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I have following sequence:

$$ f_n = -\frac{1}{n} $$

I wanted to show, that it converges to $f = 0$, but my book says, that it doesn't, because the condition

$$ \int_{\mathbb R} f_1 d\lambda > -\infty $$

is not met.


My question is - How can I compute Lebesgue integral of $f_1$ ?

$$ f_1 = - \frac{1}{1} = -1 $$

And the integral is:

$$ \int_{\mathbb R} -1 d\lambda $$


I know, that the integral is defined like this:

$$ \int_E f d\lambda = sup \left\{ \int_E s d\lambda: 0 \leq s \leq f \right\}, $$

where $s$ is considered a simple function, so its integral is defined like this:

$$ \int_E s d\lambda = \sum^k_{i=1} \alpha_i \lambda(A_i \cap E), $$

where $A_i = \{x \in \mathbb R^n: s(x) = \alpha_i \in \mathbb R^+_0\}$.

But still, unfortunately, I'm not able to "connect" all those definitions and compute the integral...

Eenoku
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    Use the fact that $$\int_{\Bbb R} 1 dx = \lambda( \Bbb R ) $$ – Tryss Jun 13 '16 at 13:22
  • @Tryss Ok, that would be great :-) ... But, how did you get this formula? I get, that $0 \leq s = 1 \leq f = 1$... I don't understand, how did you get the integral of the simple function... – Eenoku Jun 13 '16 at 13:29
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    The integral of a (positive) simple function $f(x) = \sum_{i=1}^n \alpha_i \chi_{A_i}(x)$ is defined as $\sum_{i=1}^n \alpha_i \lambda(A_i)$. And the constant function $c(x) = 1$ is indeed a simple function : $c(x) = \sum_{i=1}^1 1 \times \chi_{\Bbb R} (x)$ – Tryss Jun 13 '16 at 13:33
  • @Tryss

    Thank you very much - could you, please, add all this info as an answer, so I could accept it?

    – Eenoku Jun 20 '16 at 16:27

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