Stacie Worsley

2022-01-15

Function f with $|f|$ is Lebesgue integrable but f isn't locally Lebesgue integrable

sonorous9n

Expert

Step 1
In point of fact a function
$f:R\to R$
such that $|f|$ is integrable will be integrable (locally and/or otherwise) if and only if it is measurable.
Your f is not measurable because
$A=f-1\left[\frac{1}{2},\mathrm{\infty }\right)$,
and therefore it isn't integrable.

Beverly Smith

Expert

Step 1
Hint:
$\in {y}_{X}f\left(x\right)dx$
$={\int }_{A}f\left(x\right)dx+{\int }_{\frac{X}{A}}f\left(x\right)dx$
$={\int }_{A}1dx-{\int }_{\frac{X}{A}}1dx$

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