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I am struggling to solve an exercise in my measure theory book and any help for solving it would be appreciated:

Let $(\Omega,\mathcal{A},\mu)$ be a measure space and let $f:\Omega \to \mathbb{R}$ be measurable. Find a function $g:\Omega \to \mathbb{R}$ which is equal to $f$ almost everywhere but is not measurable.

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  • $\begingroup$ Please don't feel that I am cheating or so. It is a self-study. Thanks. $\endgroup$ – Cupitor Mar 28 '13 at 15:07
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    $\begingroup$ Are you familiar with the concept of complete measures? This doesn't sound true for arbitrary measures (under counting measure, "almost everywhere" is the same as "everywhere"). $\endgroup$ – Erick Wong Mar 28 '13 at 15:14
  • $\begingroup$ I just checked it on wiki but I don't see the relation? Well, thats why counting measure cannot generate a counterexample here! $\endgroup$ – Cupitor Mar 28 '13 at 15:16
  • $\begingroup$ If the measure space is complete then any function almost everywhere equal to a measurable function is also measurable. $\endgroup$ – Kabo Murphy Oct 27 '17 at 9:21
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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 $g:=f+\mathbf 1_{N}$, 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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  • $\begingroup$ your second case is still a little bit vague for me. So you are saying that if we are in the second case $g$ will be measurable as well? $\endgroup$ – Cupitor Mar 28 '13 at 15:52
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    $\begingroup$ I've edited in order to clarify. $\endgroup$ – Davide Giraudo Mar 28 '13 at 16:01
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    $\begingroup$ Since $f \chi_{f = g} = g \chi_{f = g}$, it is evident that $g \chi_{f = g}$ is $\mathcal A$ measurable. But why does this imply that $g \chi_{f \neq g}$ is a pointwise limit of simple functions? $\endgroup$ – el_tenedor Oct 27 '17 at 5:05
  • $\begingroup$ @el_tenedor You are right, it deserved a clarification. There is no need to use simple functions actually. $\endgroup$ – Davide Giraudo Oct 27 '17 at 8:36
  • $\begingroup$ In the first case, how can you say that $g$ and $f$ coincide almost everywhere if the set on which they disagree can't be measured? $\endgroup$ – mattecapu Jan 11 '18 at 11:53

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