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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?

  • 0
    Are there any assumptions on the measured space?2012-11-10
  • 0
    Here, we assumed that $E$ is of finite measure.2012-11-11
  • 1
    This asks both directions, so is not quite the same as [this question](http://math.stackexchange.com/q/51502).2012-11-12

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