Cauchy in measure question 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.

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.
