On measure theory
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On measure theory
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 \(\displaystyle{f=g}\) .
\(\displaystyle{f=g}\) \(\displaystyle{\,\,\,\,\,\,\,\,\lambda}\) a.e (almost everywhere),
where \(\displaystyle{\lambda}\) stands for \(\displaystyle{\rm{Lebesque}}\) measure.
Prove that \(\displaystyle{f=g}\) .
Re: On measure theory
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 .
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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