Would everyone please help me on how to prove this value of Lebesgue integral of the function $f(x)=\frac{1}{x}$ in the interval [1,5] by using approximation by simple function $f_n$ step by step?
What if the function in this example defined on the interval of $[0,\infty]$? What will the value of integral be? (Still need help for the explanation by using simple function approximation)
Thank you.
For $k=0,\cdots,4n-1$, $x\in[1+\frac{k}n,1+\frac{k+1}n)$: $$f_n(x)=\frac1{1+(k+1)/n}=\frac{n}{k+n+1},$$ $$\int_1^5f_n=\sum_{k=0}^{4n-1}\frac{n}{k+n+1}\frac1{n}=\sum_{k=0}^{4n-1}\frac1{k+n+1}=\sum_{k=n+1}^{5n}\frac1k=H_{5n}-H_n\approx\log(5n)-\log n=\log 5.$$ In $[0,\infty]$ the harmonic series (divergent) is a lower bound of the integral. Try yourself, is much easier.