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let $X$ be a finite measure space and $\{f_n\}$ be a sequence of integrable functions, $f_n \rightarrow f\text{ a.e.}$ on $ X$. I want to show if (1) holds, then (2) holds too.

$\lim_{n \rightarrow \infty}\int_X |f_n| \, d\mu=\int_X |f| \, d\mu,\tag{1}$

$\lim_{n \rightarrow \infty}\int_X |f_n-f| \, d\mu=0.\tag{2}$

My attempt:

I have proven that (2) holds for nonnegative $f$. Then for the general case, I split the set to $E^+=\{x: f \geq 0\}$ and $E^-=\{x: f \leq 0\}$:

$\lim_{n \rightarrow \infty}\int_{E^+} f_n \, d\mu-\int_{E^+} f \, d\mu -\lim_{n \rightarrow \infty}\int_{E^-} f_n \, d\mu+\int_{E^-} f \, d\mu=0$

But I don't know how to proceed from here!

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    This is possibly a duplicate of http://math.stackexchange.com/q/51502 and http://math.stackexchange.com/q/2220392012-11-01

1 Answers 1

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As Lukas Geyer points out, the result isn't true unless $f$ is integrable. To see why, consider $f_n(x) = n$. Clearly, $f(x) = \infty$. Also: $ \int_X f_n \, d\mu = n \mu(X) $

Thus: $ \lim_{n \to \infty} \int_X f_n \, d\mu = \int_X f \, d\mu = \infty $

Yet: $ \lim_{n \to \infty} \int_X |f_n - f| \, d\mu = \infty \neq 0 $

Now, assuming $f$ is integrable, we have: $ \left||f_n - f| - |f_n|\right| \le |f| $

Hence, by the dominated convergence theorem: $ \lim_{n \to \infty} \int_X \left(|f_n - f| - |f_n|\right) \, d\mu = - \int_X |f| \, d\mu $

Rearrange to get the required result.

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    @PatrickDaSilva I wanted to show that the inequality came from the triangle inequality. What I wrote clearly implies what you wrote. I also didn't want to write all steps in detail. :) As for the finiteness of the space, yeah it's not required for the proof. I used it in the counterexample though.2012-11-01