Let ${f_n}$ be a sequence of integrable functions on E for which $f_n \to f$ a.e. on E and f is integrable over E. Show that $\int_E |f-f_n| \to 0$ if and only if $\lim_{ n\to\infty} \int_E |f_n| = \int_E |f|$.
My Answer:
Let ${f_n}$ be a sequence of integrable functions such that $f_n \to f$ on E and f is integrable over E.
Let $\int_E |f_n – f| \to 0$, then $\left|\int_E |f_n| – \int_E |f| \right|$ $\leq$ $\left|\int_E|f_n| – |f|\right|$ $\leq$ $\int_E |f_n – f| \to 0$.
Does this imply what we are trying to prove? Or is more necessary? If so, then:
Conversely, suppose $\lim_{n\to\infty} \int_E |f_n| = \int_E |f|$. Can we use the General Lebesgue Dominated Convergence Theorem to show this?
Best Answer
Note that
$$0\le\left|\int_{E}|f_n|-\int_{E}|f|\right|= \left|\int_{E}(|f_n|-|f|)\right| \le \int_{E}|f_n-f|.$$
From this inequality, it follows that
$$\int_{E}|f_n-f|\to 0 \quad \Rightarrow \int_{E}|f_n|\to \int_{E}|f|.$$
Conversely, assume that $\int_{E}|f_n|\to \int_{E}|f|.$ It can be shown that the assumptions of the General Lebesgue Dominated Convergence Theorem implies that $\int_{E}|f_n-f|\to 0.$ Replacing $g_n$ by $|f_n|$ and $g$ by $|f|$, all of the assumptions of the General Lebesgue Dominated Convergence Theorem are satisfied. Hence $\int_{E}|f_n-f|\to 0.$