Use monotone convergence theorem to show that $f(x)=\frac{1}{x}$ is not summable on $(0,1)$

Question

Use monotone convergence theorem to show that $f(x)=\frac{1}{x}$ is not summable on $(0,1)$.

The hint was to consider the sequence $f_n(x)=\max \{n, f(x)\}$, $x\in(0,1)$. It is clear that $0\le f_1 \le f_2 ,\ldots $ and each $f_1, f_2,\ldots $ is measurable.

But I have no idea at all how can I proceed...

Answer 1

Read again the hint. It must be $f_n(x)=\min\{n,f(x)\}$. For such $f_n$, to calculate $\int_{(0,1)}f_n(x)\,\mathrm{d}x$, divide the interval $(0,1)$ in $(0,1/n)$ and $[1/n,1)$.