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I was wondering what is wrong in the following proof:

Proposition. Let $\{f_{n}\}$ be a sequence of integrable functions such that $f_{n}$ converges pointwise to a function $f$. Show that if:

$\lim \int |f_{n} - f| d\mu =0$ then $\int |f| d\mu = \lim \int |f_{n}| d\mu$.

Well I used the fact that $||f_{n}| - |f| | \leq |f_{n} - f|$ and integrating both sides and using the assumption that $\lim \int |f_{n} - f| d\mu =0$ I get the result. But I never used the fact that $f_{n}$ converges pointwise to $f$. Why do we need this assumption or what is wrong?


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$f_n - f$ could be $0$ except on a "moving" set of measure ${1 \over n}$ where it's $\sqrt{n}$. – Zarrax Nov 17 '10 at 15:50
You are right that you don't need the assumption of pointwise convergence to prove this statement. You do need it to prove the converse; perhaps that is what was meant – Qiaochu Yuan Nov 17 '10 at 16:28
@Qia.. Why is that? Not from Lebesgue dominated convergence theorem.. – TCL Nov 17 '10 at 18:03
@TCL: actually it follows from (a clever application of) Fatou's lemma. – Qiaochu Yuan Nov 17 '10 at 19:02

Nothing's wrong, your proof is fine.

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Zaricuse has already given the correct answer, but I want to add a few remarks.

Even if you don't assume pointwise convergence, the hypothesis $\lim \int |f_{n} - f| d\mu =0$ implies that $(f_n)$ converges to $f$ in measure, which implies that there is a subsequence that converges to $f$ pointwise almost everywhere. But as you have shown, this is not needed in the proof.

If you consider the normed space $L^1(\mu)$ of absolutely $\mu$ integrable functions with norm $\|f\|=\int|f|d\mu$, what you have shown is that the norm is continuous as a function from $L^1(\mu)$ with the norm topology to $[0,\infty)$. The same is true in any normed space.

As Qiaochu mentioned in a comment, pointwise convergence (almost everywhere) allows you to prove the converse, which follows from a generalized form of the dominated convergence theorem.

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@Jonas..By a generalized form, I guess you mean replacing the condition $|f_n(x)|\le g(x)$ in the Lebesgue dominated convergence theorem by $|f_n(x)|\le g_n(x)$, where $g_n$ satisfy $$\lim_n\int g_n =\int \lim_n\ g_n.$$ Or is there another better form ? – TCL Nov 17 '10 at 20:10
That is what I had in mind. The proof works just as for the ordinary dominated convergence, using Fatou's lemma as Qiaochu indicated above. – Jonas Meyer Nov 17 '10 at 21:02
In my comment above, $g_n$ are measurable and $$\int\lim_n g_n \lt \infty.$$ – TCL Nov 18 '10 at 18:04
@TCL: Thanks, that's what I meant, too. – Jonas Meyer Nov 18 '10 at 19:35

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