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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?

math-learner
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  • Wouldn't the integral be infinite? – Matthew Levy Nov 26 '14 at 07:24
  • Two different limits for the same quantity? Hmmm... – Did Nov 26 '14 at 07:25
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    Pretty sure this is the wrong solution. Because $f$ is strictly positive for $x>0$, the integral of $f$ over $[0,3]$ should be larger than the integral of $f$ over $[1,3]$ right? Yet if we consult the Riemann integral of $f$ over $[1,3]$ we have $$\int_1^3\frac{1}{x}dx = \ln(3)$$ – graydad Nov 26 '14 at 07:29
  • See my answer in http://math.stackexchange.com/questions/1030604/proof-of-lebesgue-integral-of-fx-frac1x-in-the-interval-1-5-equals-to/1030626#1030626. – Martín-Blas Pérez Pinilla Nov 26 '14 at 07:37
  • $f_n$ isn't defined in the interval $[0,3]$. – Martín-Blas Pérez Pinilla Nov 26 '14 at 07:39
  • I just found in a reference that this type of function is neither Riemann integrable nor Lebesgue integration. And the result turns out to be $\infty$. I just wondered before, that all functions that are not Riemann integrable are Lebesgue integrable. But it's wrong. – math-learner Nov 26 '14 at 07:48

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