We have the function $f(x)=\infty$ if $x=0$ and $f(x)=\frac{1}{x}$ if $x$ otherwise. So, in this two values of function, I made simple approximation of $f(x)$ by the help of simple function :
$f_n(x)=n$ if $f(x)\in[3n,\infty]$, and $\sum_{k=0}^{3n-1}\frac{n}{k+1}1_{\{\frac{k}{n}\leq x<\frac{k+1}{n}\}}$ if $f(x)\in[\frac{n}{k+1},\frac{n}{k})$
(I actually am not sure about the former approximation, please help to fix it.)
Then, when it comes to the Lebesgue integration, I can write
$\lim_{n\to\infty}\int_{0}^{3}f_n(x)d\mu_0=\lim_{n\to\infty}\int_{0}^{3}nd\mu_0=n\mu(0)=0$, and $\lim_{n\to\infty}\int_{0}^{3}\sum_{k=0}^{3n-1}\frac{n}{k+1}1_{\{\frac{k}{n}\leq x<\frac{k+1}{n}\}}d\mu_0=\sum_{k=0}^{3n-1}\frac{n}{k+1}\frac{1}{n}=\sum_{k=1}^{3n}\frac{1}{k}=H_{3n}=H_{3n}-H_{n}\approx ln3$
So, the lebesgue integration of our given function is $0+ln3=ln3$.
Do I have the right solution? How can we compare it to riemann integral?