Showing that $1/x$ is NOT Lebesgue Integrable on $(0,1]$ I aim to show that $\int_{(0,1]} 1/x = \infty$.  My original idea was to find a sequence of simple functions $\{ \phi_n \}$ s.t $\lim\limits_{n \rightarrow \infty}\int \phi_n = \infty$.  Here is a failed attempt at finding such a sequence of $\phi_n$:
(1) Let $A_k = \{x \in (0,1] : 1/x \ge k  \}$ for $k \in \mathbb{N}$.
(2) Let $\phi_n = n \cdot \chi_{A_n}$
(3) $\int \phi_n = n \cdot m(A_n) = n \cdot 1/n = 1$
Any advice from here on this approach or another?

I think this may be the same as what Davide Giraudo wrote, but this way of saying it seems simpler.  Let $\lfloor w\rfloor$ be the greatest integer less than or equal to $w$.  Then the function
$$x\mapsto \begin{cases} \lfloor 1/x\rfloor & \text{if } \lfloor 1/x\rfloor\le n \\[8pt]
n & \text{otherwise} \end{cases}$$ is simple.  It is $\le 1/x$ and its integral over $(0,1]$ approaches $\infty$ as $n\to\infty$.
