Proving $\mathbb{Q} \bigcap [0, 1]$ is not Riemann Integrable

Question

Let $f = \chi_A$, where $A = \mathbb{Q} \bigcap [0, 1]$. Prove that $f$ is not Riemann integrable.

From what I understand, Riemann integration is only defined for compact sets. Is $f$ not Riemann integrable because $A = \mathbb{Q} \bigcap [0, 1]$ is not compact? Or am I thinking of Riemann integration the wrong way? Any help would be very helpful. Thank you.

Answer 1

One is considering the integral of the function $f$. This function is defined by $f(x)=1$ if $x$ is rational and $0\le x\le 1$, and $f(x)=0$ otherwise. You don't specify the interval of integration, but integrating over $[0,1]$ seems reasonable, as $f$ is zero outside this interval.

But one finds that each upper Riemann sum for $\int_0^1 f(x)\,dx$ is $\ge1$ and each lower Riemann sum is $\le0$. Therefore, from first principles, one sees that $\int_0^1 f(x)\,dx$ does not exist as a Riemann integral.

Answer 2

Incorrect. You are to infer the domain of $f$ from its definition. Indeed, $f = \chi_A$ is defined on any compact set which contains $Q \cap [0,1]$. For example, $f$ could be defined on $[0,1]$ itself. Therefore, there is no issue of compactness coming into play.

Furthermore, $g = \chi_{(0,1)}$ is Riemann integrable (on $[0,1]$) although it is the indicator of an open set, so compactness has nothing to do with the set whose indicator is being taken.

The only reason why $f$ is not Riemann integrable on $[0,1]$ is because the upper and lower limits don't match!

Indeed, let $0 = x_0 < x_1 < ... < x_n = 1$ be a partition of $[0,1]$. Then, recall the definition of upper and lower sums for a partition $P$ : $$ U(f;P) = \sum_{i=0}^{n-1} (x_{i+1} - x_i) \sup_{[x_i,x_{i+1}]} f(x_i) \\ L(f;P) = \sum_{i=0}^{n-1} (x_{i+1} - x_i) \inf_{[x_i,x_{i+1}]} f(x_i) $$

Now, for any $x_{i} < x_{i+1}$, there is both a rational and an irrational between $x_i$ and $x_{i+1}$. Hence, it is always the case that $\sup_{[x_i,x_{i+1}]} \chi_{Q \cap [0,1]}(x) = 1$, and $\inf_{[x_i,x_{i+1}]} \chi_{Q \cap [0,1]}(x) = 0$.

Hence, for any partition $P$, we see that $U(f;P) = 1$ and $L(f ; P) = 0$! Hence, these can never match, therefore $\chi_{Q \cap [0,1]}$ is not integrable on any compact set containing $[0,1]$.

Answer 3

The Riemann integral exists if and only if there is a unique $I$ (called the integral) such that for all partitions $P$, $$L(P,f)\leq I\leq U(P,f)\;.$$ But since each element in any partition $P$ contains both rationals and irrationals, it follows immediately that $L(P,f)=0$ and $U(P,f)=1$. There are many $I$'s such that $0<I<1$.