h$ and $$|f_n|=|\sum_{k=1}^{n}(f_k-f_{k-1})|\le \sum_{k=1}^{n}|f_k-f_{k-1}|=h_n\le h$$ Now you should show $f\le h$, then we have $$|f_n-f|\le|f_n|+|f|\le 2h$$ Indeed $$\int_{\Omega}(f_n-f)d\mu\to 0$$