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Let $w\in L^1(\mathbb{R}^d)$, where $w>0$. Let $\{f_n\}:\mathbb{R}^d\to\mathbb{R}$ be Lebesgue measurable functions such that $$\lim_{m,n\to\infty}\int_{|f_n-f_m|>t}w(x)\,dx=0$$ for any $t>0$.

Prove that $\{f_n\}$ has a subsequence that converges almost everywhere to a measurable function $g$.

Attempt:

I am trying to show that $\{f_n\}$ is Cauchy in measure, and thus converges in measure, and thus has a convergent subsequence $f_{n_k}\to g$ a.e.

We have $\lim_{m,n\to\infty}\int w(x)\chi_{\{|f_n-f_m|>t\}}(x)\,dx=0$.

Since $|w(x)\chi_{\{|f_n-f_m|>t\}}(x)|\leq|w(x)|\in L^1$ so by Lebesgue's Dominated Convergence Theorem, $$\int w(x)\lim_{m,n\to\infty}\chi_{\{|f_n-f_m|>t\}(x)}=0$$

This means $w(x)\lim_{m,n\to\infty}\chi_{\{|f_n-f_m|>t\}}(x)=0$ almost everywhere on $\mathbb{R}$.

Since $w>0$, so $\lim_{m,n\to\infty}\chi_{\{|f_n-f_m|>t\}}(x)=0$ a.e.

However, I think we cannot conclude $\lim_{m,n\to\infty}|\{|f_n-f_m|>t\}|=0$ which is exactly what we need (Cauchy in measure). So close yet so far..

Thanks for any help.

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yoyostein
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    Look at the measure $\mu \colon A \mapsto \int_A w(x),dx$. – Daniel Fischer Dec 30 '16 at 15:14
  • I see that $f_n$ is Cauchy in measure (under $\mu$), and $\mu$ is absolutely continuous with respect to Lebesgue measure. So $f_n$ converges in (Lebesgue) measure! Is this correct? – yoyostein Dec 30 '16 at 15:21
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    Not quite. That $\mu$ is absolutely continuous with respect to the Lebesgue measure is not sufficient. You also need the converse, that $\mu$-a.e. is $\lambda$-a.e. – Daniel Fischer Dec 30 '16 at 15:27
  • We know $\lim_{m,n\to\infty}\mu({|f_n-f_m|>t})=0$. How do we go from here to conclude that $\lim_{m,n\to\infty}\lambda({|f_n-f_m|>t})=0$, where $\lambda$ is Lebesgue measure?. I know if $\mu(A)=0$, then $\lambda(A)=0$, but having trouble making things rigorous. – yoyostein Dec 30 '16 at 16:07
  • Since technically, we wish to show that there exists $N$, such that if $m,n\geq N$, then $\lambda({|f_n-f_m|>t})<\epsilon$. – yoyostein Dec 30 '16 at 16:08
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    You can't show that without further assumptions. If you take $f_n = \chi_{\mathbb{R}^d\setminus B_n(0)}$, then $(f_n)$ satisfies the assumptions, but for $t < 1$, $\lambda({\lvert f_n - f_m\rvert > t}) = c\lvert n^d - m^d\rvert$ can be arbitrarily large. The sequence need not be Cauchy in measure with respect to the Lebesgue measure. – Daniel Fischer Dec 30 '16 at 16:19
  • Yes, you are right indeed. I am a bit confused now, how does the measure $\mu(A)=\int_A w(x),dx$ help in proving the statement? – yoyostein Dec 30 '16 at 16:24

1 Answers1

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The sequence need not be a Cauchy sequence in measure with respect to the Lebesgue measure. An example of that would be $f_n = \chi_{\mathbb{R}^d\setminus B_n(0)}$. However, the sequence is Cauchy in measure with respect to the measure $\mu$, where

$$\mu(A) := \int_A w(x)\,dx.$$

Then it follows that there is a subsequence that converges $\mu$-almost everywhere to a measurable function $g$.

Now we use that $\mu$-almost everywhere is the same as $\lambda$-almost everywhere, which is guaranteed by the strict positivity of $w$, to conclude.

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