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 Post subject: On measure theory
PostPosted: Mon Dec 07, 2015 1:39 pm 
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Let \(\displaystyle{f\,,g:\mathbb{R}\longrightarrow \mathbb{R}}\) be two continuous functions such that

\(\displaystyle{f=g}\) \(\displaystyle{\,\,\,\,\,\,\,\,\lambda}\) a.e (almost everywhere),

where \(\displaystyle{\lambda}\) stands for \(\displaystyle{\rm{Lebesque}}\) measure.

Prove that \(\displaystyle{f=g}\) .


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 Post subject: Re: On measure theory
PostPosted: Wed Jul 26, 2017 4:32 am 

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
I will change the notation slightly denoting $ m $ the Lebesgue measure .

Suppose $ f=g $ on $ \mathbb{R} \setminus E $ with $ m(E)=0 $ so it's safe to assume that $ E^{\mathrm{o}}= \varnothing $ because if $ E^{\mathrm{o}} \ne \varnothing $
Then we get $ m(E)>0 $

Since $ \overline{\mathbb{R} \setminus E} = \mathbb{R} \setminus E^{\mathrm{o}}= \mathbb{R} $

Hence $ f=g $ on a dense set and due to continuity the desired result follows .


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