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 Post subject: On measure theory Posted: Mon Dec 07, 2015 1:39 pm
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Joined: Mon Nov 09, 2015 1:52 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 $\displaystyle{f=g}$ .

Top   Post subject: Re: On measure theory Posted: 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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