Showing that $f(x) = n(-1)^{n+1}$ if $\frac{1}{n+1}\lt x \lt \frac{1}{n}$ and $f(x)= 0$ if $x=0$ is not Lebesgue integrable

Question

I want that $f(x) = n(-1)^{n+1}$ if $\frac{1}{n+1}\lt x \lt \frac{1}{n}$ and $f(x)= 0$ if $x=0$ is not Lebesgue integrate-able but is improperly Riemann integrate-able.

I think I need to show that $$\sum_{a_n}^{b_n}\int \mid f(x)\mid dx$$ doesn't converge. But I don't know how to choose what $a_n$ and $b_n$ are. Thanks.

score 2 · Accepted Answer · answered Apr 07 '17 at 23:09

2

Presumably this is on $[0,1]$. $|f(x)| = n > 1/x - 1$ for $0 < x < 1$, implying it is not Lebesge integrable. But the improper Riemann integral is

$$ \lim_{a \to 0+} \int_a^1 f(x)\; dx = \sum_{n=1}^\infty n (-1)^n \left(\frac{1}{n} - \frac{1}{n+1}\right) $$

answered Apr 07 '17 at 23:09

Robert Israel

448,999

can you explain you're first comment more? why does that imply it's not lebesgue integrable? – MathNoob Apr 07 '17 at 23:30
$\int_0^1 \frac{1}{x}; dx = \infty$. – Robert Israel Apr 08 '17 at 00:09

Showing that $f(x) = n(-1)^{n+1}$ if $\frac{1}{n+1}\lt x \lt \frac{1}{n}$ and $f(x)= 0$ if $x=0$ is not Lebesgue integrable

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