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Question: Let $E$ be a set of infinite measure. Let $f$ be a integrable function over $E$, then there necessarily exists a function $g$ which is bounded and measurable such that $f=g$ a.e.

There is a theorem that asserts that a bounded measurable function defined on a set of finite measure is (Lebesgue) integrable. Determining if the above statement is true or false is what I am stuck on.

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    I am not sure, but isn't $x^{-1/2}$ integrable on $(0,1)$ in the Lebesgue sense?2012-11-29
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    If $E=(0,1)$ and $f(x)=x^{—1/2}$, then such a $g$ doesn't exist, because for all $n$, $(0,n^{—1})$ has a positive measure (so there is $x_n$ such that $n^{1/2}\geqslant x_n^{-1/2}=g(x_n)$.2012-11-29
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    You should phrase your question as a question rather than an assertion. Neither of the two statements is correct. Take $f = \sum_{n=0}^\infty n 1_{(\frac{1}{2^{n+1}},\frac{1}{2^{n}}]}$. $f$ is integrable on $[0,1]$ but not essentially bounded.2012-11-29

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