# Riemann integration of a function

I would like to ask if the function $$f(x)=\begin{cases}0\text{ if } x\in \mathbb{\mathbb{[0,1]}\cap Q}\\ x^2\text{ if } x\in \mathbb{[0,1]\setminus Q} \end{cases}$$ is Riemann Integrable in $$[0,1]$$?

My thought:

Let $$S(f,P)$$ and $$s(f,P)$$ be the upper and low sums of $$f$$ with respect to partition $$P$$ of interval $$[0,1]$$. Let $$M_i=\sup\{f(x)| x \in I_i\}$$ and $$m_i=\inf\{f(x)| x \in I_i\}$$, where $$I_i$$ is the $$i$$th interval of the partition $$P$$. Note that $$M_i=1$$ for all $$i$$, because every interval $$I_i$$ of the partition $$P$$ contains rational numbers. On the other hand, $$m_i=0$$ for all $$i$$ because every interval $$I_i$$ of the partition $$P$$ contains irrational numbers. By definition, $$S(f,P)=\sum_{i=1}^nM_i\mu(I_i)=\sum_{i=1}^n1\mu(I_i)=\sum_{i=1}^n\mu(I_i)=1-0=1$$ $$s(f,P)=\sum_{i=1}^nm_i\mu(I_i)=\sum_{i=1}^n0\mu(I_i)=0$$ Thus, $$f$$ is not Riemann integrable because the upper and lower integrals are not equal. (The upper integral is the limit of the upper sums and the lower integral is the limit of the lower sums).

• You're very close, but your calculation of $M_i$ is wrong. For instance if $I_1=[0,1/\sqrt{2}]$ then $M_1=1/2$, rather than (as you claim) $M_1=1$. – Jamie Radcliffe Aug 25 '19 at 14:04

No, it is not Riemann-integrable, but your computations are not correct. You are right when you claim that $$s(f,P)=0$$, for each partition $$P$$ of $$[0,1]$$. However, it is not true that $$S(f,P)=1$$ for each partition $$P$$ of $$[0,1]$$. Actually, it is not true that “$$M_i=1$$ for all $$i$$, because every interval $$I_i$$ of the partition $$P$$ of $$[0,1]$$ contains rational numbers”. However, it is true that, for each partition $$P$$ of $$[0,1]$$, $$S(f,P)\geqslant\frac13$$, and this is enough to prove what you want.
• What is confusing you? You wrote that each $M_i$ is equal to $1$, which is just wrong. If $P=\left\{0,\frac12,1\right\}$, then you have two intervals: $I_1=\left[0,\frac12\right]$ and $I_2=\left[\frac12,1\right]$. In this case, $M_2$ is indeed $1$, but $M_1=\frac14$. – José Carlos Santos Aug 25 '19 at 14:20