Suppose that is $f$ is a measurable function, and $g$ differs with $f$ only by a measure zero set, is $g$ measurable? I do not think this is necessarily right, but is there any counter example?
Suppose that is $f$ is a measurable function, and $g$ differs with $f$ only by a measure zero set, is $g$ measurable?
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measure-theory
measurable-functions
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0Are you talking about measurabilty on $\Bbb R^n$ with the usual measurable structure, or about abstract measurable spaces? – 2017-02-12
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0I am trying to think of an abstract measurable space, but will the conclusion differ under different situations? – 2017-02-12
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0I was just confused by your chosen tag - "real-analysis". I have changed it to suit your comment. – 2017-02-12
2 Answers
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Hint:
- The sum of two measurable functions is measurable.
- For any $E$, either $\{x: f(x)-g(x) \in E\}$ or its complement has measure $0$.
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0I am sorry but I am a little confused with your second hint. – 2017-02-12
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0I see. So the answer for the question should be yes, right? – 2017-02-12
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This depends on whether your measure is complete. Consider, for example, Lebesgue measure on the Borel subsets of the unit interval $[0,1]$. Let $f(x)=x$ for $x\in[0,1]$. Let $K\subset[0,1]$ denote the Cantor set. Choose a non-Borel-measurable subset $B$ of $K$. Define $g(x)=f(x)$ for $x\in [0,1]\setminus K$ and $g(x) =2+1_B(x)$ for $x\in K$. Then $\{x:f(x)\not=g(x)\}=K$ is a measurable set of measure $0$, but $g^{-1}(\{3\})=B$, so $g$ is not (Borel) measurable.