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I'm trying to find a function $f$ with $|f|$ is Lebesgue integrable but $f$ isn't locally Lebesgue integrable.

My approach:

Let $X = [0,1]$ and $A \subset X$ with A is the Vitali-set. So $A$ is not measurable. Define: $f(x) = \{1 \text{ if } x \in A,\,\,\, -1 \text{ if } x \notin A\}$.

I think that $f$ is not integrable since $A$ is not measureable. But how can I proof this?

And if this is correct, I think I'm finished, since if $f$ isn't integrable, it's also not locally integrable since $X$ is already compact.

And: $|f| = 1$ (the constant function)

And $\int_X |f| d\lambda = 1$, so $|f|$ is integrable.

Thanks

mathematics-and-caffeine
  • 764

2 Answers2

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You've got the right setup.

Hint:

$$\int_{X}f(x)dx = \int_{A}f(x)dx + \int_{X\setminus A}f(x)dx = \int_A 1 dx - \int_{X\setminus A} 1 dx.$$

Zim
  • 4,318
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In point of fact a function $f:\Bbb R\to\Bbb R$ such that $\lvert f\rvert$ is integrable will be integrable (locally and/or otherwise) if and only if it is measurable.

Your $f$ is not measurable because $A=f^{-1}[1/2,\infty)$, and therefore it isn't integrable.