Why is $\frac{1}{x}$ not Lebesgue integrable on $[0,1]$?

Question

My teacher said (without explaining) that $\frac{1}{x}$ is not Lebesgue integrable on $[0,1]$?

Could someone please explain why is this true?

Answer 1

$\frac{1}{x}$ is non-negative on $[0,1]$ (almost everywhere at least), so its Lebesgue integral is defined, but it turns out that $$\int_0^1\frac{1}{x}\;dx=\infty$$ This can be seen by applying the monotone convergence theorem to the sequence $f_n(x)=\frac{1}{x}1_{[\frac{1}{n},1]}(x)$.

Answer 2

Because $ \displaystyle \int_{[0,1]} \left|\,\frac 1 x\, \right| \, dx = \infty. $

To say that $f$ is Lebesgue-integrable on $A\subseteq\mathbb R$ means $\displaystyle \int_A |f(x)|\,dx < \infty$.

Answer 3

Note that $f_n(x) = \sum_{k=1}^n k 1_{({1 \over k+1} , {1 \over k}]}(x) $ satisfies $f_n(x) \le {1 \over x}$ for $x>0$ and $\int f_n = \sum_{k=1}^n {1 \over k+1}$. Since $\sum_n {1 \over n}$ is divergent, we see that $x \mapsto {1 \over x}$ is not integrable.