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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.

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    Thank you all for your help - I'll need a few minutes to think about all this2011-10-26

3 Answers 3

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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 $\mathbb R$.

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    I guess mainly I was unclear as to why denseness was needed to show $f=g$ everywhere, when all that I see needed is continuity ($f=g$ a.e. already)2011-10-27
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Consider $f-g$, which is zero almost everywhere and continuous. If $f-g$ is non-zero at some point $x$, then by continuity it is also non-zero in some open ball around $x$. Since this open ball has positive measure, it contradicts the assumption that $f-g$ is zero almost everywhere.

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If $f\neq g$ everywhere, then there exists $x_0$ such that $f(x_0)>g(x_0)$ (for example). By continuity of $f,g$ it follows that $f(x)>g(x)$ on a neighborhood $(x_0-\varepsilon,x_0+\varepsilon)$ of $x_0$. and that means that the set $[f\neq g]$ has measure at least $2\varepsilon>0$.