Prove that $f$ is not Lebesgue integrable

Question

I need a hand with the following exercise:

Prove that $f: (0,2) \to \mathbb{R}$ given by

$f(x) = \begin{cases} \frac{1}{x} & 0<x\leq 1 \\ \frac{1}{x-2} & 1< x < 2 \end{cases} $

Is not Lebesgue-integrable.

Here are my thought:

We can write $f$ as $f = f^+ - f^-$ where

$f^+(x) = \begin{cases} f(x) & \text{if} \ f(x)>0 \\ 0 &\text{otherwise}& \end{cases} $

$f^-(x) = \begin{cases} -f(x) & \text{if} \ f(x)\le0 \\ 0 &\text{otherwise}& \end{cases} $

And by definition $f$ is integrable if and only if $f^+$ and $f^-$ are both integrable, so I just need to prove that $f^+$ or $f^-$ is not Lebesgue integrable.

Now, in this particular case $f^+(x) = \begin{cases} \frac{1}{x} & 0<x\leq1 \\ 0 & 1<x<2 \\ \end{cases} $ and $f^-(x) = \begin{cases} 0 & 0<x<1 \\ -\frac{1}{x-2} & 1\leq x<2 \\ \end{cases} $

So $f = f^+-f^-$ becomes $$ f = \frac{1}{x}\chi_{(0,1]}-\left(-\frac{1}{x-2}\chi_{[1,2)}\right)$$

thus if I proove that, says, $\frac{1}{x}$ is not Lebesgue integrable on $(0,1]$ the problem is solved. But how to prove that?

Answer 1

$\int |f| \geq \int_{1/n}^{1} \frac 1 x\, dx=\log \, n \to \infty$ so $f$ is not Lebesgue integrable. I have used the fact that on $(\frac 1n, 1)$ the function is Riemann integrable which implies that it is Lebesgue intergable and the two integrals are equql.

Answer 2

For any $n$, the simple function $\sum_{k=1}^n \chi_{(0, 1/k]}$ is nonnegative and less than $f^+$, so the integral of $f^+$ is larger than $\sum_{k=1}^n \frac{1}{k}$ for any $n$.

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Edit: For $x \in (0,1]$, we have $$\sum_{k=1}^n \chi_{(0,1/k]}(x) = \#\{k \in\{1,\ldots,n\} : k \le 1/x\} = \lfloor 1/x \rfloor \le 1/x = f^+(x)$$ and thus $$\sum_{k=1}^n \frac{1}{k} = \int_0^1 \sum_{k=1}^n \chi_{(0, 1/k]}(x) \, dx \le \int_0^1 f^+(x) \, dx.$$