# Lebesgue integral and sigma algebra

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.

• Which definition of the Lebesgue integral for non-negative functions are you using? Is measurability required? – Daniel Fischer Aug 9 at 20:53
• Edited my question. I define it as the supremum of the integral of the simple functions over measurable sets that are weakly smaller than $f$. – user1691278 Aug 9 at 21:01
• 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 $\leqslant f$. – Daniel Fischer Aug 9 at 21:04
• I see. If you type it up as a solution, I'd select it. Thank you. Your explanation is very clear. – 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 $$\leqslant f$$.