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In this question (functions $f=g$ $\lambda$-a.e. for continuous real-valued functions are then $f=g$ everywhere) it is stated that if $ f,g : \mathbb{R} \to \mathbb{R} $ are continuous and $ f = g $ a.e. equal then in fact $ f(x) = g(x) $ on all of $ \mathbb{R} $.

What if $ f,g : G \to \mathbb{C} $ (continuous maps) where $ G $ is a locally compact group? Does it still hold that $ f = g $ a.e. equal implies $ f(x) = g(x) $ for all $ x\in G $?

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If you are talking about the Haar-measure on $G$, this is true.

For if there was a nonempty open set $\emptyset \neq U \subset G$ with $\mu(U) = 0$, we can assume (by translation) w.l.o.g. that $e \in U$ (the identity of $G$).

Then for every compact $K \subset G$, we could cover $K$ by finitely(!) many of the translates $xU$ for $x \in K$ (here, we use that $K$ is compact).

This implies $\mu(K) = 0$ for every(!) compact $K \subset G$.

By inner regularity of the Haar measure, i.e. by

$$ \mu(V) = \sup\{\mu(K) \mid K \subset V \text{ compact} \}, $$

we conclude $\mu(V) = 0$ for all open subsets $V \subset G$. By outer regularity, we get $\mu \equiv 0$, a contradiction.

This shows that every nonempty open set has positive measure.

By continuity of $f,g$ we know that

$$ U := \{ x \in G \mid f(x) \neq g(x) \} $$

is open. Thus it is either empty (i.e. $f \equiv g$), or has positive measure (i.e. NOT $f=g$ a.e.).

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  • $\begingroup$ Thank you, I didn't expect a proof to take this route at all, but it makes sense when you explain it. Was wondering for a moment how you could know that $ U $ must be open, but this is of course due to the uniqueness of limits. :) $\endgroup$
    – zo0x
    Commented Jun 11, 2014 at 10:56
  • $\begingroup$ You could also use that $U = (f-g)^{-1}(\Bbb{C}^\ast)$ is open as the inverse image of an open set under the continuous function $f-g$. $\endgroup$
    – PhoemueX
    Commented Jun 11, 2014 at 11:44
  • $\begingroup$ That makes sense. I assume $ \mathbb{C}^\ast = \mathbb{C}\backslash\{0\} $? Is this a standard notation? $\endgroup$
    – zo0x
    Commented Jun 11, 2014 at 12:37
  • $\begingroup$ Yes, it is standard notation for the group of units of a ring (or a field in this case). See en.wikipedia.org/wiki/Unit_%28ring_theory%29 $\endgroup$
    – PhoemueX
    Commented Jun 11, 2014 at 12:56

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