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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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What are the functions $g_n$? – Patrick Da Silva Oct 31 '12 at 5:32
Your sets $E^+$ and $E^-$ depend on $n$. Is it intended? – Did Oct 31 '12 at 7:20
@did: sorry, it was a typo. – Anita Oct 31 '12 at 12:58
You should also assume that $f$ is integrable, otherwise the result won't be true. – Lukas Geyer Oct 31 '12 at 16:18
This is possibly a duplicate of and – robjohn Nov 1 '12 at 16:19

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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You should write $$ | |f_n - f| - |f_n| | \le |f| $$ to use the dominated convergence theorem on $|f_n - f| - |f_n|$(what you wrote needs to be corrected but it's true, no worries). – Patrick Da Silva Nov 1 '12 at 6:57
You don't seem to assume that the measure space has finite measure. Does that mean you got "the wrong proof" that the exercise assumed the OP to pull off?... I'm wondering. – Patrick Da Silva Nov 1 '12 at 6:59
@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. – Ayman Hourieh Nov 1 '12 at 9:24

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