A Monotonicity property for Convergence in Measures

Question

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 =)

Answer 1

First, note that $g - f>0$ if and only if $g - f > 1/m$ for some $m \in \mathbb{N}$. Hence, $$\mu(g-f>0) = \mu(\cup_{m \in \mathbb{N}} \{g-f> 1/m\}).$$

It remains to show that $\mu(g-f>1/m) = 0$ for all $m \in \mathbb{N}$.

To that end, observe the following chain of implications. If $g - f > 1/m$, then $(g - g_n) + (f_n - f) > 1/m$ because $f_n \geq g_n$. Then, $|g - g_n| + |f_n -f| > 1/m$. Then, either $|g - g_n| > 1/2m$ or $|f_n - f| > 1/2m$. From these implications we now have $$\mu(g-f>1/m) \leq \mu(|g - g_n| > 1/2m) + \mu(|f_n - f| > 1/2m)$$ for all $n$ and $m$. Letting $n \to \infty$ on the right-hand side we conclude that $$\mu(g-f>1/m) = 0$$ for all $m \in \mathbb{N}$, and the proof is complete.

Regarding the second question, yes, your intuition is correct. To show this, simply reduce the problem to the elementary notion of a limit by considering a trivial measure space, e.g. $\Omega = \{\omega\}$. Let $f_n(\omega) = 1/2^n$, $g_n(\omega) = 1/3^n$.

Answer 2

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 $m$, $\mu(A_m)=0$.