# Lebesgue Convergence using The General Lebesgue Dominated Convergence Theorem

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 $\lim_{ n\to\infty} \int_E |f_n| = \int_E |f|$.

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?

• Are there any assumptions on the measured space? – Davide Giraudo Nov 10 '12 at 10:57
• Here, we assumed that $E$ is of finite measure. – Juniven Nov 11 '12 at 1:22
• This asks both directions, so is not quite the same as this question. – robjohn Nov 12 '12 at 2:33

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 $\int_{E}|f_n-f|\to 0.$

• What is a dominating function? – Davide Giraudo Nov 10 '12 at 14:30
• For each $n$, $f_n$ is dominated by $|f_n|$. Am I right? – Juniven Nov 10 '12 at 14:49
• But we need a bound independent on $n$. – Davide Giraudo Nov 10 '12 at 14:50
• If I may so, one of the assumptions of the General Lebesgue Dominated Convergence states that for each $n\ge 1$, $|f_n|\le g_n$. – Juniven Nov 10 '12 at 14:57