# Proving a version of Dominated Convergence Theorem for $L^p$

First, please note the difference between this and the question asked here: $L^p$ Dominated Convergence Theorem

Let $$f_n$$ be a sequence of measurable functions on $$E$$ and $$p\ge 1$$. Suppose $$f_n\to f$$ in measure on $$E$$ and there exists an $$L^p$$-integrable function $$g$$ such that $$|f_n|\leq g$$. Prove that:

(1) $$f_n$$ and $$f$$ are $$L^p$$ integrable.

(2) $$f_n\to f$$ on $$E$$ in $$L^p$$.

Here is my proof:

I first prove that convergence in measure implies that there exists a subsequence $$f_{n_k}$$ converging to $$f$$ pointwise a.e.

$$\mu\{|f_n-f|\ge\frac{1}{k}\}\leq\frac{1}{2^k}$$. Define $$E_k=\{|f_{n_k}-f|>\frac{1}{k}\}$$ and $$H_m=\bigcup_{k=m}^{\infty}E_k$$.

It is clear that $$\mu(E_k)<\frac{1}{2^k}$$ and $$\mu(H_m)\leq\sum_{k=m}^{\infty}\frac{1}{2^k}=\frac{1}{2^{m-1}}$$.

Let $$Z=\bigcap_{m=1}^{\infty}H_m$$, then $$\mu(Z)\leq\mu(H_m)\leq\frac{1}{2^{m-1}}\to 0$$ as $$m\to\infty$$.

If $$x\notin Z$$, $$x\notin H_m$$, $$x\notin E_k$$ for all $$k\ge m$$, i.e. $$|f_{n_k}-f|\leq\frac{1}{k}$$ for all $$k\ge m$$. That means $$f_{n_k}\to f$$ for all $$x\notin Z$$ and $$\mu(Z)=0$$, thus pointwise convergence is proved.

To prove (1), since $$|f_n|\leq g$$ and $$p\ge 1$$,

$$\int_{E}|f_n|^pd\mu\leq\int_{E}g^pd\mu<\infty$$, hence $$f_n$$ is $$L^p$$-integrable.

By Fatou's lemma, $$0\leq\int_{E}f^pd\mu\leq\liminf_{n\to\infty}\int_{E}f_n^{p}d\mu<\infty$$, hence $$f$$ is $$L^p$$-integrable.

To prove (2), since pointwise convergence a.e. is proved, $$|f_n|\leq g$$ a.e. implies $$|f|\leq g$$. Since $$(f_n-f)^p\to 0$$ a.e., and $$|f_n-f|^p\leq 2^p(|f|^p+|f_n|^p)\leq 2^{p+1}|g|^p$$. So by Dominated Convergence Theorem for $$L^1$$, $$\int_{E}|f_n-f|^p\to 0$$.

Can anyone help me to check my proof? My question is whether the subsequence converges pointwise implies the sequence converges pointwise, and whether I can use DCT. I doubt this step.

Thanks!

• Can anyone help me verifying the proof?
– user758472
May 9 '20 at 8:29

You have shown that if $$(f_{n})$$ converges to $$f$$ in measure, then there exists some subsequence $$(f_{n_{k}})$$ that converges to $$f$$ in $$L^{p}$$. To complete the proof suppose the original sequence $$(f_{n})$$ does not converge to $$f$$ in $$L^{p}$$, then there exists some subsequence $$(f_{n_{k}})$$ and positive $$\epsilon$$, such that

$$\int{ \lvert f_{n_{k}}-f \rvert^{p} > \epsilon}\textit{, for all k}$$

Thus the subsequence has no subsequence that converges to $$f$$ in $$L^{p}.$$However, since the subsequence $$(f_{n_{k}})$$ also converges to $$f$$ in measure we have a contradiction.

• Thanks! So you proved that the subsequence converges pointwise a.e. implies the original sequence converges pointwise a.e.? And can you also help me check the rest of the proof is correct? And is it valid to apply the original DCT for $L^1$ here?
– user758472
May 10 '20 at 7:13
• I have not shown that the original sequence converges pointwise, only in $L^{p}$. Your proof of $(2)$ looks good. To prove integrability just note $\lvert f_{n} \rvert \leq g$. Also where you use Fatou and dominated convergence you should have the subsequence.
– Alex
May 10 '20 at 7:36
• So how to apply what you proved to my original proof? DCT requires $f_n\to f$ pointwise a.e.. Can you add more details? Thanks.
– user758472
May 10 '20 at 7:57
• It is not true in general that $(f_{n})$ converges to $f$ pointwise almost everywhere, since convergence in measure does not imply pointwise convergence a.e. What I have shown is that every subsequence of $(f_{n})$ has a subsequence that converges to $f$ in $L^{p}$. This is enough to show $f_{n} \to f$ in $L^{p}$.
– Alex
May 10 '20 at 8:30
• I checked Folland. There is a theorem saying that convergence in measure implies a subsequence converging a.e.. So I think I don't need pointwise convergence here. If I understand you correctly, to prove $f_n\to f$ in $L^p$, I need every subsequence $f_{n_k}\to f$ in $L^p$. Argue by contradiction that there exists such subsequence that does not converge in $L^p$. Since convergence in measure implies $f_{n_k}\to f$ a.e., so $\int|{f_{n_k}-f}|^p=0$, which is a contradiction. For the last step, we cannot apply DCT, right, since DCT requires $f_n\to f$ pointwise a.e.?
– user758472
May 10 '20 at 10:02