I know the formal definition of a Lebesgue integral: it is essentially an approximation through simple functions, but simple functions are defined over sets in the associated $\sigma$-algebra. In all the examples that I have encountered, the $\sigma$-algebra is the collection of Borel subsets. However, consider this example: the measure space is $(\mathbb{R}, \mathcal{S}, \lambda)$, where $\mathcal{S} = \{\emptyset, \mathbb{R}\}$ and $\lambda$ is the Lebesgue measure. Consider $f(x) = \chi_{[0,1]}(x) \times x$ ($\chi$ is the indicator function). What is $\int_{\mathbb{R}} f(x)$?

Normally, when the $\sigma$-algebra is the Borel subsets, the answer is $1/2$. Right now the $\sigma$-algebra only contains the whole set and the empty set, I'm not sure how to build simple functions to approximate. The "closest" approximation that I can find is the zero function, but that doesn't sound right to me.

I use the definition of Lebesgue integral of non-negative functions here.

  • $\begingroup$ Which definition of the Lebesgue integral for non-negative functions are you using? Is measurability required? $\endgroup$ – Daniel Fischer Aug 9 at 20:53
  • $\begingroup$ Edited my question. I define it as the supremum of the integral of the simple functions over measurable sets that are weakly smaller than $f$. $\endgroup$ – user1691278 Aug 9 at 21:01
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    $\begingroup$ The wikipedia article talks only of measurable functions. Since only constant functions are $\mathcal{S}$-measurable, $\int f\,d\lambda$ doesnt exist by that definition. If the measurability assumption is dropped (I've seen that done, but I don't think that's a particularly useful way), then the result is $0$ since $0$ is the largest $\mathcal{S}$-measurable function $\leqslant f$. $\endgroup$ – Daniel Fischer Aug 9 at 21:04
  • $\begingroup$ I see. If you type it up as a solution, I'd select it. Thank you. Your explanation is very clear. $\endgroup$ – user1691278 Aug 9 at 21:08

The only $\mathcal{S}$-measurable functions (assuming that we take the Borel $\sigma$-algebra on the codomain) are the constant functions, thus $f \colon x \mapsto \chi_{[0,1]}(x)\cdot x$ isn't measurable.

Hence if the definition of the Lebesgue integral for non-negative functions demands the measurability, $\int f\,d\lambda$ is not defined.

Some authors define $$\int g \,d\mu = \sup \: \biggl\{ \int s\, d\mu : 0 \leqslant s \leqslant g, s \text{ simple}\biggr\}$$ even for non-measurable functions $g \geqslant 0$. With that definition we would have $\int f\,d\lambda = 0$ since $0$ is the largest simple measurable functions $\leqslant f$.

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