I am supposed to show that if $f$ and $g$ are continuous, real-valued functions on $\mathbb{R}$, then if $f=g\;\;$, $\lambda$-a.e., then $f=g$ everywhere.
So I have been reading and I think that this question is saying that these two functions are mappings like, $$f,g:=(\mathbb{R},\mathcal{L}) \stackrel{\lambda^{\ast}}{\to} (\mathbb{R},\mathcal{B}(\mathbb{R})).$$
I started my argument by saying:
If $f,g$ are $\lambda$-a.e. on $\mathbb{R}$ then $\exists$ a set $E:=\{x \in \mathbb{R} : f \neq g\}$. But the claim incorporates that $f=g$ everywhere and so there can't exist a set on $\mathbb{R}$ that has Lebesgue outer measure zero, i.e. all sets on $\mathbb{R}$ must have positive Lebesgue measure….
I think the reason there is no set of zero Lebesgue measure is because the preimage of all similar type sets in the Borel algebra, $\mathcal{B}(\mathbb{R})$, is included in a similar set in the algebra of Lebesgue measurable sets $\mathcal{L}$. But, aren't there Lebesgue measurable sets of zero Lebesgue measure – the countable $\mathbb{N}$ for instance? Or does it not count because the preimage is coming from a Borel algebra, even though a Borel set has measure zero too?
Thanks for any help in clearing this up for me.
Best Answer
If $f$ and $g$ are a.e. equal, show that the set $\{x\in\mathbb R:f(x)= g(x)\}$ is dense. The using the fact that $f$ and $g$ are continuous, show that $f=g$ on all of $\mathbb R$.
Altenratively, show that if $f\neq g$, then $f$ and $g$ differ on a whole open interval, so that the set $\{x\in\mathbb R:f(x)= g(x)\}$ is not almost $\mathbb R$.