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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).

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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.
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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.
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