Can a function be Riemann integrable, but not Lebesgue integrable?

Question

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?

Answer 1

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 $$ \int_{[0,5]} f\,\mathrm d\lambda=\lim_{n\to\infty}\frac{5(5n\pm1)}{2n}=\frac{25}2. $$

Answer 2

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.

Answer 3

An intuitive understanding of Lesbesgue Integration?

Look at the cover of Schilling's book.