TITLE: A Monotonicity property for Convergence in Measures
QUESTION [5 upvotes]: Let $\{ f_n\}_{n\in\mathbb{N}}$ and $\{ g_n\}_{n\in\mathbb{N}}$ are two sequences of real valued measurable functions on the measure space $(\mathscr{X},\mathcal{A},\mu)$ such that $f_n\overset{\mu}\longrightarrow f$ and $g_n\overset{\mu}\longrightarrow g$ and $f_n \ge g_n$ $\mu$-a.e. for each $n\in\mathbb{N}$, then I believe that we should also have $f\ge g$ $\mu$-a.e.
It seems intuitive to me that this result should hold true...but I've been trying to prove this statement to myself for quite a while and haven't gotten anywhere! 
Also I was thinking....if the above statement is true, if I change the condition to $f_n > g_n$ $\mu$-a.e. for each $n\in\mathbb{N}$ would it be correct to say that this DOES NOT guarantee $f> g$ $\mu$-a.e. as I have the feeling that the strict inequality is not preserved under a limiting argument?
Thanks =)

REPLY [4 votes]: The accepted answer is slightly incorrect, since there might not exist an $\epsilon$ such that $\{\omega \in \mathcal{X}: g(\omega)-f(\omega) > 0\} = \{\omega \in \mathcal{X}: g(\omega)-f(\omega) > \epsilon\}$. For example, let $\mathcal{X}=\mathbb{N}$, $f \equiv 0$ and $g(n)=n^{-1}$. However, the argument can be fixed.
Let $A_{m}=\{g-f > m^{-1}\}$. 
If, for every $m$ $\mu(A_m)=0$, then
\begin{align*}
 \mu(\{g -f > 0\})
 &= \mu(\cup_{m=1}^{\infty}\{g-f > m^{-1}\}) \\
 &= \mu(\cup_{m=1}^{\infty}A_m) = 0
\end{align*}
The previous answer shows that, for every $m$, $\mu(A_m)=0$.