Examples of Lebesgue-integrable, but not Riemann-integrable functions

The standard example of this is the characteristic function of the rationals. However this is somewhat pathological as this function is zero almost everywhere. What are other examples that differ from a Riemann-integrable function on more than a set of measure zero?

• I think the characteristic function of a fat cantor set works. – Thorgott Apr 20 at 13:44
• Are there any examples that aren't pathological? – Jam Apr 20 at 18:24

Take an enumeration $$\{x_1,x_2,\dots,x_n,\dots\}$$ of the rationals in the interval $$[0,1]$$. For each $$n \in \mathbb{N}$$, define $$G_n=\left(x_n-\dfrac{1}{2^{n+2}},x_n+\dfrac{1}{2^{n+2}}\right).$$

Note that $$\lambda(G_n)=\frac{1}{2^{n+1}}$$.

Now define the set $$F=[0,1]\cap(\cup_{n=1}^\infty G_n)^c$$.

$$F$$ is clearly Lebesgue mensurable, and, because

$$\lambda\left(\bigcup_{n=1}^\infty G_n \right) \le \sum_{n=1}^{\infty} \lambda(G_n) = \sum_{n=1}^{\infty}\dfrac{1}{2^{n+1}}=\dfrac{1}{2},$$

we have $$\lambda(F)>\dfrac{1}{2},$$

and $$\int \chi_F d\lambda>\dfrac{1}{2}.$$

$$\chi_F$$ is the charachteristic function of $$F$$.

However, $$\chi_F$$ is not Riemann integrable. Note that $$F$$ contains no interval, because it doesn't contain any rationals, so any interval will contain points that are not in $$F$$ $$F$$. Therefore, the minimum of $$\chi_F$$ in any interval will be $$0$$, and $$\underline{\int}\chi_F dx=0.$$

• @Jam about your last comment: It depends on what you mean by "pathological". A bounded function defined in an limited interval is Riemann-integrable iff it is almost everywhere continuous. This means that all the "nice" functions are Riemann-integrable. The Lebesgue integral was developed to extend the concept of integral to functions that are not "nice". So, in this sense, there aren't "non-pathological" examples. – Célio Augusto Apr 20 at 20:33