5 added 677 characters in body edited Oct 27 '17 at 8:35 Davide Giraudo 137k1717 gold badges174174 silver badges292292 bronze badges Two cases: There is a set $$N$$ contained in a measurable set of zero measure which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. There is a set $$N$$ contained in a measurable set of zero measure which is not measurable. In this case, take $$g:=f+\mathbf 1_{N}$$, where $$\mathbf 1_N$$ is the indicator function of $$N$$. It's not a measurable function, because otherwise so would be $$\mathbf 1_{N}$$. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ hence $$g\chi_{f=g}$$ is $$\mathcal A$$-measurable. Since $$g\chi_{f\neq g}$$ is a pointwise limit of simple function of $$\mathcal A$$ measurable set, $$g$$ is $$\mathcal A$$ measurable. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, there is no such function $$g$$. Indeed, assume that $$f$$ is a measurable function and that $$g\colon\Omega \to\mathbb R$$ is such that $$f(x)=g(x)$$ for almost every $$x$$. We shall show that $$g$$ is $$\mathcal A$$-measurable. The set $$D=\left\{x\in\Omega\mid f(x)\neq g(x) \right\}$$ is contained in a set of measure zero hence is measurable. Since $$g(x)=f(x)\mathbf 1_{\Omega\setminus D}(x) +g(x)\mathbf 1_D (x),$$ it suffices show that the function $$h\colon x\mapsto g(x)\mathbf 1_D (x)$$ is $$\mathcal A$$ measurable. To this aim, let $$A_t :=\left\{x\in\Omega\mid h(x)\lt t \right\}$$. If $$t\leqslant 0$$, then $$A_t\subset D$$ hence $$A_t$$ is contained in a set of measure zero and is $$\mathcal A$$-measurable. If $$t\gt 0$$ then $$A_t=\left(\Omega\setminus D\right) \cup\left(D\cap \left\{x\in\Omega\mid g(x)\lt t \right\}\right)$$, which is the union of two measurable sets (the second one because it is contained in $$D$$ hence contained in a set of zero measure). Two cases: There is a set $$N$$ contained in a measurable set of zero measure which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ hence $$g\chi_{f=g}$$ is $$\mathcal A$$-measurable. Since $$g\chi_{f\neq g}$$ is a pointwise limit of simple function of $$\mathcal A$$ measurable set, $$g$$ is $$\mathcal A$$ measurable. Two cases: There is a set $$N$$ contained in a measurable set of zero measure which is not measurable. In this case, take $$g:=f+\mathbf 1_{N}$$, where $$\mathbf 1_N$$ is the indicator function of $$N$$. It's not a measurable function, because otherwise so would be $$\mathbf 1_{N}$$. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, there is no such function $$g$$. Indeed, assume that $$f$$ is a measurable function and that $$g\colon\Omega \to\mathbb R$$ is such that $$f(x)=g(x)$$ for almost every $$x$$. We shall show that $$g$$ is $$\mathcal A$$-measurable. The set $$D=\left\{x\in\Omega\mid f(x)\neq g(x) \right\}$$ is contained in a set of measure zero hence is measurable. Since $$g(x)=f(x)\mathbf 1_{\Omega\setminus D}(x) +g(x)\mathbf 1_D (x),$$ it suffices show that the function $$h\colon x\mapsto g(x)\mathbf 1_D (x)$$ is $$\mathcal A$$ measurable. To this aim, let $$A_t :=\left\{x\in\Omega\mid h(x)\lt t \right\}$$. If $$t\leqslant 0$$, then $$A_t\subset D$$ hence $$A_t$$ is contained in a set of measure zero and is $$\mathcal A$$-measurable. If $$t\gt 0$$ then $$A_t=\left(\Omega\setminus D\right) \cup\left(D\cap \left\{x\in\Omega\mid g(x)\lt t \right\}\right)$$, which is the union of two measurable sets (the second one because it is contained in $$D$$ hence contained in a set of zero measure). 4 edited body edited Mar 28 '13 at 16:00 Davide Giraudo 137k1717 gold badges174174 silver badges292292 bronze badges Two cases: There is a set $$N$$ contained in a measurable set of zero measure $$N$$ which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ andhence $$g\chi_{f=g}$$ is $$\mathcal A$$-measurable. Since $$g\chi_{f\neq g}$$ shouldn't beis a pointwise limit of simple function of $$\mathcal A$$ measurable. It can't be true approximating set, $$g$$ by simple functions.is $$\mathcal A$$ measurable. Two cases: There is a set of zero measure $$N$$ which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ and $$g\chi_{f\neq g}$$ shouldn't be $$\mathcal A$$ measurable. It can't be true approximating $$g$$ by simple functions. Two cases: There is a set $$N$$ contained in a measurable set of zero measure which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set contained in a measurable set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ hence $$g\chi_{f=g}$$ is $$\mathcal A$$-measurable. Since $$g\chi_{f\neq g}$$ is a pointwise limit of simple function of $$\mathcal A$$ measurable set, $$g$$ is $$\mathcal A$$ measurable. 3 edited body edited Mar 28 '13 at 15:49 Davide Giraudo 137k1717 gold badges174174 silver badges292292 bronze badges Two cases: There is a set of zero measure $$N$$ which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$f=f\chi_{f=g}+g\chi_{f\neq g}$$$$g=f\chi_{f=g}+g\chi_{f\neq g}$$ and $$g\chi_{f\neq g}$$ shouldn't be $$\mathcal A$$ measurable. It can't be true approximating $$g$$ by simple functions. Two cases: There is a set of zero measure $$N$$ which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$f=f\chi_{f=g}+g\chi_{f\neq g}$$ and $$g\chi_{f\neq g}$$ shouldn't be $$\mathcal A$$ measurable. It can't be true approximating $$g$$ by simple functions. Two cases: There is a set of zero measure $$N$$ which is not measurable. In this case, take $$g:=f+\chi_{N}$$, where $$\chi_N$$ is the characteristic function of $$N$$. It's not a measurable function, because otherwise so would be $$\chi_{N}$$. Each set of zero measure is an element of $$\mathcal A$$ (the measure space is called complete). In this case, if $$g=f$$ almost everywhere, the set $$f\neq g$$ is measurable. So if the property was true, we could write $$g=f\chi_{f=g}+g\chi_{f\neq g}$$ and $$g\chi_{f\neq g}$$ shouldn't be $$\mathcal A$$ measurable. It can't be true approximating $$g$$ by simple functions. 2 added 57 characters in body edited Mar 28 '13 at 15:41 Davide Giraudo 137k1717 gold badges174174 silver badges292292 bronze badges 1 answered Mar 28 '13 at 15:26 Davide Giraudo 137k1717 gold badges174174 silver badges292292 bronze badges