# Proof that if $f_n\to f$ a.e., $g_n \to g$ a.e. and $g_n = f_n$ a.e. implies $f=g$ a.e.

So I am to prove that, for a measure space $(\Omega,\mathcal{A},\mu)$ and sequences of extended real-valued measurable functions $f_n,g_n$ from $(\Omega,\mathcal{A})$ to the extended real line, such that $$\label{1}\tag{1} \lim_{n\to\infty} f_n = f \text{ a.e.}, \quad \lim_{n\to\infty} g_n = g \text{ a.e.}, \quad \forall n \in \mathbb{N}: f_n = g_n \text{ a.e.},$$

we have $f=g$ a.e..

As an attempt, I let $$A := \{\omega\in\Omega : f(\omega) = g(\omega)\},$$

such that for $$A_n := \{\omega\in\Omega : g_n(\omega) = f_n(\omega)\},$$ $$B_1 := \{\omega\in\Omega : lim_{n\to\infty} f_n(\omega) = f(\omega)\},$$ $$B_2 := \{\omega\in\Omega : lim_{n\to\infty} g_n(\omega) = g(\omega)\},$$

we have $$\left(\bigcap_{n=1}^\infty A_n\right) \cap B_1 \cap B_2 \subset A,$$

i.e.

$$A^c \subset \left[\left(\bigcap_{n=1}^\infty A_n\right) \cap B_1 \cap B_2\right]^c = \left(\bigcup_{n=1}^\infty A_n^c\right) \cup B_1^c \cup B_2^c.$$

Now, by the subadditivity of measures, and by the assumptions \eqref{1}, $\mu(A^c) = 0$ since $$0 \leqslant \mu(A^c) \leqslant \sum_{n=1}^\infty \mu(A_n)^c + \mu(B_1^c) + \mu(B_2^c) = 0,$$

so it would follow that $f=g$ a.e.. However, since it is not given that $(\Omega,\mathcal{A},\mu)$ is a complete measure space, we do not know if $A^c\in\mathcal{A}$, so $\mu(A^c)$ may be undefined. Which approach would be appropriate? Thanks in advance!

• In short: even if $A^c$ isn't measurable, it will be contained in some null set which is measurable, and so we treat it as a null set. In fact null-ness can be defined without reference to full-fledged Lebesgue measure theory, as is done in Jordan measure theory. – Ian Jan 28 '18 at 15:47
• since $f_n\to f$ a.e and $g_n \to g$ a.e and $g_n, f_n$ are measurable functions it follows that the set $A=\{x\in X: f(x)=g(x)\}$ is measurable , so your proof can be complete! – dem0nakos Jan 28 '18 at 15:50
• Actually i am wrong the above statement that i told you works only if you have complete measure space . So i think Ian is right you have to go the complete measure space of $(\Omega,\mathcal{A},\mu)$ and find an $E \in \mathcal{A}$ such that $A\subset E$. – dem0nakos Jan 28 '18 at 16:01
• But if I were to follow this advice, then I could only prove that $f=g$ a.e. with respect to the completion $\bar\mu$ of $\mu$. Can I assume this is equivalent? The question is not specific about which measure it should concern, so I assume it should be simply $\mu$. – User2935032946 Jan 31 '18 at 8:29

$B_1^c$ and $B_2^c$ and each $A_n^c$ is a null set so $D=B_1^c\cup B_2^c\cup (\cup_nA_n^c)$ is a null set, and $f(x)=g(x)$ for every $x\in D^c.$
It may be that $f(x)=g(x)$ for some $x\in D$ but that is moot. We do not want to show that the complement of $S=\{x:f(x)=g(x)\}$ is null. If the measure is incomplete then $S^c$ may be a non-measurable set. What we need is a null set $D\supset S^c$.