# Can I use this trick for my proof?

I am given the following problem, with a hint that states that I should use the General Lebesgue Dominated Convergence Theorem.

Let $\{f_n\}$ be a sequence of integrable functions on $E$ for which $f_n \rightarrow f$ a.e. on $E$ and $f$ is integrable over $E$. Show that $\int_E \lvert f - f_n \rvert \rightarrow 0$ if and only if $\lim_{n \rightarrow \infty} \int_E \lvert f_n \rvert = \int_E \lvert f \rvert$.

($\Rightarrow$) Let $g_n = k_n \lvert f - f_n \rvert$ where $k_n \in \mathbb{R}$ and $\lvert f_n \rvert \leq g_n$ for all $n$. If we let $k = \text{sup}\{ k_n \}$, then $\lvert f_n \rvert \leq \frac{k}{k_n}g_n$

It seems like I should be able to do something like this using the Archimedean property and the fact that it does not change the value of the limit since $k \cdot lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert = k \cdot 0 = 0$.

I finish the forward direction of the proof by stating: since $g_n$ converges, and $f_n \rightarrow f$, it follows by the General Lebesgue Dominated Convergence Theorem that

$$\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$$

as desired.

• It's unclear what you're asking. Do you need help verifying your proof (@detnvvp helped you there), or are you seeking help with the other direction? – Jonathan Y. Oct 31 '13 at 23:36
• I needed help verifying that choosing $k$ in the way I did above was valid. – user1876508 Oct 31 '13 at 23:45
• I don't see a reason why there would be some $k$ such that $|f_n|\leq k|f-f_n|$ for all $n$, but as mentioned below, you don't need it for this direction. – Jonathan Y. Nov 1 '13 at 0:33
• Could we define a $k_n$ for each n, and have $k$ be the supremum of all such $k_n$? – user1876508 Nov 1 '13 at 2:23
• No, saying that no such $k$ necessarily exists is the same as saying the supremum might be $\infty$. – Jonathan Y. Nov 1 '13 at 8:43

Yes, but, how would you use the dominated convergence theorem there? You have to talk about the limit of the integrals of the absolute values of $f_n$, not $f_n$ themselves.
This implication is a consequence of the inequality $$\left|\int_E|f_n|-\int_E|f|\right|=\left|\int_E(|f_n|-|f|)\right|\leq\int_E||f_n|-|f||\leq\int_E|f_n-f|,$$ and the last term goes to $0$. This is also one way of getting this at the proof of dominated convergence theorem.
Edit: For the other direction, you can do the following: first, you have that $\left||f_n-f|-|f_n|\right|\leq|f_n-f-f_n|=|f|$ and $f$ is integrable. Since $f_n\to f$ almost everywhere, you have that $|f_n-f|-|f_n|\to -|f|$ almost everywhere. Therefore, from the dominated convergence theorem, $$\int_E(|f_n-f|-|f_n|)\to-\int_E|f|,$$ hence $$\int_E|f_n-f|=\int_E(|f_n-f|-|f_n|)+\int_E|f_n|\to-\int_E|f|+\int_E|f|=0.$$
• I thought all of the requirements were satisfied since $f_n \rightarrow f$, $g_n \rightarrow g$, $\lvert f_n \rvert \leq g_n$ , and $\int_E g_n \rightarrow \int_E g$. – user1876508 Oct 31 '13 at 23:51
• I see. If I have understood well, the requirements are satisfied, but you have to talk about the convergence $|f_n|\to |f|$ (not of $f_n\to f$), and the fact that $|f_n|$ is dominated. But, this is sort of an overkill; to do the first part you can do what I wrote above. – detnvvp Nov 1 '13 at 5:37
• Although, does $\lim_{n \rightarrow \infty} \int_E f_n = \int_E f$ imply that $\lim_{n \rightarrow \infty} \int_E \lvert f - f_n \rvert$ converge? – user1876508 Nov 1 '13 at 6:21