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I understand that there are functions where the Lebesgue integral exists, but they are not Riemann integrable (e.g. the Dirichlet function). Are there also functions that are Riemann integrable but not Lebesgue integrable?

If we e.g. have a function $f(x)=x$, $x\in[0,5]$ and have to find the Lebesgue integral of that function $\int_x x d\lambda$ how is the calculus then. I know it should be the same as when we integrate with respect to Riemann (25/2). But how do we calculate it if we don't use the Riemann integral?

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    $\begingroup$ "Are there functions that are riemann integrable but not lesbegue integrable?" No there are not. $\endgroup$
    – Did
    Commented Feb 24, 2014 at 12:37
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    $\begingroup$ To expand on @Did's remark, you should expect to see (whether you're going over a textbook yourself or attending a course) that whenever a function is Riemann integrable, it is also measurable (in fact, you should already know that Riemann integrability is equivalent to a.e. continuity), and both integrals agree. Think of Lebesgue integral as division by the values of $f$, instead of division by values of $x$, a la Riemann/Darboux sums. $\endgroup$ Commented Feb 24, 2014 at 12:41
  • $\begingroup$ Of course the normal PDF is (Borel) measurable. But Riemann integrability concerns bounded intervals hence to ask whether a function is Riemann integrable on the whole line needs explanations... $\endgroup$
    – Did
    Commented Feb 24, 2014 at 12:56
  • $\begingroup$ Which textbook are you following? $\endgroup$
    – Did
    Commented Feb 24, 2014 at 12:57
  • $\begingroup$ I read Rene l Schilling. I am preparing for re-exmination i measure theory. Last time i got the question about calculating the lesbegue integral of the function f(x)=x. I can not see how it can be done without using riemann. The other questions was just for intuitive understanding and thank you for your answers. $\endgroup$ Commented Feb 24, 2014 at 13:02

3 Answers 3

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There are functions which are Riemann integrable, but not Lebesgue integrable, altough one needs to generalize to improper Riemann integrals.

One can show that $$\int_0^\infty \frac{\sin(x)}{x}dx$$

exists as an improper Riemann integral, and we have $\lim_{b \rightarrow \infty}\int_0^b \frac{\sin(x)}{x}dx = \frac{\pi}{2}$.

However, since a function is Lebesgue integrable iff its absolute value is Lebesgue integrable and $\int_0^\infty \vert \frac{\sin(x)}{x} \vert dx$ is not finite, the initial integral doesn't exist in the sense of Lebesgue. The main point here is that for Lebesgue integrability, one needs to ensure finiteness of the integrals over the postive and the negative part of the integrand separately - for Riemann integrals these are allowed to cancel each other.

This is similar to the fact that the alternating harmonic series has a limit, but its not absolutely convergent and hence, one cannot arbitrarily change the summation order.

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To compute the Lebesgue integral of the function $f:[0,5]\to\mathbb R$ directly, one can note that $f_n\leqslant f\leqslant g_n$ for every $n\geqslant1$, where the simple functions $f_n:[0,5]\to\mathbb R$ and $g_n:[0,5]\to\mathbb R$ are defined by $$ f_n(x)=n^{-1}\lfloor nx\rfloor,\qquad g_n(x)=n^{-1}\lfloor nx\rfloor+n^{-1}. $$ By definition of the Lebesgue integral of simple functions, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}k\cdot\lambda([kn^{-1},(k+1)n^{-1})), $$ that is, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-2}\sum_{k=0}^{5n-1}k=\frac{5(5n-1)}{2n}. $$ Likewise, $$ \int_{[0,5]} g_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}(k+1)\cdot\lambda([kn^{-1},(k+1)n^{-1}))=\frac{5(5n+1)}{2n}. $$ Since the integrals of $f_n$ and $g_n$ converge to the same limit, $f$ is Lebesgue integrable and $$ \int_{[0,5]} f\,\mathrm d\lambda=\lim_{n\to\infty}\frac{5(5n\pm1)}{2n}=\frac{25}2. $$

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  • $\begingroup$ Thank you a lot for your answer @Did. From the book i read(Rene L. Schilling) i dont see any approach like that, and no examples of calculating aribritary functions(maybe im missing something). Can you recommend some textbook that shows your approach? $\endgroup$ Commented Feb 27, 2014 at 15:55
  • $\begingroup$ "My approach" (which is hardly "mine") is to apply the definition of Lebesgue integral, based on the fact that the integral of $\sum\limits_ka_k\mathbf 1_{A_k}$ is $\sum\limits_ka_k\lambda(A_k)$. $\endgroup$
    – Did
    Commented Feb 27, 2014 at 18:26
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An intuitive understanding of Lesbesgue Integration?

Look at the cover of Schilling's book.

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  • $\begingroup$ Yea oka i get that part. But the integration part the opposite way than on riemann is what i dont get. If you know Schillings book, can you tell me from which part i should be able understand how to calculate my example as Did. I cannot find the simple function definiton as did use. $\endgroup$ Commented Feb 27, 2014 at 16:56
  • $\begingroup$ I recommend that you look at that picture, then look at Did's functions $f_n$ and $g_n$. $\endgroup$ Commented Feb 28, 2014 at 10:59

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