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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?

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  • $\begingroup$ Are there any assumptions on the measured space? $\endgroup$ Nov 10, 2012 at 10:57
  • $\begingroup$ Here, we assumed that $E$ is of finite measure. $\endgroup$ Nov 11, 2012 at 1:22
  • 1
    $\begingroup$ This asks both directions, so is not quite the same as this question. $\endgroup$
    – robjohn
    Nov 12, 2012 at 2:33

1 Answer 1

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

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  • $\begingroup$ What is a dominating function? $\endgroup$ Nov 10, 2012 at 14:30
  • $\begingroup$ For each $n$, $f_n$ is dominated by $|f_n|$. Am I right? $\endgroup$ Nov 10, 2012 at 14:49
  • $\begingroup$ But we need a bound independent on $n$. $\endgroup$ Nov 10, 2012 at 14:50
  • $\begingroup$ If I may so, one of the assumptions of the General Lebesgue Dominated Convergence states that for each $n\ge 1$, $|f_n|\le g_n$. $\endgroup$ Nov 10, 2012 at 14:57

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