This result seems too convenient and I feel like Folland would have used this to prove proposition 2.30 if this was true.
My "proof":
Let $f_n$ be Cauchy in measure.
By definition of $\liminf f_n(x)$, for every $n \in \mathbb{N}, x \in X$ there exists an $m_{x,n}> n$ such that $|\liminf f_n(x) - f_m(x)| < \epsilon / 2$
Now consider the set {$x\in X : |f_n(x) -\liminf f_n(x)| > \epsilon$}
By triangle inequality, $\epsilon \leq |f_n(x) -\liminf f_n(x)| \leq |f_n(x) - f_{m_{x,n}}(x)| + |f_{m_{x,n}}(x) - \liminf f_n(x)|$
but since $|\liminf f_n(x) - f_{m_{x,n}}(x)| \leq \epsilon / 2$, we must have that $ \epsilon/2 \leq |f_n(x) -\liminf f_n(x)| $
therefore, the set
{$x\in X : |f_n(x) -\liminf f_n(x)| > \epsilon$} $\subset$ {$x\in X : |f_n(x) - f_{m_{x,n}}(x)| > \epsilon/2$}
but the measure of the set on the right approaches to zero as $n$ tends to infinity (since $m_{x,n}> n$)
therefore the measure of the set {$x\in X : |f_n(x) -\liminf f_n(x)| > \epsilon$} tends to infinity as $n$ tends to infinity
Is there anything wrong with this proof?
thank you